# Random variable with exponential distribution.

Let $X$ be random variable with exponential distribution $\mathcal{E}(2)$ and let $Y$ be another random variable such that $$Y=\max\left(X^2, \frac{X+1}{2} \right).$$ Find the distribution for random variable $Y$.

Distribution for $X$ is $f_X(x)= 2e^{-2x}, x>0$ and zero otherwise.

Now, for variable $Y$ we have that it's distribution is zero whenever $y \leq \frac{1}{4}$

For $y=t> \frac{1}{2}$ we have the following:

$F_Y(t)=\int_0^{2t-1} f_X(x)dx= 1- e^{2-4t}$

Similarly, for $y=t>1$ we have

$F_Y(t)=\int_0^{\sqrt{t}} f_X(x)dx= 1- e^{-2\sqrt{t}}$

But, i cannot understand what happens in case that $y$ takes random value on interval $(\frac{1}{4}, \frac{1}{2})$. It's the black line on the graph. How can i handle situations like this? Any help appreciated!

Let's compute the CDF of $Y$ firstly: \begin{align}P(Y \le y) &= P(\max\left(X^2, \frac{X+1}{2}\right) \le y) \\&= P(X^2 \le y, \frac{X+1}{2} \le y) \\&= P(X^2 \le y, X \le 2y - 1) \\&= P(-\sqrt{y} \le X \le \sqrt{y}, X \le 2y - 1) \\&= \begin{cases} P(X \le \sqrt{y}) , y > 1 \\ P(X \le 2y - 1), \frac{1}{2} \le y \le 1 \\ 0, \text{otherwise}\end{cases}\end{align} As we can see $Y$ can't be less than $\frac{1}{2}$ because $max(X^2, \frac{X+1}{2})$ always $\ge \frac{1}{2}$

• @D F so whenever i conclude that $Y$ can't take any value below (or above) some specified value $a$ then i can say that it's distribution below (or above) that value $a$ must be zero. Is this correct? – cdummie Jul 30 '18 at 11:50
• almost correct. If $Y$ cannot take any value below some value $a$ then the CDF $F_Y(y) = 0 \ \forall y \le a$ because the CDF of $Y$ is $P(Y \le y)$ hence if $Y$ cannot be below $a$ then $\forall \ y \le a \to P(Y \le y) = F_Y(y) = 0$. However, if $Y$ cannot be above some value i.e. $P(Y \ge a) = 0$ then $\forall \ y \ge a \ \to F_Y(y) = 1$ since $P(Y \ge y) = 1 - P(Y \le y) = 1 - F_Y(y)$ – D F Jul 30 '18 at 12:00

Here from $X>0$ a.s. it follows directly that $Y>\frac12$ a.s.

Consequently $F_Y(y)=0$ for $y\leq\frac12$.

Observe that this matches with the expression $F_Y(t)=1-e^{2-4t}$ for $t>\frac12$ close to $\frac12$.

If there $t$ approaches $\frac12$ then the expression approaches $0$.

That is enough already to conclude that $F_Y(\frac12)=0$ and if this is the case then automatically $F_Y(t)=0$ for every $t\leq\frac12$.

This simply because $F$ is monotonically increasing and only takes values $\geq0$ (and $\leq1$ of course, but that is not relevant here).