Let $X$ be uniform on $(-1, 2)$ and let $Y = X^2$. Find the pdf of $Y$.

So far I have noted that $F_X(x) = P(X \leq x) = \int_{-1}^x \frac{1}{3} dt = \frac{1}{3}(x+1)$.

Then, since $Y=X^2$, $y \in [0,4]$.

My initial attempt was to do the normal procedure of

$F_Y(y) = P(Y \leq y) = P(X^2 \leq y) = \begin{cases} P(X \geq -\sqrt{y}, \quad x \in [-1,0) \\ P(X \leq + \sqrt{y}, \quad x \in [0,2] \end{cases}$


$F_Y(y) = \begin{cases} F_X(-\sqrt{y}), \quad x \in [-1,0) \\ F_X(+\sqrt{y}), \quad x \in[0,2] \end{cases} = \begin{cases} \frac{1}{3}(1-\sqrt{y}), \quad x \in[-1,0) \\ \frac{1}{3}(1+\sqrt{y}), \quad x \in[0,2] \end{cases}$

I'm fairly happy that the CDF of $Y$ is continuous and well defined but I don't like the fact that I need to specify the x domain since it should after all be a function of y, right? Or is it necessary in this case since $Y=X^2$ is not one-to-one on the domain?

  • $\begingroup$ You should be able to write the final answer only using $y$ restrictions. $\endgroup$ Feb 23, 2019 at 10:39
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    $\begingroup$ It might be worth considering $F_Y(y)$ when $y$ is (a) below $0$, (b) between $0$ and $1$, (c) between $1$ and $4$, and (d) above $4$ $\endgroup$
    – Henry
    Feb 23, 2019 at 11:00

1 Answer 1


There is an error in your calculation (it seems you would get $F_Y(0) = 1/3$, but this probability should be $0$). Note that (for $y\geq 0$), $P(X^2\leq y)$ is actually equal to $P(-\sqrt{y}\le X \le \sqrt{y}) = \color{red}{F_X(\sqrt{y}) - F_X(-\sqrt{y})}$. Try and compute this red part for $y\in [0,4]$. You will need to remember that $F_X(t)$ will become $0$ if $t < -1$.

  • $\begingroup$ OK. I now get $F_Y(y) = \begin{cases} \frac{2}{3} \sqrt{y} , \quad 0 \leq y \leq 1 \\ \frac{1}{3} \sqrt{y} + 1\ \quad 1 \leq y \leq 4 \end{cases}$. Does that seem right? I took into account the restricted domain of $F_X(-\sqrt{y})$ so it makes sense mathematically but I am struggling to get intuition on this. I think we can understand it as the coefficient is twice as high in $0 \leq y \leq 1$ region since both $-1 \leq x < 0$ and $0 \leq x \leq 1$ regions are contributing, whereas for $y \in (1,4]$, only 1 part of $x$ domain is contributing. The $+1$ constant adds the area already accumulated $\endgroup$
    – user11128
    Feb 23, 2019 at 11:01
  • $\begingroup$ The part for $1\le y\le 4$ has a typo I think (needs brackets). Otherwise looks good. $\endgroup$ Feb 23, 2019 at 11:09

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