1
$\begingroup$

I have the following problem:

Suppose $Z = Y^X$ is a random variable, where $Y$ is a generic random variable and $X$ is a binary random variable which takes value $1$ with probability $p$ and value $2$ with probability $1 - p$.

I need to find the expected value $\mathbb{E}[Z]$, but I don't even understand from where to start!

$\endgroup$
2
$\begingroup$

Let Y be a generic random variable, and let X be a binary random variable such that $P(X=1)=p$ and $P(X=2)=1-p$.

$$ E[Z] = E[Y^X|X=1]P(X=1) + E[Y^X|X=2]P(X=2) = E[Y|X=1]p + E[Y^2|X=2](1-p) $$

If $X$ and $Y$ are independent, then $E[Y|X=1]=E[Y]$ and $E[Y^2|X=2]=E[Y^2]$, which means that the above equation reduces to:

$$ E[Z] = E[Y]p + E[Y^2](1-p)$$

$\endgroup$
4
  • 2
    $\begingroup$ Note that the last step assumes $Y$ and $X$ are independent, which wasn't explicitly stated in the question. $\endgroup$ – Danica Feb 18 '17 at 1:40
  • $\begingroup$ Without assuming independence, the last step should be: $~\ldots~=~\mathsf E(Y\mid X=1)~p+\mathsf E(Y^2\mid X=2)~(1-p)$ $\endgroup$ – Graham Kemp Feb 18 '17 at 6:13
  • $\begingroup$ Thanks for the comments. I've updated my answer. $\endgroup$ – Michael R Feb 18 '17 at 13:48
  • $\begingroup$ Indeed it's very tricky because it doesn't assume that Y and X are indipendent. The problem offers 4 possible solutions that made me crazy: A) E[Z] is the mean square value of Y if p = 0 B) E[Z] is the mean square value of Y if p = 1 C) E[Z] is the expected value of Y if p = 0 D) E[z] is always the variance of Y. I can't see none of them. Just E[Z] = E[Y] for p = 1, but not p = 0. $\endgroup$ – blocknius Feb 18 '17 at 18:41
2
$\begingroup$

If you know nothing about the distribution of $Y$ you can, of course, say nothing about the distribution of $Z$.

But let's say you know only that $E[Y] = \mu$ and the variance of $Y$, which is $E[(Y-E[Y])^2] = \sigma^2$, for some real $\mu$ and non-negative $\sigma^2$. Then you can in fact find $E[Z]$:

$Z$ is a variable with value which $p$ of the time is $Y$ and $(1-p)$ of the time is $Y^2$. So
$$ E[Z] = pE[Y] + (1-p) E[Y^2] $$ We know $E[Y] = \mu$ and $E[E[(Y-E[Y])^2] = E[Y^2]-(E[Y])^2 = \sigma^2$ so that $$ E[Y^2] = \sigma^2+\mu^2 $$ and $$ E[Z] = p\mu + (1-p) (\sigma^2+\mu^2) $$

$\endgroup$
1
  • 2
    $\begingroup$ Like the other answer, this assumes that $Y$ and $X$ are independent. $\endgroup$ – Danica Feb 18 '17 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.