probability-statistics

[honest trench]

hmmm

right

[frozen relic]

You already did

[frozen relic]

Smth smth law of total prob?

[honest trench]

nth comes to mind

the second part of this question is so much easier

[frozen relic]

You don’t know what law of total probability is?

[honest trench]

ii) Find the probability that Rachel wins 2 games and loses 1 game out of the first three games that they play.

[honest trench]

Does this help me solve the first question? I solved this one.

[frozen relic]

You need to know this to solve it

Or tree diagram or whatever

[honest trench]

i know tree diagram

[frozen relic]

(Which is just law of total prob drawn)

[honest trench]

that's how i solved the seond one

ohhhh

[frozen relic]

So why not use that then? You clearly have cases you need to work through

[honest trench]

how would I use that though?
From the tree diagram, I can see, that there is two probability of her losing the second round. Besides that, I can't extract much from it.

[frozen relic]

P(LS)=P(WF and LS)+P(LF and LS)?

[honest trench]

Ahhhhhhhhhh

I'm so stupid

Thanks a lot @frozen relic apprecieate it

[median hazel]

Suppose six dogs and six cats run in a race. Assume
that each of the 12! rankings of the animals are equally likely. Let X denote
the highest ranking achieved by a dog; e.g., X = 1 if the top-ranked animal
is a dog. Compute P(X = i) for i = 1, 2,...,12. Just wanted to know if I have worked the problem out correctly and if there is an easier solution to this

[hardy badge]

Just a minor notational thing, but

It's probably easier to write 6Px to count the number of ways for cats to come before the first dog.

Otherwise this looks correct.

I also don't think there's an easier way to do this. You could do it using combinations instead of permutations but it's the same thing.

[median hazel]

Got it. Thanks a lot @hardy badge

[hardy badge]

(To elaborate... there are 12!/6!6! equally likely combinations of cats and dogs, disregarding individual identity. Let's consider X = 3 as an example. With two cats and one dog in the first three positions, there are 9!/5!4! ways to arrange the remaining animals. You can verify that this leads to exactly the same answer as your method.)

[median hazel]

Ahh... I get it. Thank you

[high copper]

really sorry to ping you but can you please say if this is right? @frozen relic

[frozen relic]

Or kendall, yes

[high copper]

thanks a lot scapeprof

[lusty drift]

This might be advanced, but will place here to see, this is for a real event I'm trying to figure out

I have p1 = 0.50, but only for two trials
I have p2 = 0.65 but only for three trials

What exactly is the binomial probability distribution for Y=y trial success

So how do I figure out the probability for example, that 4 trials is a success

I guess Y = X1 + X2, where X1 = success for p1, X2 = success for p2

Hmmmmm, I do have combinations, and I set up an excel sheet for each combination, but I still need to know like, if there is a formula or modification to the formula

Edit: It was basic multiplication since its just 2

They are independent, hmmmmm, I think I got it, but I still want to know the case if there was a p3 that exist, so more than 3 probabilities

I'm guessing with more p_i, and so long as they are independent, its basically this but in higher dimensions of combinations of success

[glacial vale]

your working out sounds good

and yeah finding the distribution of a sum of random variables is a bit of pain

the fancy name for the operation you just performed is called convolution

https://en.wikipedia.org/wiki/Convolution_of_probability_distributions

[lusty drift]

So I thought, if there was an X3, and I have Y = X1 + X2 + X3, I can first make Y1 = X1 + X2, DO THIS, then set up Y = Y1 + X3, using this method

if thats on the right track

also thank you! will look into this

[glacial vale]

that would work yeah

[glacial vale]

there's not a lot in the wiki article itself but at least you'll recognize the term if you see it

[glacial vale]

yeah convolving (X1+X2) with X3

[lusty drift]

I do, and happy to see there is a continuous form for this set up.

[nocturne granite]

Hello, I fitted a linear regression model using Python's statsmodels.regression.linear_model.OLS.fit

I know what a hypothesis testing and p-value is from textbooks, but what is the t-value of every feature?

[frozen relic]

Its testing if said parameter can be left out

(So testing coef equal to 0)

[nocturne granite]

OK, thank you, I understand it now.

Can anyone please explain this code to me

yhat = res.fittedvalues
ehat = res.resid
# Studentized residuals
ehatStudentized = res.get_influence().resid_studentized_internal
# Asolute squared studentized residuals
ehatStudentizedAbsSqrt = np.sqrt(np.abs(ehatStudentized))
# absolute residuals
ehatAbs = np.abs(ehat)
# leverage:
xleverage = res.get_influence().hat_matrix_diag
# Cook's distance:
modelCooks = res.get_influence().cooks_distance[0]

fig = plt.figure(figsize=[13, 11.5])  

#-- QQ PLOT:
# This one shows how well the distribution of residuals fit the normal distribution.
# This plots the standardized (z-score) residuals against the theoretical normal quantiles.
# Anything quite off the diagonal lines may be a concern for further investigation.

ax = fig.add_subplot(2, 2, 1)
QQ = ProbPlot(data=ehatStudentized, fit=True)
QQ.qqplot(line='45', alpha=0.5, color='#4C72B0', lw=2, ax=ax)
ax.set_title('Normal Q-Q')
ax.set_xlabel('Theoretical Quantiles')
ax.set_ylabel('Studentized Residuals');

