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I have two random variables: $x = N(0, \sigma^2)$ and $y =U[0, b]$. I need to compute $E(x/(1+y))$. How does one go about doing this? They are independent so the joint pdf is just the product of the two pdfs but can the integral be computed in closed form or is this something that should just be done numerically?

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    $\begingroup$ Is this homework? Hint: Try to reason by symmetry… $\endgroup$
    – Neil G
    May 29, 2012 at 17:48
  • $\begingroup$ 1. No this is not homework. 2. That's true, I should have been more specific: The equation in my model is actually $\frac{1}{1+y}$ so it does not diverge. $\endgroup$
    – Alex
    May 29, 2012 at 17:51

2 Answers 2

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From your description (and comments) you're trying to calculate $E \left( \frac{X}{1+Y} \right)$ where $X \sim N(0,\sigma^2)$ and $Y \sim {\rm Uniform}(0,b)$. By independence,

$$E \left( \frac{X}{1+Y} \right) = E(X) \cdot E \left( \frac{1}{1+Y} \right)$$

We know $E(X) = 0$, therefore $$E \left( \frac{X}{1+Y} \right) = 0$$

as long as $ E \left( \frac{1}{1+Y} \right) $ is finite. We know that $\frac{1}{1+Y}$ is bounded within $\left(\frac{1}{1+b},1 \right)$ with probability 1, therefore its mean also is.

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    $\begingroup$ Isn't $Y\sim U(0,b)$ and, therefore, $1/(1+Y)$ bounded by $(1/(1+b),1)$ with probability 1? Because $b>0$, the expectation is finite anyways as you said. $\endgroup$
    – Néstor
    May 29, 2012 at 18:01
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    $\begingroup$ You're right. I was assuming $U(0,1)$. Will fix that. $\endgroup$
    – Macro
    May 29, 2012 at 18:02
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Since it's not homework, then the expectation is zero by symmetry. How could there be an argument suggesting the answer is $k>0$ without a similar argument suggesting $-k$?

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  • $\begingroup$ I presume you mean the symmetry under $x \to -x$. But you do need to establish that the expectation even exists before you can apply that ... :-). $\endgroup$
    – whuber
    May 29, 2012 at 17:58
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    $\begingroup$ @whuber: Yes, you're right. I initially wrote it up as the expectation is either zero or doesn't exist, but then he edited his question and it was clear that it did exist. Macro has a nice explicit answer. $\endgroup$
    – Neil G
    May 29, 2012 at 17:59
  • $\begingroup$ The problem would have been more interesting if X had a mean different from 0. Then you would have to calculate E(1/(1+y)) $\endgroup$ May 29, 2012 at 19:23

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