Question: Suppose that random variables $X$ and $Y$ are independent, $X$ is uniformly distributed on $[1,2]$ and $Y$ has the following density function:
$f_Y(y) = \begin{cases} \cfrac{2y}9: & 0 \le y \le 3 \\ 0: & \mathrm{elsewhere} \end{cases}$
Find the distribution function $F(t)$ of the random variable $T=\max(X,Y).$
Attempt:
$F_X(x) = \begin{cases} x-1: & 1<x<2 \\ 0: & \mathrm{otherwise}\end{cases}$
$F_Y(y) = \begin{cases} \cfrac{y^2}9: & 0<y<3 \\ 0: & \mathrm{otherwise}\end{cases}$
I know that $T=\max(X,Y) \implies F_T(t)=F_X(t)\cdot F_Y(t)$, but domain of first distribution function is different from domain second distribution function. Thus, if I multiply them directly, result is nonsense (e.g. not monotonic increasing function). In this case what should I do? Thanks in advance.
Best Answer
You can rewrite $F_X$ as $$F_X(x) = (x-1) \mathbf{1}_{(1,2)}(x) + \mathbf{1}_{[2,\infty)}(x)\tag{$x\in\mathbb{R}$}$$ and $F_Y$ as $$F_Y(x) = \frac{x^2}{9} \mathbf{1}_{(0,3)}(x) + \mathbf{1}_{[3,\infty)}\tag{$x\in\mathbb{R}$}$$ to see how they are both defined on $\mathbb{R}$. Then you can multiply them and simplify the expression, using that $\mathbf{1}_A\cdot \mathbf{1}_B = \mathbf{1}_{A\cap B}$.
Here is an illustration (plot made in Mathematica):