Covariance/Independence

[steel fox]

Following on from a post i made previously, after finding the covariance as -93750 from the table provided, a follow up question which after looking back has given me trouble too, is "are x and y independent" the method i used was finding the correlation coefficient using the second formula provided to get a value between -1 and 1, however since looking back i think ive also done this incorrectly by using a formula for normal distribution to obtain the variance/standard deviations of both x and y. Can anyone show me the correct method of obtaining the SD for x and y to use in the formula given?

[hollow remnant]

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[jaunty thunder]

As usual, for discrete distributions we have:
E(X) = x(1) p(x(1)) + x(2) p(x(2)) + ...
E(X^2) = x(1)^2 p(x(1)) + x(2)^2 p(x(2)) + ...
Var(X) = E(X^2) - E(X)^2
σ(X) = √(Var(X))
cov(X, Y) = E(XY) - E(X)E(Y)
And you already have the formula for ρ(X, Y).

I recommend writing the distributions of X, Y and XY separately first.

You'll be able to calculate E(X), E(Y) and E(XY) easier that way.

[steel fox]

Var(X) = E(X^2) - E(X)^2

for this calculation specifically, doing var(X) first, would i be doing (85000x1/12)^2 + (85000x1/4)^2 ...... + (110000x1/12)^2 for the E(X^2) part and then - 97500^2 for the E(X)^2 part

[jaunty thunder]

You need to calculate E(X^2) first.

Also, that's not correct. It sums over x(k)^2 p(x(k)), not (x(k) p(x(k)))^2.

Probability isn't squared in that calculation.

[steel fox]

ah okay i think that is where my error came from, im struggling with stats a bit since i keep using continuous formulas for discrete variables etc for some reason, i think this is the solution but can i leave this chat open until i have calculated it to check

[jaunty thunder]

Sure!

As I said, better to find the marginal distributions of X, Y and XY first.

Then you can easily automate it in Excel.

[steel fox]

i wouldnt be able to use excel in an exam so i was just doing everything by hand for practice but i used the same method as this but with the ^2 in the correct places, resulting in 156250000 for the var(X), would you be able to verify this part was done correctly, as the rest of the question would just be doing this for Var(y) so i should be able to finish from here assuming that part was done correctly

also just to add, would you have any online formula sheets to recommend which may include descriptions of which formulae to use with discrete vs continuous data, i know that it should be pretty clear but its an area i seem to be making a lot of mistakes in atm, but if you dont know of one, no worries, thanks for the help

[jaunty thunder]

Well, the only difference is you use sums for discrete data and integrals for continuous data. That's about it.

For example, if X is discrete:
E(X) = Σ(xp(x), x ∈ supp(X))
And if X is continuous:
E(X) = ∫(xf(x)dx, supp(X))

As you can see, pretty much the same.

[steel fox]

would using this formula for finding var and SD be for continuous only, im a bit confused because i dont tend to see integrals in most of these types of equations

[steel fox]

also was the new method i described here correct to obtain 156250000 for var(X), this question didnt have any answer options so its a bit more difficult to verify if the method was wrong

[jaunty thunder]

No, that's a competely different thing. That's sample variance.

You don't need it here it all.

[steel fox]

even if it was the same but only over n instead of n-1? for population var

[jaunty thunder]

Again, that's related to the sample of a distribution.
You are doing probability theory for now, not statistics. Forget about the sample parameters.

[steel fox]

oh okay, i think thats what im confused with sometimes i dont really know why.

[steel fox]

was the method described here the correct way as far as you can see then? im just trying to note down on my formula sheets which situations to use what

[jaunty thunder]

Let me check.

[steel fox]

it was with the ^2 inside the bracket to clarify sorry**

[jaunty thunder]

Hm. Well, I'm getting cov(X, Y) = 0.

So, no need to find the variances, as ρ(X, Y) will also be 0 anyway.

[steel fox]

you attempted this question a few days ago i believe and got -93750 for cov

i also repeated it after and got that too

[jaunty thunder]

Ah, really? Hm...

Can you show how we did it?

This is what I'm getting now.

Ah, wait.

I see the problem, hold on.

[jaunty thunder]

Yeah, here you go. This should be good.

Did you get the same result?

[steel fox]

yea thank you you helped a ton

[jaunty thunder]

Great, you're welcome!

Also, in this question it would probably be better to work with X' = X/1000 and Y' = Y/10. The values would be smaller, except for ρ(X, Y), which would stay the same.

Do you see why?

[steel fox]

oh okay i havent done stats for years so ill have to try it again to see probably, bit slow at it now

[jaunty thunder]

Remember - this is still just probability theory, not statistics.

[steel fox]

yeah i couldnt remember what to call it or google so i was struggling with finding similar examples

[jaunty thunder]

Ah, no worries. Just make sure to practice.

[steel fox]

i will thanks for helping, im sure ill be back here at some point anyway hahah so maybe ull see me again

[jaunty thunder]

As for what I meant here, here are some properties. Here a, b, a', b' are constants.
E(aX + b) = aE(X) + b
Var(aX + b) = a^2 Var(X)
σ(aX + b) = |a|σ(X)
cov(aX + b, a'Y + b') = aa' cov(X, Y)
And if a, a' ≠ 0:
ρ(aX + b, a'Y + b') = sgn(aa')ρ(X, Y)