CDF of $\max(X,X^2)$ for $X$ uniformly distributed on $[-1,1]$

Question

$X$ is uniformly distributed on $[-1,1]$. And $Y=max(X,X^2)$.

What is $F_{Y}(t)$ , the CDF of $Y$?

My attempt:

I tried to graph it, but I think I found wrong. I found the joint pdf $5/6$. Is this correct? And what can I do next?

Answer 1

Note that $0\leqslant Y\leqslant1$ almost surely and that $[Y\leqslant y]=[-\sqrt{y}\leqslant X\leqslant y]$ for every $y$ in $(0,1)$. The length of the interval $[-\sqrt{y},y]$ is $y+\sqrt{y}$ hence $$F_Y(y)=P(Y\leqslant y)=\tfrac12(y+\sqrt{y}).$$

Answer 2

Using the joint cdf may not be the easiest way.

If $X\le 0$, then $Y=X^2$. If $X\gt 0$, then $Y=X$. Let us find the cdf of $Y$. The only interesting part is for $0\le y\le 1$.

With probability $\frac{1}{2}$, we have $X\le 0$. Conditional on $X\le 0$, the distribution of $X$ is uniform on $[-1,0]$. Thus
$$\Pr(Y\le y|X\le 0)=\Pr(X^2\le y|X\le 0)=\Pr(-\sqrt{y}\le X\le 0|X\le 0)=\sqrt{y}.$$

With probability $\frac{1}{2}$ we have $X\gt 0$. Conditional on $X\gt 0$, the distribution of $X$ is uniform on $[0,1]$. Thus $$\Pr(Y\le y|X\gt 0)=\Pr(X\le y|X\gt 0)=y.$$ We conclude that $$F_Y(y)=\frac{1}{2}(\sqrt{y})+\frac{1}{2}y$$ for $0\le y\le 1$.