Question: I have two independent random variables (say $X$ and $Y$) such that $X \sim U[0,1]$ and $Y \sim$ Exp$(1)$, and I want to find the PDF of $Z=X+Y$.

My attempt: I know $f_X(x)=1$ for $x \in[0,1]$, and $f_Y(y)=e^{-y}$ for $y \in [0, \infty)$.

I also know that, due to their independence, $f_Z(z)=(f_X * f_Y)(z)$ where $(f_X * f_Y)(z)$ is the convolution of $f_X$ and $f_Y$.

Furthermore, $(f_X * f_Y)(z)=\int^{\infty}_{-\infty} f_X(z-y)f_Y(y) dy = \int^{\infty}_{-\infty} f_Y(z-x)f_X(x) dx$.

However, I am unsure of a few things:

  • Can I use a convolution approach even though $f_X$ and $f_Y$ are not defined for all real numbers?
  • If I can, how would I determine the bounds of the integral given $f_X$ and $f_Y$ are defined for different subsets of the real numbers ($x \in[0,1]$ and $y \in [0, \infty)$ respectively)?

Context: Ultimately, I need to compute $P(Z>z)$ for two different cases (when $z\in[0,1]$ and when $z>1$), so I planned on integrating $f_Z(z)$ to get the CDF for $Z$.

Any help would be greatly appreciated.


1 Answer 1


When you say $f_X(x)=1$ for $0<x<1$ what you really mean is $f_X(x)=1$ for $0<x<1$ an d $f(x)=0$ for all other $x$. All density functions are defined on the entire real line. So there is no problem in using the convolution formula.

In this case $(f_X*f_Y)(z)=\int_{-\infty}^{\infty} f_X(z-y)f_Y(y) dy$. [This is the general formula for convolution]. Let $z >0$. Note that $f_Y(y)=0$ if $y <0$ and $f_X(z-y)=0$ if $z-y \notin (0,1)$ i.e., if $y \notin (z-1,z)$. Hence integration is over all positive $y$ satisfying $z-1<y<z$. In order to carry out this integration you have to consider two cases: $z >1$ and $z <1$. In the first case the integration is from $z-1$ to $z$. In the second case it is from $0$ to $z$.

  • $\begingroup$ Can you please clarify the sentence "Note that $f_Y(y)=0$ if $y <0$ and $f_X(x-y)=0$ of $z-y \notin (0,1)$ is if $y \notin (z-1,z)$."? $\endgroup$
    – Viv4660
    May 20, 2020 at 6:08
  • $\begingroup$ Typos have been corrected. The exponential density is $0$ on $(-\infty,0)$ and uniform density is $0$ outisde the interval $(0,1)$. Carry out the integration over the portion where $f_X(z-y)f_Y(y)$ is not zero. @Viv4660 $\endgroup$ May 20, 2020 at 6:14
  • $\begingroup$ I thought $f_X(x)=1$ for $x\in [0,1]$ including $0$ and $1$, and so I have been working without strict inequalities. Is this incorrect? $\endgroup$
    – Viv4660
    May 20, 2020 at 6:28
  • $\begingroup$ Also, considering the two cases $z>1$ and $z<1$, what about when $z=1$? $\endgroup$
    – Viv4660
    May 20, 2020 at 6:29
  • $\begingroup$ For the purpose of integration values at the end points have no effect. Changing the values of a function at a finite number of points does not change the value of the integral. For this reason I am completely ignoring the end points. ( In fact density functions are defined only up to sets of measure $0$). @Viv4660 $\endgroup$ May 20, 2020 at 6:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .