If discrete random variable $X$ take certain values, then calculate pdf of $Y=(X-2)^2$, mean and dispersion.

Question

So if discrete random variable $X$ takes values: $-2, 1, 2, 3, 6$, then I have to find density function $f(y)$ for a variable $Y=(X-2)^2$. Each of these variables have a probability of $0.2$. So for $X$ the $F(x)$ is $$F(x)=\begin{cases} \frac{1}{5}, & \text{ if } -2\leqslant x<1 \\ \frac{2}{5}, & \text{ if } 1\leqslant x<2 \\ \frac{3}{5}, & \text{ if } 2\leqslant x<3 \\ \frac{4}{5}, & \text{ if } 3\leqslant x<6 \\ {1}, & \text{ if } x\geqslant 6. \end{cases}$$ But I don't know how to get density function for $Y$, neither how to find mean and dispersion. Can someone help?

Answer 1

We have $$ \mathbb P(Y=k) = \mathbb P((X-2)^2=k) = \begin{cases} \frac15,& k=0\\ \frac25,& k=1\\ \frac25,& k=16. \end{cases} $$ It follows that $$ \mathbb E[Y] = \sum_k k\cdot\mathbb P(Y=k) = 0\cdot\frac15+1\cdot\frac25+16\cdot\frac25 = \frac{34}5, $$ $$ \mathbb E[Y^2] = \sum_k k^2\cdot\mathbb P(Y=k) = 0^2\cdot \frac15 + 1^2\cdot\frac25 + 16^2\cdot\frac 25 = \frac{514}5, $$ and hence $$ \operatorname{Var}(Y) = \mathbb E[Y^2] - \mathbb E[Y]^2 = \frac{514}5 - \left(\frac{34}5\right)^2 = \frac{1414}{25}. $$