Let $X$ be a random discrete variable with probability mass function (pmf) of $p_X(x) = P(X = x)$. Let $Y = g(X)$ (from $\mathbb{R}$ to $\mathbb{R}$). Then, why is it that: $$p_Y(y) = \sum_{x \in g^{-1}(y)}p_X(x)$$


Let $X \in \{1,2,3,4,5,6\}$ be six sided dice and binary $Y\in\{0,1\}$ with $Y=1$ if and only if $X$ is even.

$g^{-1}(y)$ is the inverse image which is the set of all the values for $X$ such that $g(x) = y$ wiki definition of inverse image.

Then $g^{-1}(y)\lvert_{y=1} = \{x \lvert x \ is \ even \} = \{2,4,6\}$ is the event that $X$ is even and this is by definition of $Y$ the event that $Y=1$. The events are the same so the probability is the same. The probability that $X \in \{2,4,6\}$ is off course

$$P_X(2) + P_X({4}) + P_X(6) = \sum_{x \in \{2,4,6\}}P_X(x) = \sum_{x \in g^{-1}(1)}P_X(x) = P_Y(y)$$.

So to summarize: You want to find the probability $P_Y(y)$ for a given value $y$ of the stochastic variable $Y$ that is defined as a function $Y=g(X)$. For $Y$ to take the value $y$ it must be the case that $X$ has taken some value $x$ such that $g(x) = y$ and the set of all the values that can bring the particular value $y$ about is $g^{-1}(y)$ by definition of the inverse image.

  • $\begingroup$ Wow, thanks! I understand it now. $\endgroup$ – user12055579 Dec 12 '19 at 2:44

Because $Y =y$ iff $g(X)=y$ iff $X \in g^{-1}(\{y\})$.


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.