Let $Y=\frac{X^{2}}{2}$, Determine $F_{Y}(y)$

Question

https://i.stack.imgur.com/KGHPW.jpg

I have figured out that $F_{Y}(y)=1/8$ for $0<y<8$. The only problem is that shouldn't P(Y=2)=P(X=2)+P(X=-2). If i sub the values into their respective densities, they give unequal values. Why is this the case?

Answer 1

• First of all it is the pdf $f_Y(y)$ rather than the cdf $F_Y(y)$. The complete pdf is

  $f_y(y)=\begin{cases} \frac18, \ \text{if} \ 0<y<8 \\ 0, \ \text{elsewhere} \end{cases}$

• In general $P(X=x)=0$ if X is continuous.

• But $F_Y(2)=F_X(2)$. $F_X(-2)=0$ since the cdf of $X$ is

  $F_X(x)=\begin{cases} 0, \ \ \text{if} \ x<0 \\ \frac1{16}x^2, \ \text{if} \ 0\leq x\leq 4 \\ 1, \ \ \text{if} \ x>4 \end{cases}$

$F_X(2)=\frac1{16}\cdot 2^2=\frac14$

And the cdf of Y is

$F_Y(y)=\begin{cases} 0, \ \ \text{if} \ y<0 \\ \frac1{8}y, \ \text{if} \ 0\leq y\leq 8 \\ 1, \ \ \text{if} \ y>8 \end{cases}$

$F_Y(2)=\frac18\cdot 2=\frac14$