Finding the Pdf of two RVs

Question

I have the following density function

\begin{equation} f_{X, Y}\left(x,y\right)=\:\frac{1}{x},\ \text{for} \ 0 \ \leq \ y \ \leq \ x \ \leq 1 \end{equation}

I need to find the probability density function for \begin{equation} Z = X+Y \end{equation}

Here is my works so far

\begin{equation}f_{X}\left(x\right)=\:\int _0^x\frac{1}{x}dy\:=\:1\:\\f_{Y}\left(y\right)=\:\int _y^1\:\frac{1}{x}dx\:=\:-ln\left(y\right)\: \end{equation} Now \begin{equation} {f_Z\left(z\right)=\:\int _{-\infty }^{\infty }\:f_Y\left(y\right)f_X\left(z-y\right)dy\:=\:\int _{-\infty \:}^{\infty \:}\:\left[-ln\left(y\right)\right]dy\:} \end{equation} Which am stuck now on the limits of integration for the Z pdf. How do I find those please help.

Answer 1

In general, we have: $$f_Z(z)=\int_{-\infty}^\infty f_{XY}(w,z-w)dw$$ What you've written (the convolution formula) is true when $X$ and $Y$ are independent only, which enables you to factorize the above joint PDF.

Here, to take this integral, you need to plot the region of support, as well as $x+y=z$ line. Only then, it'll be obvious that we need to take this integral for $0\leq z\leq 1$ and $1< z \leq 2$ cases. $$0\leq z \leq 1\rightarrow f_Z(z)=\int_{z/2}^z\frac{1}{w}dw=\ln w\bigg\vert_{z/2}^z=\ln2$$ $$1<z\leq2\rightarrow f_Z(z)=\int_{z/2}^1\frac{1}{w}dw=\ln w\bigg\vert_{z/2}^1=\ln2-\ln z$$

Edit: Region of support (RoS) is the region between the lines $y=0,x=1,y=x$. When you intersect $x+y=z$ line with RoS (draw it), if $0\leq z\leq1$, the intersection line segment will start at $y=x$ and end at $y=0$. The boundaries in the integral are the x coordinates of these points. When $1<z\leq 2$, it'll again start at $y=x$ but end at $x=1$.