Integration help for expectation of a function of a random variable

In summary, the expectation of a function g of a random variable X can be found by using the formula E[g(X)] = \int^{\infty}_{-\infty} g(x)fx(x)dx, where fx is the pdf of X. This can often involve integrals with a term containing e, which can be solved by using techniques such as integration by parts or switching to polar coordinates. However, there may be cases where these methods do not work and a complementary error function may need to be used.
  • #1
trance_dude
3
0

Homework Statement



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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  • #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.
 
  • #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.
 
  • #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.
 
  • #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.
 

1. What is integration in relation to expectation of a function of a random variable?

Integration is a mathematical concept that involves finding the area under a function curve. In the context of expectation of a function of a random variable, integration is used to calculate the average value of a function over all possible outcomes of a random variable.

2. How is integration used to find the expectation of a function of a random variable?

Integration is used to find the expectation of a function of a random variable by taking the integral of the function multiplied by the probability density function (PDF) of the random variable. This results in a single value that represents the average value of the function over all possible outcomes of the random variable.

3. What is the difference between discrete and continuous random variables in terms of integration for calculating expectation?

Discrete random variables have a finite or countably infinite number of possible outcomes, while continuous random variables have an infinite number of possible outcomes. Integration for calculating expectation is used for continuous random variables, while discrete random variables use summation instead.

4. Can you provide an example of using integration to calculate the expectation of a function of a random variable?

One example is finding the expected value of rolling a fair six-sided die. The random variable is the number rolled, and the function is the square of the number rolled. The integral of x^2 from 1 to 6, multiplied by the PDF of a fair die (1/6), results in an expected value of 15/6 or 2.5.

5. Are there any special techniques for integrating functions of random variables?

Yes, there are special techniques such as the substitution method, integration by parts, and u-substitution that can be used to solve more complex integrals involving functions of random variables. It is important to understand the properties of the random variable and its PDF in order to determine the appropriate integration technique to use.

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