# Annotations for the largest residuals:
yPointsTop3 = np.flip(np.argsort(np.abs(ehatStudentized)), 0)[:4]
for r, i in enumerate(yPointsTop3):
    xy = (1.01 * np.flip(QQ.theoretical_quantiles, 0)[r], 1.01 * ehatStudentized[i])
    ax.annotate(i, xy=xy, fontsize=14);
    
#-- LEVERAGE PLOT:
# This plot shows if any outliers have influence over the regression fit.
# Anything outside the group and outside “Cook’s Distance” lines, may have
# an influential effect on model fit.

ax = fig.add_subplot(2, 2, 2)
plt.scatter(xleverage, ehatStudentized, alpha=0.5)
sns.regplot(xleverage, ehatStudentized, scatter=False, 
            ci=False, lowess=True,
            line_kws={'color': 'red', 'lw': 3, 'ls':'--', 'alpha': 0.8},
            ax = ax);
ax.set_xlim(np.min(xleverage) - 0.001, 1.05*np.max(xleverage))
ax.set_ylim(1.05*np.min(ehatStudentized), 2*np.max(ehatStudentized))
ax.set_title('Residuals vs Leverage')
ax.set_xlabel('Leverage')
ax.set_ylabel('Studentized Residuals')

# Annotations for the points with the largest Cook's distance:
yPointsTop3 = np.flip(np.argsort(modelCooks), 0)[:4]
for i in yPointsTop3:
    ax.annotate(i, xy=(1.01*xleverage[i], 1.01*ehatStudentized[i]), fontsize=14)

It's beyond me

I only know it's for drawing QQ-plot to find the outliers.

What is a ehatStudentized and why ehatStudentizedAbsSqrt ?

[frosty field]

The first is a measure of the studentized residual, which is just a residual divided by the estimate of its standard deviation

The second is the sqrt of the absolute value of said studentized residual

[nocturne granite]

OK, thank you, but what's the purpose of the second variable? I can understand the first one.

I think the first variable is already normalized by dividing its standard deviation?

[frosty field]

I have no idea, I've never seen that before

Some quantities we like to square root (for example, variance) because it makes the units match those of the original observations

But the units of residuals already do match, so I don't know

[nocturne granite]

Thank you very much.
The second question is that xleverage = res.get_influence().hat_matrix_diag

I see the hat matrix $X(X^TX)^{-1}X^T$ is the projection matrix. But its diagram is the leverage? What is a leverage here?

[vivid meadow]

runoob

[frosty field]

Leverage is a measure that tells you how much influence a point has on the regression line

ie if you were to remove that observation, how strong a change would occur in the regression

Points with too high a leverage can sometimes be outliers, so it has utility in outlier detection

[nocturne granite]

I can see its definition from wiki, but I still cannot understand why it's related to the diagram of the hat matrix.

Thank you. I will keep searching.

[frosty field]

Alright, good luck

[honest trench]

Three friends, Rick, Brenda and Ali, go to a football match but forget to say which entrance to the ground they will meet at. There are four entrances, A, B, C and D. Each friend chooses an entrance independently.
The probability that Rick chooses entrance A is 1/3. The probabilities that he chooses entrance B, C or D are all equal.
Brenda is equally likely to choose any of the four entrances.
The probability that ALi chooses entrance C is 2/7 and the probability that he chooses entrance D is 3/5. The probability that he chooses the other two entrances are equal.

i) Find the probability that at least 2 friends will choose entrance B.

For the Q (i).
Is this how we do q(i)

\text{ let c = choose, eB = entrance B } \
$P(\text{ 2 friends choosing entrance B } ) = $ \ $P(\text{ Rick c eB } \cap \text{ Brenda c eB } ) + P(\text{ Rick c eB } \cap \text{ Ali c eB } ) + P(\text{ Ali c eB } \cap \text{ Brenda c eB })$

am I correct or is this wrong?

[vivid meadow]

∫Inheritanc-e ♦

[urban lodge]

I think you have to also exclude the case where three friends pick entrance B

[honest trench]

oh

nope

don't work

[urban lodge]

Wait

@honest trench you’re asking two different questions

[honest trench]

i am?

[urban lodge]

The problem statement you quoted said find the probability of at least 2 friends

But your latex just says “2 friends…”
I.e. “exactly two friends”

There’s a difference between exactly two friends and at least two friends

[honest trench]

oh

well how would at least two friends be written

ohhh

hmm

didn't work

could u please help me

i can't figure out how to represent at least in probability

[urban lodge]

The sum you have minus two times the intersection of all three events

[honest trench]

Why two times

oh

that worked

but i don't get why it worked

[urban lodge]

Here's a visual explanation

when you are calculating the intersection of two friends, you will notice you will overcount the intersection of all three two times over

so you subtract the intersection of all three times 2 to account for that

if you were finding the probability of exactly two friends, then you would have that sum - 3*intersection of all three

[honest trench]

ah

[urban lodge]

I can give you a more set-theoretic explanation if you want

involving inclusion-exclusion

[honest trench]

i think i get the gist of it

[honest trench]

@urban lodgeI am extremely sorry to disturb you again, but is it okay if you can explain the concept of subtracting twice and thrice again. I encountered a similar problem and when I did subtract thrice then I got the correct answer, but I don't understand why that would give me a correct answer. I would really aprpecieate your help.

