# time between poisson events

The number of calls during a one hour shift period has a Poisson distribution with a mean of 8. This question is: let Y be the time in hours between the arrival of the first and second call. Find P(Y< 0.1).

The density function of the time between two consecutive events in a Poisson process is exponentially distributed, given by λexp(-λt). But when I substitute 8 for λ and 0.1 for t, I get a value > 1, so this can't be correct.

Not sure what I am doing wrong, could anyone pl. help.

• If the mean of the Poisson distribution is $8$, then the distribution of $Y$ is $$\frac18e^{-t/8}$$ on $t>0$, not the PDF $8e^{-8t}$ used in the (accepted) answer below. – Did Jul 25 '18 at 10:13

If you want the probability that $Y$ is less than $0.1$, then you should be integrating the distribution over possible values of $Y$ less than $0.1$: $$\Bbb P(Y<0.1)=\int_0^{0.1}\frac{1}{8}e^{-t/8}\;dt=1-e^{-0.0125}\approx 0.0124$$
EDIT (in response to a comment): If you want the distribution of a sum of i.i.d. exponential random variables, you can do a convolution. For example, if $X_1, X_2$ are i.i.d. exponentially distributed and you want to find $P(X_1+X_2 = t)$, this is the probability that $X_1 = \tau$ and $X_2 = t-\tau$, for $0\leq\tau\leq t$: $$P(X_1+X_2 = t) = \int_0^t\lambda ^2 e^{-\lambda\tau}e^{-\lambda(t-\tau)}\;d\tau=\lambda^2te^{-\lambda t}$$ Similarly, for the sum of three i.i.d. exponentials, we have: \begin{align} P(X_1+X_2+X_3 = t) &= \int_0^t(\lambda^2\tau e^{-\lambda \tau})(\lambda e^{-\lambda (t-\tau)})\;d\tau\\ &=\int_0^t \lambda^3e^{-\lambda t}\tau \;d\tau\\ &=\frac{\lambda^3}{2}t^2e^{-\lambda t} \end{align} So, for example, if you want to find the probability that the time until the third call is $\leq 0.5$, you would do: \begin{align} P(X_1+X_2+X_3 \leq 0.5) &= \frac{\lambda^3}{2}\int_0^{0.5}t^2e^{-\lambda t}\;dt\\ &=1 - e^{-0.5\lambda}-\frac{\lambda}{2}e^{-0.5\lambda}-\frac{\lambda^2}{8}e^{-0.5\lambda} \end{align}