The text I'm using on questions like these does not provide step by step instructions on how to solve these, it skipped many steps in the examples and due to such, I am rather confused as to what I'm doing.

Here is the question: Let $X$ be an exponential random variable with parameter $λ$ and $Y$ be an exponential random variable with parameter $2λ$ independent of $X$. Find the probability density function of $X + Y$.

Now, I know this goes into this equation: $\int_{-\infty}^{\infty}f_x(a-y)f_y(y)dy$
What I tried to do is $=\int_{-\infty}^{\infty}\lambda e^{-\lambda (a-y)}2\lambda e^{-\lambda y}dy$ but I quite honestly don't think this is the way to go. Can anyone give me a little insight as to how to actually compute $f_x(a-y)$ in particular?

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    $\begingroup$ You are proceeding correctly, but note the exponential distribution is only non-zero for positive arguments so the limits of integration will be from $0$ to $a$. Also, the second factor is missing a 2 in the exponent $2 \lambda e^{-2\lambda y}$. You should end up with a linear combination of the original exponentials. $\endgroup$ – A. Webb Mar 6 '17 at 21:37
  • $\begingroup$ @A.Webb Thank you! By doing this and then taking the derivative with respect to a I was able to get the right answer. I didn't think I was doing it right, but apparently the integral really does suck that much. $\endgroup$ – Heavenly96 Mar 7 '17 at 0:33
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    $\begingroup$ @A.Webb why the limit of the integration will be from 0 to $a$ ? And not from 0 to infinite? $\endgroup$ – Laura Jan 27 at 19:01
  • $\begingroup$ @Laura, the value $t - x$ of the exponential r.v. is only nonnegative in the range $0 \leq x \leq t$. $\endgroup$ – gwg May 16 at 19:42

$$\begin{align*} \Pr[X + Y \le t] &= \int_{x=0}^\infty \Pr[Y \le t - x \mid X = x] f_X(x) \, dx \\ &= \int_{x=0}^t (1 - e^{-2\lambda(t-x)}) \lambda e^{-\lambda x} \, dx \\ &= \lambda \int_{x=0}^t e^{-\lambda x} - e^{-2\lambda t} e^{\lambda x} \, dx \\ &= \left[ -e^{-\lambda x} - e^{-2\lambda t} e^{\lambda x} \right]_{x=0}^t \\ &= 1 + e^{-2\lambda t} - 2e^{-\lambda t}. \end{align*}$$

The probability density is then found by differentiation with respect to $t$.

  • $\begingroup$ when I differentiate that I end up with $2\lambda e^{-2\lambda t}(e^{\lambda t} -1)$ which is not the answer. However it is very close, the answer is: $2\lambda e^{-\lambda t}(1-e^{-\lambda t})$ so maybe I differentiated wrong? $\endgroup$ – Heavenly96 Mar 7 '17 at 0:28
  • $\begingroup$ @Heavenly96 $$f_{X+Y}(t) = \frac{d}{dt}\left[1 + e^{-2\lambda t} - 2e^{-\lambda t}\right] = -2\lambda e^{-2\lambda t} - 2(-\lambda) e^{-\lambda t} = 2\lambda ( e^{-\lambda t} - e^{-2\lambda t} ) = 2\lambda e^{-\lambda t} (1 - e^{-\lambda t}),$$ as claimed. Your answer is actually equivalent. You merely pulled out a factor of $e^{-2\lambda t}$ instead of $e^{-\lambda t}$. If you distribute your answer and the answer you were given, you will find they are identical. $\endgroup$ – heropup Mar 7 '17 at 0:40

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