This is one of my homework problems. Consider two random variables $X$ and $Y$. $X \sim U(1,2)$, and the distribution of $Y$ given $X$ is $\operatorname{Exp}(X)$. Given that $X \leq 1.5$ what is the joint distribution of $X$ and $Y$?
I know that the joint distribution of $X$ and $Y$ is just $xe^{-xy}$ for $1 \leq x \leq 2, y > 0$ and $0$ otherwise, but I am not sure how to find the joint distribution conditioned on $X$.
On
You have the joint probability density function, so just apply the definition of conditioning to obtain:
$$\begin{align}f_{X,Y\mid X\leq 1.5}(x,y) &=\dfrac{f_{X,Y}(x,y)\,\mathbf 1_{x\leq 1.5}}{\mathsf P(X\leq 1.5)}\\[1ex] &= 2x\mathrm e^{-xy}\,\mathbf 1_{1\leq x\leq 1.5}\,\mathbf 1_{0\leq y}\end{align}$$
Likewise the conditional joint cumulative distribution function shall be $$F_{X,Y\mid X\leq 1.5}(x,y) =\dfrac{F_{X,Y}(x,y)\,\mathbf 1_{x\leq 1.5}+F_{X,Y}(1.5,y)\,\mathbf 1_{1.5\lt x}}{\mathsf P(X\leq 1.5)}$$
You have $\Pr(X\le1.5) = 0.5.$ The conditional joint density given the event $X\le1.5$ is just the marginal joint density divided by that probability.
Consider for example $\Pr(X\le x\ \&\ Y\le y\mid X\le 1.5),$ where $1\le x\le 1.5.$
\begin{align} & \Pr(X\le x\ \&\ Y\le y\mid X\le 1.5) \\[8pt] = {} & \frac{\Pr(X\le x\ \&\ Y\le y\ \&\overbrace{\ X\le 1.5\,\,}\,)}{\Pr(X\le 1.5)} \\ & \text{(but, $x\le1.5,$ by hypothesis, the part under} \\ & \phantom{\text{(}}\text{the } \overbrace{\text{overbrace}} \text{ is redundant)} \\[10pt] = {} & \frac{\Pr(X\le x\ \&\ Y\le y)}{\Pr(X\le 1.5)}. \end{align}