# Thread: Convolution of Exponential Random Variables

1. ## Convolution of Exponential Random Variables

Let $\displaystyle T_1, T_2$ be independent exponential variables with parameters $\displaystyle \lambda_1, \lambda_2$ respectively. Find the density function of $\displaystyle R = T_1 - T_2$.

I get a piecewise density function (for $\displaystyle r > 0, r \leq 0$):

$\displaystyle f_R (r) = \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2} e^{{-\lambda_2}r}, r>0$

$\displaystyle f_R (r) = \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2} e^{{\lambda_1}r}, r\leq 0$

The answer given however is simply that the density function is exponential without further information. I was wondering if my answer is correct?

P.S. I can't seem to do piecewise using normal LaTeX code; is it different in these forums?

2. The first one should have $\displaystyle e^{-\lambda_1 r}$ I think but other than that it seems fine.

If you have problems with latex, post it in the latex subforum, 'tis working fine for me.

3. Originally Posted by h2osprey
Let $\displaystyle T_1, T_2$ be independent exponential variables with parameters $\displaystyle \lambda_1, \lambda_2$ respectively. Find the density function of $\displaystyle R = T_1 - T_2$.

I get a piecewise density function (for $\displaystyle r > 0, r \leq 0$):

$\displaystyle f_R (r) = \left\{\begin{array}{c}\frac{\lambda_1\lambda_2}{\ lambda_1+\lambda_2} e^{{-\lambda_2}r}, r>0\\ \frac{\lambda_1\lambda_2}{\lambda_1+\lambda_2} e^{{\lambda_1}r}, r\leq 0\end{array}\right.$

The answer given however is simply that the density function is exponential without further information. I was wondering if my answer is correct?

P.S. I can't seem to do piecewise using normal LaTeX code; is it different in these forums?
There is a mistake in the exponents: it is $\displaystyle e^{-\lambda_1 r}$ when $\displaystyle r>0$ and $\displaystyle e^{-\lambda_2 r}$ else.

How to find the result with little computation (but some nontrivial knowledge): first, the probability that $\displaystyle T_1>T_2$ is $\displaystyle \frac{\lambda_2}{\lambda_1+\lambda_2}$ (classic useful computation); then, by the memoryless property of exponential distribution, conditionally to $\displaystyle T_2$ and to the event $\displaystyle \{T_1>T_2\}$, the distribution of $\displaystyle T_1-T_2$ is exponential with parameter $\displaystyle \lambda_1$ (note that since this distribution does not depend on $\displaystyle T_2$, it was unnecessary to condition by $\displaystyle T_2$). This gives the "$\displaystyle r>0$" part: $\displaystyle \frac{\lambda_2}{\lambda_1+\lambda_2}\lambda_1 e^{-\lambda_1 r}$, and symmetry gives the other part.

4. Originally Posted by Laurent
There is a mistake in the exponents: it is $\displaystyle e^{-\lambda_1 r}$ when $\displaystyle r>0$ and $\displaystyle e^{-\lambda_2 r}$ else.

How to find the result with little computation (but some nontrivial knowledge): first, the probability that $\displaystyle T_1>T_2$ is $\displaystyle \frac{\lambda_2}{\lambda_1+\lambda_2}$ (classic useful computation); then, by the memoryless property of exponential distribution, conditionally to $\displaystyle T_2$ and to the event $\displaystyle \{T_1>T_2\}$, the distribution of $\displaystyle T_1-T_2$ is exponential with parameter $\displaystyle \lambda_1$ (note that since this distribution does not depend on $\displaystyle T_2$, it was unnecessary to condition by $\displaystyle T_2$). This gives the "$\displaystyle r>0$" part: $\displaystyle \frac{\lambda_2}{\lambda_1+\lambda_2}\lambda_1 e^{-\lambda_1 r}$, and symmetry gives the other part.
Yes, turns out I switched the numbers while typing the problem: the original question was $\displaystyle R = T_2 - T_1$. However, regarding the pdf when $\displaystyle r \leq 0$ shouldn't there be a negative sign? If not it gets very, very large indeed..

and thanks for the LaTeX code!

5. Originally Posted by h2osprey
However, regarding the pdf when $\displaystyle r \leq 0$ shouldn't there be a negative sign? If not it gets very, very large indeed..
Yes, sure, it's $\displaystyle e^{\lambda_2 r}$ for $\displaystyle r<0$; the only mistake was with indices 1 and 2, due to switch them in the definition.