# Lambda and mean of sum of 2 independent exponential random variable

I have a restaurant with service time being exponentially distributed. Let's say the food-serving time has a mean of 30 minutes, and the checkout time has a mean of 10 minutes.

Would $$\lambda$$ in this problem $$=\frac{1}{30+10}$$ or $$=\frac{1}{30} + \frac{1}{10}$$?

• @orange right so I was looking into that as well but Gamma has too many parameters that it confuses me – PTN Mar 28 at 7:36
• math.stackexchange.com/questions/635443/… – badatmath Mar 28 at 7:48
• If you want the mean, you can get it easily: the mean of a sum is the sum of the means, so the mean here is $30+10=40$ minutes. – Minus One-Twelfth Mar 28 at 8:05

Given two independent exponential random variables $$X$$ and $$Y$$ with distributions $$f_X(t) = \mu e^{-\mu t}$$ and $$f_Y(t) = \lambda e^{-\lambda t}$$, the distribution of the sum $$X+Y$$ is given by the convolution: \begin{align} f_{X+Y}(t) &= \int_0^t \mu e^{-\mu \tau}\lambda e^{-\lambda (t-\tau)}\;d\tau\\ &=\lambda\mu e^{-\lambda t}\int_0^t e^{-(\mu-\lambda)\tau}\;d\tau\\ &=\frac{\lambda\mu}{\lambda-\mu}\left[e^{-\mu t} - e^{-\lambda t}\right] \end{align}
The mean is given by \begin{align} \frac{\lambda\mu}{\lambda-\mu}\int_0^\infty t\left[e^{-\mu t} - e^{-\lambda t}\right]\;dt&= \frac{\lambda\mu}{\lambda - \mu}\left[\frac{1}{\lambda^2} - \frac{1}{\mu^2}\right]\\ &= \frac{\lambda+\mu}{\lambda\mu} \end{align} which in your case is $$40$$ minutes.