given pdf fx(x) = $\frac{2x}{9}$ when 0<x<3 and its 0 otherwise and
pdf for y given x is
fY|X(y|x) = $\frac{3y^2}{x^3}$ for 0<y<x
How do i find let's say P(Y<2)
I'm guessing i have to integrate as such
$$\int_{0} ^{3}P(Y<2|X=x)\frac{2x}{9}dx$$
But how do i find what P(Y<2|X=x) equals to for my integration? How do i go about doing this or generally just these type of questions?
The joint density of $(X,Y)$ is the product of the conditional density of $Y$ given $X$ and the marginal density of $X$: \begin{align} f_{X,Y}(x,y) &= f_{Y\mid X=x}(y\mid x) f_X(x)\\ &= \left(\frac{3y^2}{x^3}\cdot \mathsf 1_{(0,x)}(y) \right)\left(\frac{2x}9\cdot \mathsf 1_{(0,3)}(x) \right)\\ &= \frac23\left(\frac yx\right)^2\cdot\mathsf 1_{(0,3)\times(0,x)}(x,y). \end{align} To find the marginal density of $Y$, we integrate the joint density over the range of $X$: \begin{align} f_Y(y) &= \int_{\mathbb R} f_{X,Y}(x,y)\ \mathsf dx\\ &= \int_y^3 \frac23\left(\frac yx\right)^2\ \mathsf dx\\ &= \frac29y(3-y),\quad 0<y<3. \end{align} It follows then that for $y\in(0,3)$ we have \begin{align} F_Y(y) &:= \mathbb P(Y\leqslant y)\\ &= \int_0^y f_Y(t)\ \mathsf dt\\ &= \int_0^y \frac29t(3-t)\ \mathsf dt\\ &= \frac{1}{27} (9-2 y) y^2. \end{align} When $y=2$, this quantity is $\frac{20}{27}$.