Joint Density Functions Let $X$ have density $f(x)=2x$ for $0<x<1$ and let $Y$ be uniform on the interval $(1,2)$. Assume $X$ and $Y$ independent.
a) Give the joint density function of $(X,Y)$. Calculate the probability $P(Y-X>1.5)$
Since $X$ and $Y$ are independent, can I just say that $f(x,y)$ is just equal to $2x$? And how would I do the second part of the problem. I know that it is a double integral, but what am I integrating over? Thanks.

Since $Y$~$U([1,2])$, calling $g$ its density, we have that $$g(y)=\begin{cases}1 \, \text{if} \, y\in [1,2] \\ 0 \, \text{if} \, y\not \in [1,2]\, \end{cases}$$ 
By independence, letting $h(x,y)$ be the joint density function: 
$$h(x,y)=f(x)g(y)=\begin{cases}2x \, \text{if} \, (x,y)\in [0,1]\times [1,2]  \\ 0 \, \text{if} \, (x,y)\not \in [0,1]\times [1,2]\, \end{cases}$$
To integrate we want the intersection of $A=\left \{ (x,y) : y-x>1.5 \right \}$ and $B=[0,1]\times[1,2]$, since we know that the joint density is zero outside of $B$. Draw the line $y=x+1.5$ and the set $B$ to find $A\cap B$. If $y=2$ then $x=0.5$, therefore: 
$$P[Y-X>1.5]=\iint_{A}h(x,y)dxdy=\iint_{A\cap B}h(x,y)dxdy=\int_0^{0.5}\left [\int_{x+1.5}^22xdy\right ]dx$$
I leave the calculation to you.
