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 $=\frac{1}{30} + \frac{1}{10}$?

  • $\begingroup$ @orange right so I was looking into that as well but Gamma has too many parameters that it confuses me $\endgroup$ – PTN Mar 28 at 7:36
  • $\begingroup$ math.stackexchange.com/questions/635443/… $\endgroup$ – badatmath Mar 28 at 7:48
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    $\begingroup$ 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. $\endgroup$ – 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.


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