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?

$\endgroup$
  • $\begingroup$ 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 May 31 at 2:12
  • $\begingroup$ By simulation, $P(T > 3000) \approx 0.32.$ $\endgroup$ – BruceET May 31 at 2:23
  • $\begingroup$ Thanks again, BruceET, you're awesome! $\endgroup$ – João Vitor Gomes May 31 at 14:16
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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you again, Gunes! $\endgroup$ – João Vitor Gomes May 31 at 14:17

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.