Convolution of Exponential Random Variables

**h2osprey** · February 10th 2010, 12:39 PM

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

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

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

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

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?

**Focus** · February 10th 2010, 01:03 PM

The first one should have $e^{-\lambda_1 r}$ 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.

**Laurent** · February 10th 2010, 01:11 PM

Originally Posted by h2osprey

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

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

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

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 $e^{-\lambda_1 r}$ when $r>0$ and $e^{-\lambda_2 r}$ else.

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

**h2osprey** · February 10th 2010, 04:01 PM

Originally Posted by Laurent

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

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

Yes, turns out I switched the numbers while typing the problem: the original question was $R = T_2 - T_1$ . However, regarding the pdf when $r \leq 0$ shouldn't there be a negative sign? If not it gets very, very large indeed..

and thanks for the LaTeX code!

**Laurent** · February 11th 2010, 12:34 AM

Originally Posted by h2osprey

However, regarding the pdf when $r \leq 0$ shouldn't there be a negative sign? If not it gets very, very large indeed..

Yes, sure, it's $e^{\lambda_2 r}$ for $r<0$ ; the only mistake was with indices 1 and 2, due to switch them in the definition.

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