[honest trench]

can't notice that

ohhh

did notice it

[honest trench]

Here, don't we overcount it thrice?

how did u come to the conclusion that we overcount twice? Could you explain that

[urban lodge]

We are probably thinking of different meanings for the word "overcount"

$\text{Rick c eB } \cap \text{ Brenda c eB } \cap \text{ Ali c eB }$ is counted three times in $(\text{ Rick c eB } \cap \text{ Brenda c eB }) + (\text{ Rick c eB } \cap \text{ Ali c eB }) + (\text{ Ali c eB } \cap \text{ Brenda c eB })$

[honest trench]

ye

right that's what i was thinking

so why did u say 2

[vivid meadow]

ALPH2H

[honest trench]

because when I did 2 it was correct

[urban lodge]

We want to count it once but we "go over" twice

hence why I said two

We just had different meanings for "overcount"

semantic difference

[honest trench]

Could u elaborate a bit more

oooo

i see

ok

then in which scenario do we subtract thrice?

cause when I did a similar question to the friends choosing the entrance, i subtracted thrice instead of twice and got the correct answer

but the only difference was

the question was asking exactly two

instead of at least two

[urban lodge]

Notice that
$\text{Rick c eB } \cap \text{ Brenda c eB } \cap \text{ Ali c eB } \subset{\text{ Rick c eB } \cap \text{ Brenda c eB }} \ $
$\text{Rick c eB } \cap \text{ Brenda c eB } \cap \text{ Ali c eB } \subset {\text{ Rick c eB } \cap \text{ Ali c eB }} \ $
$\text{Rick c eB } \cap \text{ Brenda c eB } \cap \text{ Ali c eB} \subset{\text{ Ali c eB } \cap \text{ Brenda c eB }}$

rip

[honest trench]

xd

[honest trench]

I get it going over twice. if that's what you were trying to explain. I am now wondering when it goes over thrice

if that's supposed to belongs to

then yes

i can see that

[urban lodge]

Idk why it's not going to the next line

[honest trench]

try splitting the $ line $ for each line

maybe then it will

[urban lodge]

We have $\text{Rick c eB } \cap \text{ Brenda c eB } \cap \text{ Ali c eB } \subset{\text{ Rick c eB } \cap \text{ Brenda c eB }} \ $ and
$\text{Rick c eB } \cap \text{ Brenda c eB } \cap \text{ Ali c eB } \subset {\text{ Rick c eB } \cap \text{ Ali c eB }} \ $ and
$\text{Rick c eB } \cap \text{ Brenda c eB } \cap \text{ Ali c eB} \subset{\text{ Ali c eB } \cap \text{ Brenda c eB }}$

[vivid meadow]

ALPH2H

[urban lodge]

good enough

[urban lodge]

So as you can see, the intersection of three sets is a subset of the 3 different intersections of two sets

hence, we will be counting it 2 extra times

we want to count it only once

so we have to subtract 2 times the intersection of three sets

[honest trench]

Right.

[urban lodge]

I am now wondering when it goes over thrice
It goes over when we are summing the intersection of two sets

[honest trench]

o

[urban lodge]

does that make sense?

[honest trench]

yes

[honest trench]

This part

uh kinda

wdym summing the intersection of two sets

the before part i get

[urban lodge]

$\mathbb{P}(R \cap B) + \mathbb{P}(R \cap A) + \mathbb{P}(B \cap A)$

[vivid meadow]

ALPH2H

[honest trench]

hmmm

[urban lodge]

Going back to the visual venn-diagram explanation

here's one intersection

[honest trench]

yes

[urban lodge]

here's another

and here's the last one

@honest trench notice how this space is covered in all 3

that's where the overcount happens

The general formula for at least m events occurring given a set of events is the "Generalized Inclusion-Exclusion Principle"

[lofty portal]

@honest trench https://youtu.be/N3I2ZLbh6zQ?list=PLUl4u3cNGP60hI9ATjSFgLZpbNJ7myAg6&t=374

I guess the general formula is a bit different from this show but in principle You need to be aware of overlaps so you are not doubling the probabilities on the intersections (or tripling if there are multiple overlaps at once)

[honest trench]

Ok. Thnks guys

apprecieate ya'll a lot

[lofty portal]

are joint densities easy?

[frosty field]

yes

[lofty portal]

even if you are lackluster at the multivariable stuff :v

It has been a minute since I have touched multivariable integration and differentiation

[urban lodge]

I mean, difficulty is all relative

[lofty portal]

True

[frosty field]

yeah mechanically it's more complicated but idk if it's harder to understand

[lofty portal]

so the P_XY(x,y) is just the probablity that the two events happen at the same time?

lets say you have a deck of cards (52): 4 suits, and your (2-A)

outcome card draw

X(s): 1 = Heart, 2 = Club, 3 = Diamond, 4=Spade
Y(s)= 2=2, 3=3, 4=4,... A=13?

P_XY(Heart, 2) = 1/52?

[urban lodge]

Joint density is just the intersection so yeah

well it doesn't have to be at the same time

[lofty portal]

P_XY(X(2 of Hearts) = 1, Y(2 of Hearts) = 2) = P_XY(1,2)

maybe that is what I should have said

= 1/52

[lofty portal]

wdym by this? It is just a numerical value on the same outcome is it not?

