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help a CS brainlet with a simple math question.

Anonymous Fri 22 Nov 15:26:28 2019 No.11168335 View Reply Original Report

Quoted By: >>11168342 >>11168347 >>11168356 >>11168398

I am writing a paper, and I am stuck in an analysis of what I have done close to the end.

According to my intuition the expectation of any term multiplied by something that is 0 in expectation is again 0 in expectation.
(pic only holds for independent RVs)

But I struggle to find any proof for this.

Note that I want to do this without guaranteeing anything about the other term (most notably independence).

Is my intuition wrong? Or can someone guide me towards a proof?

Would appreciate the help.

I will put 4chan anon into sources and post photo once the paper is published if someone can help out.

Anonymous

Anonymous Fri 22 Nov 2019 15:29:23 No.11168342 Report

Quoted By: >>11168371

>>11168335
>According to my intuition the expectation of any term multiplied by something that is 0 in expectation is again 0 in expectation.
wrong

Anonymous

Anonymous Fri 22 Nov 2019 15:30:25 No.11168347 Report

Quoted By: >>11168371

>>11168335
Read the sticky you tard.
>>>/adv/

Anonymous

Anonymous Fri 22 Nov 2019 15:33:50 No.11168356 Report

Quoted By: >>11168359

>>11168335
Consider a random variable that's 1 with probability 0.5, and -1 with probability 0.5.
E[X]=0. Let Y be 1 with probability 3/4, and 2 with probability 1/4. Let Y be X. Then E[XY] = 1.

Anonymous

Anonymous Fri 22 Nov 2019 15:34:53 No.11168359 Report

Quoted By: >>11168400

>>11168356
Forgot to delete one sentence. Should read:
Consider a random variable that's 1 with probability 0.5, and -1 with probability 0.5.
E[X]=0. Let Y be X. Then E[XY] = E[X^2 ] = e= 1.

Anonymous

Anonymous Fri 22 Nov 2019 15:39:16 No.11168371 Report

Quoted By: >>11168388

>>11168342
Ty for the answer. Can you point me towards any reasoing/counterexample. Would be very helpfull.

>>11168347

This is neither homework, nor career advice.

I feel like talking about something that was not discussed in university stochastic courses is fairly on topic. More then many other threads..

Anonymous

Anonymous Fri 22 Nov 2019 15:44:28 No.11168388 Report

Quoted By: >>11168418

>>11168371
You're writing a paper for Uni. That qualifies as homework. Clearly you should be taking more English courses.

Anonymous

Anonymous Fri 22 Nov 2019 15:48:20 No.11168398 Report

Quoted By: >>11168418

>>11168335
Expectated values are nothing but integrals. All the same rules apply.

$E[g(X,Y)] = \int_{\Omega} g(X,Y) dP \\ = \int_{\mathbb{R}^2} g(x,y) dP_{X,Y} (x,y) \\ = \int_{\mathbb{R}^2} g(x,y) \frac{dP_{X,Y} (x,y)}{d(x,y)} d(x,y) \\ = \int_{-\infty}^\infty \int_{-\infty}^\infty g(x,y) f_{X,Y}(x,y) dx dy $

If you want $E[X,Y] = E[X] E[Y]$ to hold then you must also be able to separate the joint density function $f_{X,Y}$ into the product $f_X f_Y$ then you just get

$E[XY] = \int_{-\infty}^\infty \int_{-\infty}^\infty xy f_{X,Y}(x,y) dx dy \\ = \int_{-\infty}^\infty \int_{-\infty}^\infty xy f_X(x) f_Y(y) dx dy \\ = \int_{-\infty}^\infty x f_X(x) dx \int_{-\infty}^\infty y f_Y(y) dy \\ = E[X] E[Y]$

Anonymous

Anonymous Fri 22 Nov 2019 15:48:31 No.11168400 Report

Quoted By: >>11168448 >>11168475

>>11168359
Thanks taking the effort to reply.

I have to admitt I can't follow your argument.

how is E[X^2]=1?

Dont we have 2 possible outcomes:
+1 (if both are one, or both are 1)
-1 (first negative and second positive or vise versa)

each with probability 1/2, and thus E[X^2]=0

fuck me I feel like a brainlet, maybe im just not used to the notations, it has been years since I had a stochastics clas..

Anonymous

Anonymous Fri 22 Nov 2019 15:56:37 No.11168418 Report

Quoted By:

>>11168388
Writing a paper is very different then homework in my experience. I guess we will have to let mods judge. And yes, I am sorry if my english is bad. It is my 3rd language, and I have to admit that I am not proofreading my 4chan posts multiple times. Hope its understandable.

>>11168398
I understand that. Thats the result I find everywhere googling.

However I think it should be possible to find another result that does not require independence, but only holds in the very specific case, that one of the expected values is 0. It seems very obvious thats the case. But I cant find any proof

Anonymous

Anonymous Fri 22 Nov 2019 16:11:46 No.11168448 Report

Quoted By: >>11168528

>>11168400
X takes values in {-1,1}, hence X^2 takes values in {1}, so the expectation of X^2 is 1.

Anonymous

Anonymous Fri 22 Nov 2019 16:26:12 No.11168475 Report

Quoted By: >>11168528

>>11168400
>I have to admitt I can't follow your argument.

suppose X is drawn from a mean-centered bell curve, then E[X^2] is variance, which isn't zero, right?

Anonymous

Anonymous Fri 22 Nov 2019 16:29:45 No.11168481 Report

Quoted By:

E[(XY)^n]=E[X^n]E[Y^n] is the same as saying that the covariance of X^n and Y^n is zero.

Since the covariance is unaffected by a constant shift, there's no reason to expect anything useful to come out of fixing the expectation of X^n (fixing E[X] gives you a nasty linear combination of the moments E[X^i], but with one degree of freedom you can eliminate at most one of them, e.g. by treating the linear combination as a depressed polynomial in X).

As for when E[(XY)^n] = E[X^n]E[Y^n] is guaranteed given constraints only on X, it's sufficient that X = 0 almost surely (i.e. that the sample space corresponds to the event X=0 almost everywhere). There may be possible ways to relax this condition, but I'm not aware of any of them.

Anonymous

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Anonymous Fri 22 Nov 2019 16:59:41 No.11168528 Report

Quoted By:

>>11168448
>>11168475
Thanks guys, I think I found where I went wrong.
I think I understand how I have to reword my argument now.

pic is my false thought process

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