Finding convolution of exponential and uniform distribution- how to set integral limits?

Question

I am studying the convolution method for creating the density function of two independent random variables and I am struggling with understanding how the bounds for integrals are created.

There is one example of a problem where I don't get the right answer:

Let $X$ be an exponential random variable with parameter $\lambda$ and $Y$ be a uniform random variable on $[0,1]$ independent of $X$. Find the probability density function of $X + Y$:

so I have two marginal functions

$\mathbf{\\ f_X(x)=\begin{cases} \lambda e^{-\lambda x} & x\geq 0\\ 0 & \text{otherwise} \end{cases}} \ \hspace{20pt} \ \mathbf{f_Y(y)=\begin{cases} 1 & 0\leq x\leq 1\\ 0 & \text{otherwise} \end{cases}}$

I am looking for $Z=X+Y$. My understanding is that two cases should be considered. First for $0\leq z\leq 1$ and second for $z>1$

for $\mathbf{0\leq z\leq 1\\ (f_X*f_Y)(z)=\int_{0}^{z} f_Y(z-y) \ f_X(y) \ dy = \int_0^z \lambda e^{-\lambda y} dy = [-e^{-\lambda y}]_{0}^{z} = 1 - e^{-\lambda z}}$

It seems that I got it correct to this point (in line with the given answer).

for $\mathbf{z>1}\\ \mathbf{(f_X*f_Y)(z)= {\color{Red} \int_1^z f_Y(z-y) \ f_X(y) \ dy = \int_1^z \lambda e^{-\lambda y} \, dy = [-e^{-\lambda y}]_{1}^{z} = 1 - e^{-\lambda z}=e^{-\lambda}(1-e^{-z})}}$

The result I got for $z>1$ is incorrect. I guess it is because I apply incorrect bounds to the integral and that's because I don't quite get the whole concept. I am looking for help with this example and more for general hint how to construct those intervals.

The correct answer should be:

$\mathbf{f_{X+Y}(z)=\begin{cases} 1-e^{-\lambda z} & 0 \leq z \leq 1 & \\ e^{-\lambda z}(e^\lambda -1) & z \geq 1 & \\ 0 & \text{otherwise} & \end{cases}}$

Answer 1

If $z > 1$, we also require that $0 \le z-y \le 1$, or equivalently, $$z \ge y \ge z-1.$$ Thus your lower limit of integration is not correct: clearly, for a convolution integral of a uniform distribution with width $1$, your interval of integration must also have a width of $1$, and your integral limits do not satisfy this basic property.

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Note that you would not be led astray if you expressed the densities in terms of indicator functions: $$f_X(x) = \lambda e^{-\lambda x} \mathbb{1}(x \ge 0), \quad f_Y(y) = \mathbb{1}(0 \le y \le 1).$$ Then our convolution is $$\begin{align*} f_Z(z) &= \int_{x = -\infty}^\infty f_X(x) f_Y(z-x) \, dx \\ &= \int_{x=-\infty}^\infty \lambda e^{-\lambda x} \mathbb{1}(x \ge 0) \mathbb{1}(0 \le z-x \le 1) \, dx \\ &= \int_{x = 0}^\infty \lambda e^{-\lambda x} \mathbb{1}(0 \le z-x \le 1) \, dx \\ &= \int_{x=0}^\infty \lambda e^{-\lambda x} \mathbb{1}(z \ge x \ge z-1) \, dx \\ &= \mathbb{1}(0 \le z \le 1) \int_{x=0}^z \lambda e^{-\lambda x} \, dx + \mathbb{1}(z > 1) \int_{x=z-1}^z \lambda e^{-\lambda x} \, dx . \end{align*}$$

The key point here is that we have a density $f_Y(z-x)$ which is nonzero only when $z-x \in [0,1]$. This is equivalent to saying that $x \in [z-1, z]$. But $x$ must also be nonnegative, because otherwise $f_X(x)$ would be zero. So in order for both densities to be positive, we must require $x \in [0,z]$ if $z \le 1$, and $x \in [z-1, z]$ when $z > 1$. We have to take the lower endpoint to be whichever is the larger of $0$ and $z-1$. This is precisely what the last expression above does. Now it is straightforward to evaluate the integrals to obtain $$\begin{align*} f_Z(z) &= \mathbb{1}(0 \le z \le 1) (1 - e^{-\lambda z}) + \mathbb{1}(z > 1) (e^{-\lambda (z-1)} - e^{-\lambda z}) \\ &= \begin{cases} 0, & z < 0 \\ 1-e^{-\lambda z}, & 0 \le z \le 1 \\ e^{-\lambda z}(e^\lambda-1) & z > 1. \end{cases} \end{align*}$$

Answer 2

Comment: I believe the answer for the density function is correct. Following the brief simulation below with $\lambda = 1/2$, I plot this density function (green) against the histogram of simulated values. Also, you can use your density function to verify that $E(Z)$ and $V(Z)$ have the obvious values.

The empirical cumulative distribution function (ECDF) of the simulated sample is also shown along with the CDF (green) corresponding to this density function.

 m = 10^4;  lam = 1/2
 x = rexp(m, lam);  y = runif(m, 0, 1);  z = x + y
 mean(z);  sd(z)
 ## 2.505842  # E(Z) = 2 + 1/2 = 2.5
 ## 2.02484
 sqrt(4 + 1/12)
 ## 2.020726

Notes: (a) $F_Z(z) = 1 - e^{-\lambda z}(e^\lambda-1)/\lambda,$ for $z \ge 1.$ The CDF must be continuous at $z = 1.$ (b) The number $m = 10000$ of iterations is relatively small to keep the file size for the ECDF within reasonable bounds, so the simulated values of $E(Z)$ and $SD(Z)$ are barely accurate to two places.

Answer 3

$$ (f_X * f_Y)(x) = \int_{-\infty}^\infty f_X(y) f_Y(x-y)\,dy = \int_A f_X(y) f_Y(x-y)\,dy $$ where $A$ is the set of values of $y$ for which both $$y\ge0\tag1$$ and $$0\le x-y \le 1.\tag2$$ Rearranging $(2)$ we get $$ y\le x \le y+1 \tag 3 $$ and that is equivalent to $$ x-1\le y \le x \tag 4 $$ since the first inequality in $(4)$ is equivalent to the second inequality in $(3)$ and vice-versa.

But we also need $(1)$, so we have $$ \max\{0,x-1\}\le y \le x. $$

Therefore we have $$ (f_X * f_Y)(x) = \begin{cases} \displaystyle \int_{\max\{0,x-1\}}^x f_X(y) f_Y(x-y)\,dy & \text{if }x\ge0, \\[10pt] 0 & \text{otherwise}, \end{cases} $$ or, if you like $$ (f_X * f_Y)(x) = \begin{cases} \displaystyle \int_{x-1}^x f_X(y) f_Y(x-y)\,dy & \text{if }x\ge1, \\[10pt] \displaystyle\int_0^x f_X(y) f_Y(x-y)\,dy & \text{if }0 < x < 1, \\[10pt] 0 & \text{otherwise}, \end{cases} $$ Since $f_Y(x-y)=1$ whenever $y$ is within those bounds, these can be trivially simplified. And after that, the integrals can easily be computed in closed form.