[urban lodge]

Yes it is

But that “same outcome” doesn’t necessarily have to be at the same time

Usually it is but it doesn’t have to be

[rustic stirrup]

Hello everyone, if X and Y are i.i.d r.v following normal distribution with the same standard deviation. Is is true that X+Y follow normal distribution with mean as sum of means of X and Y and standard deviation as of X and Y
or is standard deviation of X+Y is square root of sums of squares of standard deviations of X and Y

[urban lodge]

Yep 👍

The mean will be the sum of means

The variance will be the sum of variances

[marble fog]

or is standard deviation of X+Y is square root of sums of squares of standard deviations of X and Y
yes, this

[nimble geyser]

Bin A contains five red balls and three blue balls. Bin B contains three red and two blue balls. One ball is drawn at random from each bin. Find the probability P that one is red and one is blue.

[nimble geyser]

can someone plz interpret this problem and which method i have to apply to get the result

Solution is 19/40

[frozen relic]

Law of total prob

[urban lodge]

There are two possibilities

Either you pick a red ball from Bin A and a blue ball from Bin B

or you pick a red ball from Bin B and a blue ball from Bin A

probability of the first case is $\frac{5}{8} \cdot \frac{2}{5}$

[vivid meadow]

ALPH2H

[urban lodge]

probability of the second case is $\frac{3}{8} \cdot \frac{3}{5}$

[vivid meadow]

ALPH2H

[urban lodge]

and these two events are disjoint

so you can sum the two products to get the final answer without sieving away anything

[nimble geyser]

Thank you so much @urban lodge

Is there any best course to take on this as it's says prerequisite required is the probability theory

@urban lodge plz tell me bruh where can I find the best resource to learn probability I mean there may be certain YT channels or websites it could save a lot of time I would stick on to

[lofty portal]

#probability-statistics message

you can also try khan academy but might be enough

[urban lodge]

^

[lofty portal]

why do they need to be independent?

,w convolution

[vivid meadow]

Results provided by WolframAlpha

Click here to refine your query online
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Requested by Brandon7716#6524

[surreal abyss]

Because in order to determine the distribution of W, you must first determine the joint distribution of X and Y. If they are independent, then we know f(x,y) = f(y)f(x), but if they are not independent then the joint distribution must be given

[surreal abyss]

You can also easily check that these theorems hold if you remember how to do substitutions in a multivariable function (dividing by the Jacobian), so you create variables W = X + Y and U = X. Then, after you find the joint distribution of W and U, f(u,w), you can take the marginal distribution of W by integrating (or summing, in the case of discrete variables) over the entire range of X relative to W.

[lofty portal]

yeah i don't remember the whole substitutions in a multivariable function

[vivid meadow]

pewdssssssss

[pliant rivet]

Assuming independence moment

[frosty field]

claim: all random variables are independent
proof: https://www.youtube.com/watch?v=r7l0Rq9E8MY

[ivory scroll]

Does the variable k have some special meaning in statistics or at least in regards to Probability Density Functions? I am asked to find the expectation of this value, I know the answer is 4/3, but I am not sure how you get that from a PDF of kx with state space 0 to 2.

[marble fog]

no, k is just k

you're supposed to first find the value of k knowing that f_X is a density function (and so its integral over the entire real line is 1)

[ivory scroll]

Okay thank you

[grand dome]

about statistics i am not sure how they can formulate such you end on sigma bigger then 1,96b/root(3n)

I already got to the upper equation where you find the 2*normaldistribuation-1 > 95% but the lower part confuses me how can you find that value out.

[frozen relic]

Its delta not sigma

And you just mean from the 1st inequality to the 2nd?

Then its just isolating delta?

Multiply by b and divide by sqrt(3n) and you have isolated delta

[grand dome]

okay it could be maybe just me being confuse so forgive me but

i would have to isolate it from this integral and when i tried to integrate it the function was pretty unusual for my case(mostly because this kind of question may encounter me in my exam and it looks pretty time intensive.)

[grand dome]

no i mean trying to get that inequality in the first place $delta\sqrt{3nb^{-2}}\geq 1960$

[vivid meadow]

The Unknown Cho

[grand dome]

shoot if it didnt worked with what i gave you the full context is finding the confidence intervall of a certain problem. The problem in question is a baker basically bakes bread and the possibility of the weight of the bread is not always the same rather its uniform distributed in the intervall (a-b,b+a). I have to find a confidence intervall in which the probability is 95 percentage that weight of the bread is in that intervall.

In this problem the random variable is the weight of the baked bread. I got up to following part (let me photograph it and sent it real quick

[frozen relic]

I mean is it not just using z score 1.96 gives 0.975

And 2*0.975-1=0.95

So hence it must be at least 1.96

[grand dome]

well you are right.

You just numerically tested what value came out or rather checked the table. I was just hoping there would be a more methodical approach to it thats not fight dealing with the integrals or making dealing with the integral easier. But yeah you are right you could just solve it that way.

[frozen relic]

I mean more like just know that value since its a common one

And there is a reason tables existed back in the day (because no nice antiderivative) (no reason to still use tables when everyone has phone/pc)

[grand dome]

fair okay then

[lofty portal]

the Z= X+Y ,Z= X/Y, Z= XY, a bit confusing

[frosty field]

What's confusing about it?

[lofty portal]

I just really didn't get on of the derivations 😦

So I feel like if I see any problem it is going to be me plugging it into the established theorem

[frosty field]

Which derivations are giving you trouble?

