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.

  • 1
    $\begingroup$ There are two basic techniques to tackle such integration problems. (1) Draw a picture of the domain of integration. (This domain is complicated, making the picture a useful tool.) (2) Make the density function explicit. This method usually uses the indicator function $\mathcal{I}_A(x,y),$ equal to $1$ when $(x,y)\in A$ and $0$ otherwise. Thus, one fully correct expression of the density is $$f_{X,Y}(x,y) = \frac{1}{x}\,\mathcal{I}_{0\le y}(x,y)\mathcal{I}_{y\le x}(x,y)\mathcal{I}_{x\le1}(x,y).$$ The rules of algebra here substitute for the geometric insight afforded by the picture. $\endgroup$
    – whuber
    Oct 29, 2019 at 13:39

1 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 $x=1$.

  • $\begingroup$ Can you show how you derived the limits of integration when z is between 0 and 1 and 1 and 2? Thank you $\endgroup$
    – JordenSH
    Oct 28, 2019 at 21:11
  • $\begingroup$ Draw the region of support, and highlight its intersection with x+y=z line. Range of x in the highlighted segment is the integral limits. Do this for both cases. $\endgroup$
    – gunes
    Oct 28, 2019 at 21:15
  • $\begingroup$ What do you mean by region of support? $\endgroup$
    – JordenSH
    Oct 28, 2019 at 21:41
  • $\begingroup$ Sorry am also just having hard time understanding why we have two cases for Z between [0,1] and [1,2] whereas x and y are limited between [0,1] $\endgroup$
    – JordenSH
    Oct 28, 2019 at 21:51
  • $\begingroup$ @Raykh I've added some explanation. Also, keep in mind that $0\leq z\leq 2$. $\endgroup$
    – gunes
    Oct 29, 2019 at 6:01

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