# Continuous Convolution – A Guided Example

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Thia post is a guided example of a practice problem (Problem 15A found in Exam P Practice Problem 15 – Still More Convolution Practice). The exposition is to make clear the thought process on how to set up and evaluate the integral to find the pdf of an independent sum. The following is the statement of the problem:

Problem 15A
Find the pdf of $Z=X+Y$ where $X$ and $Y$ are independent random variables such that $X$ is uniformly distributed on the interval $0 and $Y$ is exponentially distributed with mean 10.

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Setting up the Scene

The pdfs of $X$ and $Y$ are:

$\displaystyle f(x)=\frac{1}{10} \ \ \ \ 0

$\displaystyle g(y)=0.1 \ e^{-0.1 \ y} \ \ \ \ y>0$

Since $X$ and $Y$ are independent, the joint pdf of $X$ and $Y$ is $f_{X,Y}(x,y)=f(x) \ g(y)$. The following figure shows the support of this joint distribution.

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Figure 1 – Support of the Joint Distribution

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Let $h(z)$ denote the pdf of the independent sum $Z=X+Y$. To calculate $h(z)$, we need to sum the joint density $f_{X,Y}(x,y)$ over the entire green region in Figure 1. Each value of $h(z)$ is the sum of the joint density $f_{X,Y}(x,y)$ on a particular line $x+y=z$ (see Figure 2 below).

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Figure 2

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For each point of the line $x+y=z$, the joint density $f_{X,Y}(x,y)$ at that point is $f(x) \ g(z-x)$ or $f(z-y) \ g(y)$. Thus we have:

$\displaystyle h(z)=\int_{-\infty}^{+\infty} f(x) \ g(z-x) \ dx \ \ \ \ \ \ \ \ (1)$

$\displaystyle h(z)=\int_{-\infty}^{+\infty} f(z-y) \ g(y) \ dy \ \ \ \ \ \ \ \ \ (2)$

Let’s focus on integral $(1)$. Since the support of the $X$ variable is a bounded interval, there are two cases to consider: $0 and $10. So we calculate $h(z)$ separately in each case. Let's look at the following figures.

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Looking at Two Cases

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Figure 3 – Case 1

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Figure 4 – Case 2

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Case 1 covers all the lines $x+y=z$ from the red line to the blue line in Figure 3. Case 2 covers all lines $x+y=z$ above the blue line in Figure 4. These considerations will dictate the limits for the integral in $(1)$.

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Setting Up Integrals

Here’s the integrals for the two cases.
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Case 1

$0
$\displaystyle h(z)=\int_{0}^{z} f(x) \ g(z-x) \ dx \ \ \ \ \ \ \ \ \ \ \ (3)$

Case 2

$10
$\displaystyle h(z)=\int_{0}^{10} f(x) \ g(z-x) \ dx \ \ \ \ \ \ \ \ \ \ (4)$

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The following two figures show how integrals $(3)$ and $(4)$ are set up. For a typical line $x+y=z$ in Case 1 (Figure 5), the range for $x$ is $0. For a typical line $x+y=z$ in Case 2 (Figure 6), the range for $x$ is $0.

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Figure 5 – Typical Line in Case 1

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Figure 6 – Typical Line in Case 2

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Calculation

Case 1

\displaystyle \begin{aligned}h(z)&=\int_{0}^{z} f(x) \ g(z-x) \ dx \\&=\int_0^z \frac{1}{10} \ 0.1 \ e^{-0.1 \ (z-x)} \ dx \\&=\int_0^z \frac{1}{10} \ e^{-0.1 \ z} \ 0.1 \ e^{0.1 \ x} \ dx \\&=\frac{1}{10} \ e^{-0.1 \ z} \ (e^{0.1 \ z}-1) \\&= \frac{1}{10}\ (1-e^{-0.1 \ z})\\&=0.1-0.1e^{-0.1 \ z} \end{aligned}

Case 2

\displaystyle \begin{aligned}h(z)&=\int_{0}^{10} f(x) \ g(z-x) \ dx \\&=\int_0^{10} \frac{1}{10} \ 0.1 \ e^{-0.1 \ (z-x)} \ dx \\&=\int_0^{10} \frac{1}{10} \ e^{-0.1 \ z} \ 0.1 \ e^{0.1 \ x} \ dx \\&=\frac{1}{10} \ e^{-0.1 \ z} \ (e-1) \\&=\frac{e-1}{10} \ e^{-0.1 \ z} \end{aligned}

Thus we have the following pdf:

$\displaystyle h(z)=\left\{\begin{matrix}0.1-0.1 \ e^{-0.1 \ z}& \ \ \ \ \ \ 0

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