Let $X$ and $Y$ be two random variables with the joint probability density function (PDF) given by:
$$ f_{X,Y}(x,y) = \begin{cases} \frac{2}{3}(2x + y) & \text{if } 0 \leq x \leq 1 \text{ and } 0 \leq y \leq 1 \\ 0 & \text{otherwise} \end{cases}$$ How can I find the PDF of the random variable $U = X + Y$?. I can find the pdf if $X,Y$ are independent. But I have proved they aren't in fact. Could someone help me?
Let $Z = X+Y$, and notice that $Z$ takes on values in $[0,2]$.
For any fixed value of $z$, $$F_Z(z) = P\{Z \leq z\} = P\{X+Y \leq z\}\\ = \int\int_{x+y\le z} f_{X,Y}(x,y)\,\mathrm d(x,y) =\int_{-\infty}^{\infty}\left[ \int_{-\infty}^{z-x} f_{X,Y}(x,y)\,\mathrm dy\right]\,\mathrm dx$$ and so, using the rule for differentiating under the integral sign (see the comments following this answer if you have forgotten this) $$\begin{align*} f_Z(z) &= \frac{\partial}{\partial z}F_Z(z)\\ &= \frac{\partial}{\partial z}\int_{-\infty}^{\infty}\left[ \int_{-\infty}^{z-x} f_{X,Y}(x,y)\,\mathrm dy\right] \,\mathrm dx\\ &= \int_{-\infty}^{\infty}\frac{\partial}{\partial z}\left[ \int_{-\infty}^{z-x} f_{X,Y}(x,y)\,\mathrm dy\right]\,\mathrm dx\\ &= \int_{-\infty}^{\infty} f_{X,Y}(x,z-x)\,\mathrm dx\tag{1} \end{align*}$$
You will need to set up the integral shown in $(1)$ for four separate cases: $z < 0$, $0 \leq z \leq 1$, $1 < z \leq 2$, and $z > 2$, and then massage them a bit, e.g. noticing that the limits can be reduced to $0$ and $1$ since $f_{X,Y}(x,z-x) = 0$ if $x<0$ or $x>1$, and $z-x$ also needs to be in $[0,1]$ which also affects the limits differently in the four cases. Then, plug in the explicit formula for $f_{X,Y}$ and away we go! I hope the integral will work out to have value $0$ in the first and last cases.
