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It is well known that the expected value for a function $f$ of a discrete random variable $\rho$ over $x$, the expected value is $Ef=\sum_{-\infty}^{\infty}f(x)P(\rho=x)$. I do not see how to derive this from(show it is equal to) the construction below.

Let $\rho$ be a discrete random variable on $x$ and define $f=f(\rho)$. I understand the motivation behind the expected value for $\rho$, given by $E\rho=\sum_{-\infty}^{\infty}xP(\rho=x)$.

Since $f$ is itself a discrete random variable taking only the values $y=f(x)$, it follows that $P(f=y)=\sum_{x|f(x)=y}P(\rho=x)$. Thus by the same motivation, $Ef=\sum_{-\infty}^{\infty}yP(f=y)$, which becomes $Ef=\sum_{-\infty}^{\infty}y\sum_{x|f(x)=y}P(\rho=x)$.

In Rozanov's Probability Theory: A Concise Course, it says this is equal to the well-known formula. How is this equal to the well-known formula? Sorry for any inaccuracies, I am a beginner in this.

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Start with $\mathbb Ef=\sum_{y=-\infty}^{\infty}y\sum_{x|f(x)=y}P(\rho=x)$. $$ \mathbb Ef=\sum_{y=-\infty}^{\infty}\color{red}{y}\sum_{x|f(x)=y}P(\rho=x) = \sum_{y=-\infty}^{\infty}\sum_{x|f(x)=y}\color{red}{y}P(\rho=x)=\sum_{y=-\infty}^{\infty}\sum_{x|f(x)=y}\color{red}{f(x)}P(\rho=x) $$ We got rid of $y$ under the sign of sum. In these two sums, you first take $y$ and then take all $x$ such that $f(x)=y$. This is the way to iterate over the all $x$ values: $\{x\in\mathbb Z\}=\bigcup_{y\in\mathbb Z}\{x: f(x)=y\}$. So this sums are equal to $$ \mathbb Ef= \sum_{x=-\infty}^{\infty}f(x)P(\rho=x). $$

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