integration help for expectation of a function of a random variable


by trance_dude
Tags: expectation, function, integration, random, variable
trance_dude
trance_dude is offline
#1
Jul9-10, 10:02 PM
P: 3
1. The problem statement, all variables and given/known data

Hello,
have a stats question I am hoping you guys can help with. The expectation of a function g of a random variable X is:

E[g(X)] = [tex]\int^{\infty}_{-\infty}[/tex] g(x)fx(x)dx

where fx is the pdf of X. For example, the particular expectation I am considering right now is:

E[g(X)] = [tex]\int^{-\infty}_{\infty}\frac{1}{1+ax^{2}}\cdot \frac{1}{\sqrt{2\pi}}[/tex][tex]e^{-x^{2} / 2}dx[/tex]

this form of integral (i.e. containing that particular e term) must happen often whenever one takes the expectation of a function which depends on a normal random variable. In general, what is the best approach to solve such integrals in closed form here? Integration by parts? I know that the normal curve itself must be integrated using a "trick" such as switching to polar coordinates. Integration by parts might help me isolate the e term to do so, but actually in this case I am not making much progress using that method because the other (first) term has x in the denominator. Any thoughts as to a general approach and/or to this specific problem are much appreciated. thanks!
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Raskolnikov
Raskolnikov is offline
#2
Jul10-10, 04:40 PM
P: 193
[tex]

\frac{d}{dx}[tan^{-1}(ax)] = \frac{a}{1 + a^2x^2}.

[/tex]

Also, since we have [tex] e^{-x^2/2} [/tex], we're going to want a [tex] -x [/tex] in the numerator. We can get this in a crafty sort of way by multiplying by [tex] \frac{-x}{-x} [/tex].

Can you finish from there?
Spoiler
Hint: You're going to have to do an integration by parts within an integration by parts.
trance_dude
trance_dude is offline
#3
Jul13-10, 05:16 PM
P: 3
Thanks for the response. Sadly, it appears that I am still stuck. I've tried it many different ways, with and without your suggested (-x / -x) term, and keep getting infinitely recursive integration by parts. I am clearly missing something. Might I ask what you are using for "U" in each of your two integrations by parts? Thanks much.

Raskolnikov
Raskolnikov is offline
#4
Jul13-10, 11:18 PM
P: 193

integration help for expectation of a function of a random variable


hmm...now that I try it fully, that integral doesn't work out. Are you sure you copied down the problem correctly? If yes, then I'm assuming there's a typo because the answer WolframAlpha is giving is:

[tex] \frac{\pi e^{\frac{1}{2a}} \ \ \ erfc(\frac{1}{\sqrt{2}\sqrt{a}})}{\sqrt{a}} [/tex], where erfc(z) is the complementary error function. It exists and I've read up on its definition; however, unless your teacher has mentioned it in class yet, I doubt it's the correct answer. Most likely, there's an error the problem you stated.
trance_dude
trance_dude is offline
#5
Jul14-10, 09:54 AM
P: 3
this isn't a homework problem - it's an actual equation I've encountered in a project I'm doing. Anyway, thanks for the response. The answer from Wolfram is helpful - I was getting close to a solution, I think, and perhaps that will get me to it.


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