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Problem:

Let $X$ and $Y$ be identically distributed independent random variable with the density function $$ f(y) = \begin{cases} e^{-t} & t \geq 0 \\ 0 & \text{otherwise} \end{cases} $$

Answer:

Let $Z = X + Y$. Now we want to find a density function for $Z$ by using the convoluation. Let $g(u)$ be the density function for $Z$. \begin{align*} g(u) &= \int_0^{ \infty }f(v)f(u-v) \,\, dv \\ g(u) &= \int_0^{ \infty } e^{-v} e^{-(u-v)} \,\, dv \\ g(u) &= \int_0^{ \infty } e^{-v} e^{-u + v} \,\, dv \\ g(u) &= \int_0^{ \infty } e^{-u} \,\, dv \\ g(u) &= \int_0^{ \infty } e^{-u}v \bigg|_{v=0}^{v=\infty} \\ g(u) &= 0 - e^{-u} \\ g(u) &= - e^{-u} \end{align*} However, the book gets: $$ g(u) = \begin{cases} ue^{-u} & u \geq 0 \\ 0 & \text{otherwise} \end{cases} $$ Where did I go wrong?

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    $\begingroup$ Hint: What is the value of $f(u-v)$ when $u-v<0$? It is not $e^{-(u-v)}$. $\endgroup$ Commented Oct 21, 2023 at 3:56

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If you had retained the restriction on the support of $X$ and $Y$, you would not have been led astray. That is, $$f(t) = e^{-t}$$ only when $t \ge 0$. Otherwise, it is zero. So when you write

$$g(u) = \int_{v=0}^\infty f(v)f(u-v) \, dv,$$

whenever $v > u$, then $f(u-v) = 0$. So the upper limit of integration should be

$$\begin{align} g(u) &= \int_{v=0}^{\color{red}{u}} f(v)f(u-v) \, dv \\ &= \int_{v=0}^u e^{-v} e^{-(u-v)} \, dv \\ &= \int_{v=0}^u e^{-u} \, dv \\ &= ue^{-u}. \end{align}$$ And the restriction on $u$ is that $u \ge 0$, since if $u < 0$, then the integral is taken over an interval that is negative.

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