1
$\begingroup$

I reason that is a random variable because Y is a random variable, thus making Px acting randomly. Example Y sample space is a roll of a die (1,2,3,4,5,6). So any of those values could be inputed in pX, thus making it act as a random variable. pX(1), pX(2), etc. Is my reasoning correct?

$\endgroup$
2
  • $\begingroup$ Maybe an example of what you mean: Consider an application in which the number of claims on a certain kind of policy made in a month is $N \sim \mathsf{Pois}(\lambda = 27)$ and then each claim results in a payout of $X_i \sim \mathsf{Exp}(\mathrm{rate} = 0.01)$ for a mean of $\mu = 100.$ What is the distribution of the total payout $T$ over a month? $E(T) = 2700,$ To find $Var(T)$ use a conditioning argument. Sometimes called 'random sum of random variables'. $\endgroup$
    – BruceET
    Commented May 31, 2020 at 2:12
  • $\begingroup$ By simulation, $P(T > 3000) \approx 0.32.$ $\endgroup$
    – BruceET
    Commented May 31, 2020 at 2:23

1 Answer 1

3
$\begingroup$

You are correct. $p_X(x)=f(x)$ is just another function and $p_X(Y)$ acts like a transformation on the random variable $Y$, thus making it a random variable.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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