Finding the pdf of $U=X+Y$

Question

$(X,Y)$ has the following joint pdf:

$f_{X,Y}(x,y)=x+y$ if $0<x<1, 0<y<1$

If $U=X+Y$, find the marginal pdf of $U$.

I have tried to do it using transformation.

I have considered the transformation $(X,Y)\rightarrow (U,Y)$ where $U=X+Y$.

Clearly, $0<U<2$.

Now, $x=u-y$.

The jacobian is $J(\frac{x,y}{u,y})=\frac{\delta x}{\delta u}=1$, so $|J|=1$.

So, joint pdf of $(U,Y)$ is:

$f_{U,Y}(u,y)=f_{X,Y}(u-y,y)|J|$

But $-1<u-y<2$, whereas $0<x<1$, so I don't think I can write $f_{X,Y}(u-y,y)=u-y+y=u$. I am getting stuck here.

Please anyone help me solve it. Thanks in advance.

Answer 1

For the purposes of the transformation we consider the map $(X,Y)\to (U, V)$ where $U=X+Y$ and $V=Y$. Following your work it follows that the joint density of $(U,V)$ is given by $$ f_{U,V}(u,v)=f_{X,Y}(u-v, v)=u\quad (0< v<1, v< u< v+1)) $$ and zero otherwise by application of the change of variables formula. Note that the joint density is supported on a parallelogram in the plane (sketch the region). To find the density of $U$ we integrate over $v$ i.e. $$ f_{U}(u)=\int_{S_{u}} f_{U, V}(u,v)\, dv $$ where $S_{u}=\{v\in\mathbb{R}\mid f_{U, V}(u,v)\neq 0\}$ is the slice of the parallelogram on which the joint density does not vanish.

For $0<u\leq 1$, we have $$ f_{U}(u)=\int_{0}^u u\, dv=u^2 $$ while for $1<u\leq 2$ $$ f_{U}(u)=\int_{u-1}^1 u\, dv=u(2-u)=2u-u^2. $$

Answer 2

$x, y\in [0,1]$ so $u=x+y\in [0,2]$. \begin{eqnarray*} F_U(u) &=& P(X+Y \leq u)\\ &=& \int_{x=0}^{x=1}\int_{y=0}^{y=1}(x+y){\bf 1}_{\{x+y \leq u\}}(x,y)dydx\\ &=& \int_{x=0}^{x=1\wedge u}\int_{y=0}^{y=1\wedge (u-x)}(x+y)dydx\\ \end{eqnarray*}

When $u\in [0,1]$ the integral becomes \begin{eqnarray*} F_U(u) &=& P(X+Y \leq u)\\ &=& \int_{x=0}^{x=u}\int_{y=0}^{y=u-x}(x+y)dydx\\ &=& \int_{x=0}^{x=u}\Big(xy+\frac{1}{2}y^2\Big)\Big|_{y=0}^{y=u-x}dx\\ &=& \int_{x=0}^{x=u}\Big(x(u-x)+\frac{1}{2}(u-x)^2\Big)dx\\ &=& \int_{x=0}^{x=u}\Big( xu-x^2+\frac{1}{2}u^2-ux+\frac{1}{2}x^2\Big)dx\\ &=& \frac{1}{2}\int_{x=0}^{x=u}\Big(u^2-x^2\Big)dx\\ &=& \frac{1}{2}\Big(xu^2-\frac{1}{3}x^3\Big)\Big|_{x=0}^{x=u}\\ &=& \frac{1}{2}\Big(u^3-\frac{1}{3}u^3\Big)\\ &=& \frac{1}{3}u^3\\ \end{eqnarray*} Thus, for $u\in [0,1]$ $f_U(u)=F'_U(u)=u^2$.

When $u\in [1,2]$ the integral becomes \begin{eqnarray*} F_U(u) &=& P(X+Y \leq u)\\ &=& \int_{x=0}^{x=1}\int_{y=0}^{y=1\wedge (u-x)}(x+y)dydx\\ &=& \int_{x=0}^{x=u-1}\int_{y=0}^{y=1}(x+y)dydx +\int_{x=u-1}^{x=1}\int_{y=0}^{y=u-x}(x+y)dydx\\ &=& \int_{x=0}^{x=u-1}\Big(xy+\frac{1}{2}y^2\Big)\Big|_{y=0}^{y=1}dx +\int_{x=u-1}^{x=1}\Big(xy+\frac{1}{2}y^2\Big)\Big|_{y=0}^{y=u-x}dx\\ &=& \int_{x=0}^{x=u-1}\Big(x+\frac{1}{2}\Big)dx +\int_{x=u-1}^{x=1}\Big(x(u-x)+\frac{1}{2}(u-x)^2\Big)dx\\ &=& \Big(\frac{1}{2}x^2+\frac{1}{2}x\Big)\Big|_{x=0}^{x=u-1} +\int_{x=u-1}^{x=1}\Big(xu-x^2+\frac{1}{2}u^2-xu+\frac{1}{2}x^2\Big)dx\\ &=& \Big(\frac{1}{2}(u-1)^2+\frac{1}{2}(u-1)\Big) +\frac{1}{2}\int_{x=u-1}^{x=1}\Big(u^2-x^2\Big)dx\\ &=& \frac{1}{2}u^2-u+\frac{1}{2}u +\frac{1}{2}\Big(xu^2-\frac{1}{3}x^3\Big)\Big|\int_{x=u-1}^{x=1}\\ &=& \frac{1}{2}u^2-\frac{1}{2}u +\frac{1}{2}\Big(u^2-\frac{1}{3} - (u-1)u^2+\frac{1}{3}(u-1)^3\Big)\\ &=& u^2-\frac{1}{3}-\frac{1}{3}u^3 \end{eqnarray*} Thus, for $u\in [1,2]$ $f_U(u)=F'_U(u)=2u-u^2$.

Hence, overall, the density is $f_U(u) = u^2{\bf 1}_{\{0\leq u\leq 1\}}+(2u-u^2){\bf 1}_{\{1< u\leq 2\}}$