Question

X is randomly chosen from interval (0,2). If X=x, Y is randomly chosen from (x/2, x). Find probability distribution and probability density function of Y.

Answer 1

\[P(Y\le y)=\int\limits_{0}^{2}F_{Y|X}(y|x)f_X(x)dx\]

\[=\int\limits_{0}^{y}\frac{1}{2}dx+\int\limits_{y}^{\min(2y,2)}\frac{y-x/2}{x-x/2}\cdot\frac{1}{2}dx\]

Answer 2

\[\text{If $X = x$, the distribution of $Y$ is $f_{Y|X}(y|x) = \frac{1}{x - x/2}\mathbf{1}
_{[x/2, x]}(y) = \frac{2}{x}\mathbf{1}
_{[x/2, x]}(y)$. Then, }\]\[\text{$f_{X,Y}(x,y) = f_{Y|X}(y|x)f_X(x) = \frac{2}{x}\mathbf{1}_{[x/2,x]}(y)\mathbf{1}_{[0,1]}(x)$ and we get}\]\[\text{$f_Y(y) = \int_{-\infty}^{\infty} f_{X,Y}(x,y)dx = 2\int_0^1 \mathbf{1}_{[x/2,x]}(y)dx.$}\]\[\text{If $\frac{x}{2} \leq y \leq x$ then $y \leq x \leq 2y$ so that integral becomes}\]\[f_Y(y) = 2\int_y^{\min\{2y, 1\}}\frac{1}{x} = \begin{cases}2 \log{2} & \mbox{ if } 0 \leq y \leq 1/2 \\ 2 \log(1/y) & \mbox{ if } 1/2 \leq y \leq 1 \\ 0 & \mbox{ otherwise }\end{cases}.\]\[\text{By computing one more integral, we get $F_Y(y) = \begin{cases}2y \log{2} & \mbox{ if } 0 \leq y \leq 1/2 \\ 2y + 2\log(1/y) - 1 & \mbox{ if } 1/2 \leq y \leq 1 \\ 1 & \mbox{ if} y \geq 1 \\ 0 & \mbox{ otherwise }\end{cases}$.}\]

Answer 3

Why do you have \[F_Y(y)=1\text{ for } y\ge 1\]? (among other things)