Need to find P(Y < 2) by conditioning on X where Y is uniformly distributed over (0,x)

Question

The density of $X$ is $f(x) = xe^{-x}$, for $x > 0$. Given $X = x$, $Y$ is uniformly distributed on $(0, x)$.

I need to find $P(Y < 2)$ by conditioning on $X$.

Please explain step by step... I am new to conditional distribution

Answer 1

The joint density is the product of the density of $X$ with the conditional density of $Y$ given $X=x$. $$\begin{align}f_{X,Y}(x,y) &= f_X(x) \cdot f_{Y \mid X=x}(y) \\[1ex]&= x\mathrm e^{-x} \mathbf{1}_{(0,\infty)}(x) \cdot \frac{1}{x} \mathbf{1}_{(0, x)}(y)\\[1ex]&=\mathrm e^{-x}\,\mathbf 1_{(0,\infty)}(x)\,\mathbf 1_{(0,x)}(y)\end{align}$$

Then, integrate this joint density over the region of the $(x,y)$ plane specified by $y < 2$. $$\begin{align} P(Y < 2) &= \iint_{y < 2} f_{X,Y}(x, y) \,\mathrm dx \,\mathrm dy \\[1ex] &= \int_{-\infty}^\infty \int_{-\infty}^2 f_{X,Y}(x,y) \,\mathrm dy \,\mathrm dx \\[1ex] &= \int_0^\infty \mathrm e^{-x} \int_0^{\min\{2, x\}} \,\mathrm dy \,\mathrm dx \\[1ex] &~~\vdots \end{align}$$