[lofty portal]

okay like this one

#2

seems they start by looking at the CDF

I get the drawing

w= x +y

but I am a little fuzzy on the first part

The bounds

I see we integrate with respect to y and then x

we have w=x+y

so we could have

also w refers to the line right?

I just am not sure how the dx bound is formed

I get by defintion of cdf we got from -inf to inf

then based on the picture

idk... im fuzzy at this calc stuff

[frosty field]

Yeah w is some fixed number so x+y = w is a line

[lofty portal]

so we want the area to the left of the line

[frosty field]

Yes

[lofty portal]

but idk exactly how these bounds are setup. I can see the y upper bound as w-x

by rewriting the equation

but we won't have that same bound for x?

is that because we will have it in terms of x upon intergration of the first integral?

[frosty field]

One of them needs to be free of any restriction of the other

Both variables go from -inf to inf, but we need a restriction on how one of them gets there

Since there's a region that we aren't interested in (the one to the right of the line)

[lofty portal]

okay and the reason we have f_X(x)*f_Y(y) is because of independence?

[frosty field]

Looks like it

If they weren't, we could just use the joint density

[lofty portal]

so they can pull the fX(x) out because it isn't part of the variable being integrated?

[frosty field]

It's independent of y yeah

[lofty portal]

But I don't get how we lose the lower bound term of F_Y(-inf)

[frosty field]

It's 0 by definition

[lofty portal]

ohh

yeah

okay.

okay how does this derivative work again

d/dw 2nd part

[frosty field]

x is fixed, so when you take the derivative wrt w it's the same as usually taking the derivative of any cdf

[lofty portal]

I guess I am missing something. I don't see how the steps work. We are integrating with respect to x and then taking the derivative with respect to w

[frosty field]

Try using the chain rule

Set z = w-x and find d/dw F(z)

[lofty portal]

im still confused on the inside integral at the same time...

[frosty field]

Confused about what

[lofty portal]

f(w-x)*(d/dw(w-x))?

f(w-x)*1?

[frosty field]

No and no

[lofty portal]

wait...

[frosty field]

d/dw (w-x) = 1

[lofty portal]

oh ...

lol

d/dx [f(g(x))]= f'(g(x))*g'(x), this is the chain rule right?

[frosty field]

Yes

[lofty portal]

oh I guess I also didn't read
Assume that the integrand in the above equation is sufficiently smooth so that differentiation and integration can be interchanged

wait would I not need to do the multiplication rule?

[frosty field]

That's the case for most RVs so it's a good assumption

Multiplication rule on what?

[lofty portal]

like if I move the d/dw

inside

[frosty field]

But f(x) doesn't depend on w so its derivative is 0

[lofty portal]

yeah

[frosty field]

That term will cancel out

[lofty portal]

that is what I was thinking okay

and then d/dx F_Y(w-x) = f_Y(w-x)* (d/dw)(w-x) = f_Y(w-x)(1) = f_Y(w-x)

okay I think I got it

although I followed the steps but still not sure what it "intuitively" means

other than I guess I can now integrate withrespect to 1 variable to get f_W(w)?

[frosty field]

Notice how the sum of the arguments is w; it's basically adding up all points where the two variables sum to w (not really since continuous but essentially)

Or computing the probability that the two independent variables take on a sum of w at every possible pair of points where it's possible

[lofty portal]

okay

yeah I kinda get what it is saying or at least what it is suppose to compute but didn't have much of a mental picture about it

so is computing the cdf of it hard?

[frosty field]

It depends on the distribution

[lofty portal]

like what is the CDF of W=X +Y . We didn't go over it :v

or is there not enough information to really care out anything given it is in general?

[frosty field]

Perhaps that'll be covered later

Don't overwork yourself

This is the super technical part of intro probability

[lofty portal]

other than throwing another integral

[frosty field]

It'll partially depend on X and Y since that'll determine the support of W

[lofty portal]

and so in general these new variables are suppose to have some physicsal meaning and context with units and such?

like X/Y or X+Y or X*Y would have units attached?

like maybe X and Y refer to cm so then you might have cm^2 or something

[frosty field]

yes

[urban lodge]

Do you want a intuitive explanation of this

[lofty portal]

Sure

[urban lodge]

Alright let's focus on the summation term first

this formulation

of X = x

and Y = w - x

guarantees that W = w

X + Y
x + (w - x) = w

so we have a way to guarantee that for a valid value of X, random variable W equals w

we pass these values to the pdf to get their marginal probabilities

[lofty portal]

oh

yo

[urban lodge]

we know that when X = x and when Y = w - x, we can guarantee that W = w. Now we want to know the probability of W actually being w.

[lofty portal]

The marginal probabilities are just $P_X(x) \text{ and } P_Y(w-x)$

[urban lodge]

so let's get the marginal probabilities

[vivid meadow]

Brandon7716

[urban lodge]

and since we know the random variables are independent we can just multiply them

makes sense so far?

[lofty portal]

okay

I guess

but what are we multiplying them anyways?

[urban lodge]

This part is important

[lofty portal]

that is what I kinda didn't get

[urban lodge]

why are we multiplying them?

[lofty portal]

is it not one outcome we are looking at

X(s), Y(s) , W(s) = X(s)+Y(s)

?

