# Why is the probability mass function of a transformed discrete random variable summed over the inverse values of the function?

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.

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

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