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Integration help for expectation of a function of a random variable

  1. Jul 9, 2010 #1
    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!
     
  2. jcsd
  3. Jul 10, 2010 #2
    [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?
    Hint: You're going to have to do an integration by parts within an integration by parts.
     
  4. Jul 13, 2010 #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.
     
  5. Jul 13, 2010 #4
    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.
     
  6. Jul 14, 2010 #5
    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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