[urban lodge]

Well yeah, we just want the probability that our random variable W will equal 50 say

so how many ways are there to do that?

this is where the summation comes in

Let's so over every possible value of X(s)

[lofty portal]

well the summation is just iterating over all the values

I think

[urban lodge]

note we have a way to guarantee if X(s) is say 12, W will equal 50. Why does that matter? Well we explicitly want to know the probability that W will equal 50

so we are going over every possible way W can equal 50, and getting the probability of that way

and then summing them

[lofty portal]

so W is dependent but they are both independent?

[urban lodge]

they are independent of each other.

W is dependent on them

[lofty portal]

okay

oh I think I see now. w=x+y, and we force an x, then w-x = y

[urban lodge]

you may ask why we don't need to sieve anything away? I.e. Apply some inclusion-exclusion

[lofty portal]

as we are iterating through all the x's and indirectly choosing y along the way

[urban lodge]

Well it's because these "events" are disjoint

[lofty portal]

how are these events disjoint in every case?

w=50, x=25, y=25

[urban lodge]

Let A be the event where after our experiment, X is 12 and Y is 38
Let b be the event after our experiment, X is 13 and Y is 37

A and B can't both happen. X can't be 12 and 13 after our experiment. it can be one or the other. They are mutually exclusive

[urban lodge]

You got it

[lofty portal]

okay I see how this makes sense so by disjoint you meant among the X itself ?

Anyways I see how this works now

[urban lodge]

yes

[lofty portal]

okay am I stupid or what

,w 15 + 80

[vivid meadow]

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[lofty portal]

?

am I wrong in thinking they are complements?

[pliant rivet]

It’s conditional probability, not intersections

P(Guilty | Can discredit) and P(Guilty | Cannot discredit)

[lofty portal]

P(guilty | discredit

oh...

i see

nvm

[pliant rivet]

[lofty portal]

,w sum (0,inf) (1/2)*(1/8)^(n)

[vivid meadow]

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[lofty portal]

,w (1/2)/(1-(1/8))

[vivid meadow]

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[lofty portal]

,w (1/2)^6 = (1/8)^2

[vivid meadow]

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[lofty portal]

,w sum (0,inf) (1/2)*(1/2)^(4+3n)

[vivid meadow]

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[lofty portal]

,w (1/28)+(1/4)

[vivid meadow]

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[agile widget]

,w What is the probability of the next person you meeting having a phone number that ends in 5

[vivid meadow]

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[agile widget]

,w What is the probability of getting all heads if you flip 3 coins?

[vivid meadow]

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[lofty portal]

,w sum (0, inf) of (1/2)(1/2)^(1+3n)

[vivid meadow]

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[brave wedge]

When comparing OLS to MLE I understand that they can be shown to be the same very easily especially in the case of no intercept term
Given y = xb + e where we know e ~ N(0, sigma^2). As such e = y - xb so y - xb must be N(0, sigma^2) so when we find the MLE of the normal distribution, take the natural log, and differentiate w.r.t b we wind up with X(y - Xb) = 0 or X(e) = 0. Obviously this is the same as the OLS case

Where I am confused is on the intuition. Why would finding the maximum of the likiehood function w.r.t to beta find the beta that minimizes the euclidean distance?

[lofty portal]

,w sum(0,inf) of (1/4)(1/8)^n

[vivid meadow]

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[lofty portal]

is my logic in this problem wrong?

so I am asked to compute the number of ways of spelling "CALCULUS" from the string "ACCLLUUS"

and the question is what is the chance of arranging it to spell calculus

I wanted to initially say 1/5040

ie the 1 way of spelling CALCULUS

and (8!/2!*2!*2!) = 5040

in total

but I then though what if we allow the letters to be "distinct" for the duplicates

just in the chance of spelling

actuall nvm

I was thinking you could have 8 cases of spelling CALCULUS correctly

so idk which one to go off

[cinder brook]

think more

[lofty portal]

well im picturing something shuffling letters around as objects

and I guess in my numerators considered all the cases where you allow for the L's, C's, and U's to be different

[cinder brook]

i mean you said is my logic wrong but you haven't picked which one you're more convinced by yet

[lofty portal]

idk I could see both

Because it is asked about arranging the letters to spell a word

[cinder brook]

it should give the same answer

[lofty portal]

basically I was adding more chance because you could "accidently" spell it right with your letters 8 different ways but then not considered it in the denomiator as we are not only interested in unique words but I think this would be wrong beacuse if I allow that for every word the probabilites will sum to greater than one

[cinder brook]

we want to spell "OOOOOK" from the letters "OOOOOK", we should get 1 in 6 chance whether we bother shuffling the Os or we count the words we can make

[lofty portal]

okay so I guess my flaw was basically saying we allow "O's" to be distinct in the top numerator but not in the words spelled. I see now

[cinder brook]

yeah i think so

[lofty portal]

,w (7!/2!)

