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?

  • $\begingroup$ 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

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.

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

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