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Let $X$ be the outcome of a fair die roll with probability mass function given by $$f_{X}(x)=\begin{cases}\frac{1}{6}&\text{ if }x\in\{1,2,3,4,5,6\}\\0&\text{ otherwise }\end{cases}$$ If $Y=(X−3)^2$, find the probability mass function of $Y$, $f_{Y}(y)$.

I already know the answer, but I don't understand how to solve the question at all. Here's the explanation in the answer sheet, below.

The probability mass function of $Y$ is expressed in terms of the known probabilities of $X$ as: $f_Y(y)=\mathbb{P}[Y=y]=\mathbb{P}[X\in\{x:(x-3)^2=y\}]$. We can calculate $f_Y(y)$ for each $y\in\{0,1,4,9\}$ as follows: (...the answer)

How did the domain of $X$ become $\{0,1,4,9\}$? What steps would be taken next?

Any help would be greatly appreciated, thank you.

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    Please put all relevant info from the image into the question itself, as much as you are able. – 2017-02-07

1 Answers 1

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Think about it in this way. Domain of $X$ is $\{1,2,3,4,5,6\}$, each with the same probability.The transformation to $Y=(X-3)^2$ does this $$\begin{aligned}X&\rightarrow Y\\ 1&\rightarrow(1-3)^2=4\\ 2&\rightarrow(2-3)^2=1\\ 3&\rightarrow(3-3)^2=0\\ 4&\rightarrow(4-3)^2=1\\ 5&\rightarrow(5-3)^2=4\\ 6&\rightarrow(6-3)^2=9\\ \end{aligned}$$ so that $\mathbb{P}[Y=0]=\frac{1}{6}$, $\mathbb{P}[Y=1]=\frac{2}{6}$ and you can calculate the rest.