# Convolution of Mixed Variables over Unique Domains

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.

When you say $$f_X(x)=1$$ for $$0 what you really mean is $$f_X(x)=1$$ for $$0 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 $$z-1. In order to carry out this integration you have to consider two cases: $$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$$.
• 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 $y \notin (z-1,z)$."? May 20 '20 at 6:08
• 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 May 20 '20 at 6:14
• 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? May 20 '20 at 6:28
• Also, considering the two cases $z>1$ and $z<1$, what about when $z=1$? May 20 '20 at 6:29
• 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 May 20 '20 at 6:31