[vivid meadow]

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[lofty portal]

,w 2520*2

[vivid meadow]

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[austere aspen]

are discrete random variables part of pns or discrete math?

probability & statistics

ohh

aight

and how do u get the "advanced" role

[marble fog]

#get-advanced-access

[lofty portal]

,w 1- (99999/100000)^(10000)

[vivid meadow]

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[lofty portal]

,w (99999/100000)^(10000)

[vivid meadow]

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[austere aspen]

ty

[lofty portal]

,w (10,000 choose 1) (1/100,000)(99,999/100,000)^(9999)

[vivid meadow]

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[lofty portal]

,w 1- (99,999/100,000)^(10,000)

[vivid meadow]

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[lofty portal]

duhh

,w sum (1,10000) of (10000 choose n)(1/100,000)^(n)(99,999/100,000)^(10,000-n)

wait

[vivid meadow]

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[lofty portal]

,w \sum_{k=0}^{10000}\frac{\left(n!\right)}{k!\left(n-k\right)!}\cdot\left(p\right)^{k}\cdot q^{\left(n-k\right)}

[vivid meadow]

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[lofty portal]

,w \sum_{k=0}^{10000}\frac{\left(n!\right)}{k!\left(n-k\right)!}\cdot\left(p\right)^{k}\cdot q^{\left(n-k\right)} where n=10000, p=1/100000, q=1-p

[vivid meadow]

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[junior heath]

Hiii guys

How to find constants here ?

I dun know what is the other equation

I just know the sum integral is 1

So I only have one equation for 2 variables

[lofty portal]

is that the entire question?

if so maybe you are just suppose to be it in terms of possible ranges of values but I would think there would be many in many

,w integral (-2,0) xdx

[vivid meadow]

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[lofty portal]

-2c1+(0.5)c2=1

,w integral (0,1) xdx

[vivid meadow]

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[lofty portal]

so maybe you just give a vector of solutions

[junior heath]

I do this

This is full question

[lofty portal]

I would have thought so maybe for the first question it would be okay to have it arbitrary but now it sounds like a pain for the rest even if it is just a constant

[junior heath]

Hold on f need to be continue at x=-2

So c1 is 0

Am I right ?

[marble fog]

a pdf need not be continuous

just integrable

[lofty portal]

,w sum (0,15) k*(15 choose k)(7/8)^(k)*(1/8)^(15-k)

[vivid meadow]

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[lofty portal]

np

,w (15)(7/8)

[vivid meadow]

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[dense nova]

Is the assumption, that for large enough samples we can usually estimate anything (minus its mean, divided by its variance) as being a std based on the "physical" world or on maths?

Because, although I am an engineer and I use it everyday, confidence intervals still give me the creeps.

[urban lodge]

are you talking about law of large numbers?

[dense nova]

Is that the thing that let me turn a lot of things into a t-distribution?

[urban lodge]

Sorry I’m not really familiar with t-distributions

Law of large numbers states that you can estimate the population mean through the sample mean

The more samples, the more accurate your estimation

[frosty field]

LLN wouldn't let you estimate anything, but indeed the mean would be fair game

[dense nova]

Right, but this is just based on an observation people have made IRL?

[urban lodge]

There is a proof of it

https://en.wikipedia.org/wiki/Law_of_large_numbers#Proof_of_the_weak_law

[frosty field]

and this is an ok proof of the strong LLN https://towardsdatascience.com/proof-of-the-law-of-large-numbers-part-2-the-strong-law-356aa608ca5d

[dense nova]

Alright, will definetly look into those.

[lofty portal]

Do I learn about making inferences on data on a second semester stats course?

[lofty portal]

Kinda interesting in how you can interpret and extrapolate

Also curious are just knowing the moments of the data enough to get a picture of what the Data looks like?

[frosty field]

it depends, I learned it in my first but yours seems more probability and math focused than stats focused

[lofty portal]

yeah but i guess im somewhat naive that 1 course is going to be enough to "get" stats

[frosty field]

probability and stats are different entities

but it's more useful to have an understanding of prob. going into stats than it is the other way around

[lofty portal]

so... what is stats?

[frosty field]

stats uses some of the tools developed in probability to understand and explain trends in data

[junior heath]

Guys, is the mean value of this is 500 ?

And hmmmm what is the variance here ?

[frosty field]

how did you get a mean of 500

[lofty portal]

E(X^2) + E(X)^2 if we assume such expectations are defined.

at least that is the formula I have been using

which idk if this is the same thing

[frosty field]

that's not particularly helpful for this problem

but it is true in general

[lofty portal]

well what the normal distributions might tell us

[frosty field]

we'll get to that

i'm first interested in where the mean of 500 came from

[lofty portal]

I feel like this is a basic stats question and I have no clue how I would do it. I feel like it relies heavy on the definition of "normal distribution"

not sure how you can compute a mean when all the salaries are given as a range of values (unless the normal stuff is important here)

[frosty field]

it is important, yes

you can get the mean and variance from the info given

[lofty portal]

,w normal distribution

...

[vivid meadow]

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[lofty portal]

,w erfc

[vivid meadow]

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[lime pawn]

Would the concept of percentile be useful there? (Or is it applicable?) If 50 employees have a salary of less than $200, then $200 would be like the 50/1000 or 5% or 5th percentile of salaries.

[junior heath]

I get it wrong by doing (200+800)/2

[junior heath]

So I compute the probability without finding the pmf ?

[frosty field]

it says salary follows a normal distribution, which doesn't have a pmf (it has a pdf, which you also don't need)

vkurt is definitely on the right track

[lofty portal]

welp chat

I am going to be back with more probability questions

[junior heath]

What should I name X ?

I dunno what to let it be formally here

Let X be the continuous r.v for the distribution of the salary of the employees

What do you think?

