How to compute conditional probability in this case?

[shrewd flare]

I believe I know how to compute, it is just the integral from 2log2 to +inf of f_X|Y (x, log2) dx
But there should it also be 0 since for y taking a specific actual value like log2 should be 0, right? And again another reason for it be 0 is that P(X|Y) = P(X intersect Y) / P(Y) but P(X intersect Y) should be 0 again since it is the area of a ray which should be 0

So what is it? And can you fix my erronous thinking then?

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[orchid harbor]

Hello, no that's not true. The conditional probabilities are defined using the densities and not the probability measures directly

[shrewd flare]

I was thinking that too since I've been doing exercises with these for the past hour but this thought came in my head and I couldn't remove this possibility too

[orchid harbor]

That is, by definition: $f_{X \vert Y} (x \vert y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}$

[knotty ginkgo]

Nagi

[orchid harbor]

Now, if you work on discrete probabilities, you have atoms

That is, the probability of a singleton {y} will not be zero

However, that is a quite specific case

[shrewd flare]

So the argument of P(A inter B) = P(A|B) * P(B) is only something to discrete probability spaces?

[orchid harbor]

No, that always applies.

Assuming that A and B are events

Ah, wait a sec

[shrewd flare]

You get what I mean

[orchid harbor]

Sorry, not that one

But P(A inter B) = P(A | B) P(B)

This is always true for any event A and B

[shrewd flare]

oh yeah my badf

[orchid harbor]

The issue is that if P(B) is zero, there is little you can conclude

[shrewd flare]

Ahhh

That does explain it

Since we are in 0 = P(A|B) * 0 and that is why we can't conclude anything

[orchid harbor]

But that's mainly the heuristics that motivate the definition of conditional probability through densities

So again, in your case, fall back to the definition that involves densities

[shrewd flare]

I see

I didn't learn this in the book Im reading

That conditional probabilities ought be defined through densities and abolish the measure directly

[orchid harbor]

I'll check my own lecture notes just in case, if I have them

[shrewd flare]

Thank you very much for the help here

+rep @orchid harbor

where is the bot when you need it

[orchid harbor]

Hmm I believe you need to close the post to give points

But before you do, I will just check my lecture notes

to make sure I am not telling you bullshit

[shrewd flare]

Nah it does make sense

It explains everything I thought wrong

[orchid harbor]

Ok yes, so to recap:
P(A | B) = P(A inter B)/P(B) is well-defined ONLY when P(B) is not zero. Since we established that in the case P(B) is zero, we can't do anything here.
In order to work with conditional probabilities, one needs to go through a wide arsenal of measure theory results

First, prove that:

For any $G \in \mathcal{G}$, there exists a measurable function, denoted $x \mapsto P_{Y | X=x}(G)$, defined uniquely $P_X$-almost everywhere, such that:
$$\forall H \in \mathcal{E}, P_{X, Y} (H \times G) = \int_{H} P_{Y \vert X = x}(G) dP_X(x)$$

[knotty ginkgo]

Nagi

[orchid harbor]

And this is already a huge endeavor

And only then, in the case of random variables that have a density, you can define:
$$f_{Y \vert X = x}(y) = \frac{f_{X,Y}(x,y)}{f_X(x)}$$
and show that for $P_X$-almost any $x$, the function $f_{Y \vert X =x}$ is well-defined, and $P_{Y \vert X=x}$ is of density $f_{Y \vert X = x}$

[knotty ginkgo]

Nagi

[orchid harbor]

So as you can see, this is not simple at all. But the takeaway is very clear: you define conditional probas using densities whenever you can do so

[shrewd flare]

@orchid harbor Just a moment, so this means that this problem is not something necessarly for multiple r.v. but for continous r.v.

[orchid harbor]

Sorry? Can you elaborate?

[shrewd flare]

Even in the case of a single continous random variable uniform on [0,1] then P( X in Q inter [0,1/2] ) should be problematic since it would be a division by 0/0

Right?

There is nothing wrong with what you said, I just want to check if this thinking is correct

[orchid harbor]

It depends on the measure on the ambient space, here it's a bit tricky

because A = {X in Q} is in fact an event, that is measurable with respect to the probability measure of X

but B = [0, 1/2], idk what that is, and it might not be measurable with respect to the same probability measure

Does that make sense?

$A = {\omega \in \Omega : X(\omega) \in \mathbb{Q}}$

[knotty ginkgo]

Nagi

[orchid harbor]

Oh, wait

Am I interpreting this wrong

$A = { X \in \bQ}, B = { X \in [0, 1/2]}$

[knotty ginkgo]

Nagi

[orchid harbor]

Then yes, here since P(A) is zero, you cannot define P(B | A) (assuming X follows a uniform distribution)

but you can define P(A | B), since P(B) is nonzero

It just happens that P(A | B) is still zero anyway

[shrewd flare]

Wait

I see, so a correct (general) definition of conditional probabilitties can be only through densities, it is merely a condicidence that it cna be done with the measures directly in the discrete case

[orchid harbor]

That is a good interpretation

[shrewd flare]

Is this a good conclusion I should leave with?

A ok

[orchid harbor]

Yes, at the moment, it is fine

You can prove that the property using densities will be equivalent to what you'd get using the basic definition with sets

since ultimately, discrete probability distributions do have a density

just defined on a different measure, contrary to continuous rvs which have a density with respect to the Lebesgue measure

The definition using densities is robust even if you use some mixtures of both discrete and continuous

[shrewd flare]

Thank you again very much

I wish my teacher took his time to explain this like you

[orchid harbor]

You are welcome

[shrewd flare]

+close

[cedar sundial]

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[orchid harbor]

It seems that you have to press both the blue and the red button @shrewd flare

[cedar sundial]

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