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Let $X$ and $Y$ be independent random variables with densities $$f_{X}(x)= \begin{cases} \gamma e^{-\gamma x} & \text{, } x\geq0 \\ 0& \text{, } x<0 \end{cases}$$

$$f_Y(y)= \begin{cases} \mu e^{-\mu x} & \text{, } y\geq0 \\ 0& \text{, } y<0 \end{cases}$$ where $\gamma$ and $\mu$ are positive constants.

a) Evaluate the density of the sum $Z=X+Y$.
b) Repeat in the case where $\mu = \gamma$.

My solution so far:
a)

$$f_Z(z) = \int_0^z f_X(x)f_Y{(z-x)} \, dx=\int_0^z \gamma e^{-\gamma x}\mu e^{-\mu (z-x)}\,dx$$

$$= \gamma\mu \int_{0}^{z}e^{-\gamma x-\mu z+\mu x}\,dx$$ $$= \gamma\mu e^{-\mu z}\int_{0}^{z}e^{x(\mu-\gamma)}\,dx$$ $$= \gamma\mu e^{-\mu z} \left. \left [ \frac{e^{(\mu-\gamma)x}}{\mu-\gamma} \right ]\right|_0^z$$ $$=\frac{\gamma\mu e^{-\mu z}}{\mu-\gamma}\left [ e^{(\mu-\gamma)z}-1 \right ].$$

So, $$f_Z(z)= \begin{cases} \frac{\gamma\mu e^{-\mu z}}{\mu-\gamma}\left [ e^{(\mu-\gamma)z}-1 \right ] & \text{, } z\geq0 \\ 0 & \text{, } z<0 \end{cases} $$

b)

For the case where $\mu = \gamma$, $f_{Y}(y)=\begin{cases} \gamma e^{-\gamma y} & \text{, } y\geq 0 \\ 0& \text{, } y<0 \end{cases}.$

\begin{align} f_Z(z) & = \int_0^z f_X(x)f_Y{(z-x)} \, dx=\int_0^z \gamma e^{-\gamma x} \gamma e^{-\mu (z-x)} \, dx \\[10pt] & = \gamma^2 \int_0^z e^{(-\gamma x+\gamma x-\gamma z)} \, dx \\[10pt] & = \gamma^2 \int_0^z e^{-\gamma z} \, dx \\[10pt] & = \gamma^2 e^{-\gamma z}\int_0^z 1\,dx \\[10pt] & =\gamma^2 e^{-\gamma z}[x]\Big|_0^z \\[10pt] & =\gamma^2 e^{-\gamma z}z. \end{align}

So, $$f_Z(z)= \begin{cases} \gamma^{2}e^{-\gamma z}z & \text{, } z\geq0, \\ 0 & \text{, } z<0. \end{cases}$$

I just wanted to make sure that what I have is correct. Thanks for answering!!

beepboopbeepboop
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1 Answers1

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My answers agree with yours (except for the typo where you have $\mu-z$ where you need $\mu-\gamma$.)

Note that both of these densities start with value $0$ at $z=0,$ unlike the exponential density, and that makes sense when you consider that in a tiny interval $[0,\varepsilon],$ the second arrival is not likely to be seen because the first arrival hasn't happened yet.

Michael Hardy
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  • Thanks for going through the problem, @Michael Hardy! I was afraid people would get scared off because it looks like intimidating at first glance. Very good eye with the $z$. – beepboopbeepboop May 28 '18 at 00:02
  • In your answer above, what are you meaning to say exactly with the exponential density? I'm a bit confused by your note there. Also, does this affect the answer in any way. Thanks for answering. – beepboopbeepboop May 28 '18 at 00:03
  • @beepboopbeepboop : By the exponential density I mean either of the two densities you start with. The values of those densities at $0$ are positive. Suppose $e^{-\mu x} , (\mu , dx),$ for $x\ge0,$ is the distribution of the time until the first arrival, so the average arrival rate is $\mu$ per unit of time. Suppose, for example, that $\mu$ is $1$ per hour. Then the probability that the first arrival happens within the first two minutes is about twice as big as the probability that it's within the first minute. However, the second arrival$,\ldots \qquad$ – Michael Hardy May 28 '18 at 17:28
  • $\ldots,$ is distinctly more than twice as likely to happen in the first two minutes as in the first minute, because during the second minute there is a higher probability that the first arrival has already happened, than there is in the first minute. The density of the waiting time until the second arrival starts at $0$ and grows. – Michael Hardy May 28 '18 at 17:29