[hard glacier]

You enter a contest where one ball must be drawn from each of three bags. The first bag contains 3 green balls and 2 red balls. The second bag contains 5 green balls and 2 red balls. The third bag contains 2 green balls and 4 red balls. You win if you select exactly two green balls. What is the probability that you will win? The picture above shows what I have done so far. I tried multiplying the probabilities for each bag together but I don't think that's the right answer. Can someone check my work please?

you're welcome @sonic solar

[lime pawn]

I think you should include the probability of drawing a red ball with each product, because you are drawing three balls. The winning draw could look like: (G,G,R), (G,R,G), (R,G,G).

So the first one should be (3/5)(5/7)(4/6).

[lofty portal]

@lime pawn what are tips for found bounds on these questions 😦 (sorry for ping)

[frosty field]

sure you can call it that

let X represent the salary of a randomly chosen employee, $X \sim N(\mu, \sigma^2)$

[vivid meadow]

Steakanator

[frosty field]

you know that Pr(X < 200) = 50/1000 = 0.05
and Pr(X < 800) = 1-200/1000 = 0.8

how can you use this information to find the mean and variance?

[lofty portal]

,w graph y < 3x

[vivid meadow]

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[lime pawn]

I would graph the support together with the inequality given and check where the bounds would be.

[lofty portal]

,w graph f(x,y) = 2e^(-x-y)

[vivid meadow]

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[lofty portal]

idk if this is what you meant though

it already states this

I just was / am not sure how to do P(Y < 3X)

[lime pawn]

Yes. Graph those, together with y<3x.

No need to plot the actual density.

[lofty portal]

oh

i see

[junior heath]

Guys what does E(Y|X=2) equal to ?

[frosty field]

impossible to answer without more info

[lofty portal]

,w graph y < 3x, 0 < x< y, 0 < y

[frosty field]

lmao

[vivid meadow]

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[lofty portal]

this is worthless anyways

does it matter the order of bounds of integration

ie

dydx or dxdy

I don't think so

[frosty field]

it depends on how you set your bounds

[lofty portal]

oh

yeah

true

[junior heath]

Also is P(X>1; Y<2) = P(X>1 and Y<2) ?

[frosty field]

the region isn't rectangular so the bounds change if the order of integration changes

[frosty field]

probably

[junior heath]

And hmm what is this one ?

E(Y and X=2)/E(X=2) ?

[lofty portal]

bruh im forgetting basic calc 😦

,w int 2e^-x

[vivid meadow]

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[lofty portal]

[frosty field]

by definition, $E[Y | X=2] = \sum_y y P(Y=y | X=2)$

[vivid meadow]

Steakanator

[lofty portal]

I know you use the u sub stuff but I use to be able to do this quicker

[frosty field]

i mean integral of e^x is just e^x so just use the reverse chain rule and you're good

[junior heath]

Is my marginal correct?

[frosty field]

yes

[junior heath]

I was worry since I got the same thing

[frosty field]

if you look at the joint mass function again, it's symmetric across the diagonal

so it makes sense that you'd get the same thing

[junior heath]

I see

[lofty portal]

,w ti

[vivid meadow]

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[lofty portal]

,ti

[vivid meadow]

The current time for Brandon7716 is 01:00 AM (EDT) on Fri, 22/07/2022.

[lofty portal]

[frosty field]

gn

[lofty portal]

bruh didn't think prob would be so much calc

:v

more so with all the infinities showing up

although I guess probabilities and areas go very much hand and hand.

Maybe I need to think about this stuff all from a calculus standpoint

Start looking at the graphs or something idk

[junior heath]

At (e), what happen if Y=1 like this ?

[junior heath]

My solution so far

Could you guys check my domain also ?

[frozen relic]

You need 0<x<1<2x

So x can’t be 0 or 0.1 for example

[junior heath]

Like this ?

I put the marginal in the integral

[frozen relic]

Did you even listen to what I said

All values less then 3/4 of x aren’t valid (well a subset of that just have 0 prob of happening)

[junior heath]

Ohhh sorry I just come here and paste my old paper

Let me see

[junior heath]

Hmmm why ?

1/2<x<3/4 is valid ?

[frosty field]

1 < 2x

[junior heath]

Yes I take 1/2<x from that

[frosty field]

oh sorry i misread

[junior heath]

Yes yes

I thought u go to sleep with Brandon

[frozen relic]

Yes 1/2<X<3/4

[frosty field]

nah

[junior heath]

Also I shouldn't put the marginal inside the integral right?

I have no solid reason at all, I thought I plug value in marginal then put in the integral

And yeahh what is P(Y=1) ?

[frozen relic]

Find the conditional PDF of X given Y=y

Given by joint divided by marginal of y

Then integrate conditional on region of interest

[junior heath]

So $P(Y=1)=f_Y(1)$ right?

[vivid meadow]

Potato

[frozen relic]

P(Y=y)=0 for any continious

[junior heath]

Hmmmm doesn't it mean I get divide by zero

[frozen relic]

What?

Just do what I said above

[junior heath]

$P(X<\frac{3}{4})|Y=1)=\frac{P(X<\frac{3}{4}, Y=1)}{P(Y=1)}$

[vivid meadow]

Potato

[frozen relic]

That is only formula for denom>0

[junior heath]

( dun mind my latex )

[frozen relic]

I legit said the 2 steps you needed to do

What don’t you understand from that?

[junior heath]

Denom ?

[frozen relic]

Obv not well defined when P(Y=1)=0

So that makes no sense to write

[junior heath]

Let me try