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.

  • $\begingroup$ 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. $\endgroup$ – 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}

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  • $\begingroup$ Thanks very much. Just having some difficulty applying this concept to problems..... $\endgroup$ – Denson Feb 15 '18 at 15:54
  • $\begingroup$ A follow-up question: if the question above was the time between the second and third (instead of 1st and 2nd) and to find P(W<0.1) we would still get 0.55. $\endgroup$ – Denson Feb 15 '18 at 16:09
  • $\begingroup$ Or to take another example, suppose Y denotes the time between the start of the shift and the arrival of the third call. Find P(Y<0.5). Based on the answer above, I would integrate the density function between 0 and 0.5. But if the question had stated Y denotes the time between the start of the shift and the fourth call- I would be still integrate over the same limits (0, 0.5) - this would give the same answer. This does not seem correct- but I am not sure what is the flaw in my understanding. . $\endgroup$ – Denson Feb 15 '18 at 16:10
  • $\begingroup$ @Denson Yes to your first question. As for your second, no, you do not integrate the distribution of your original, exponential Y. Y’s distribution is for a single wait time. If you want to evaluate probabilities involving a sum of wait times, you need to integrate the distribution of the sum of exponential random variables. Do you know this distribution or how to find it? $\endgroup$ – bames Feb 15 '18 at 19:19
  • $\begingroup$ Hi Bames I am not very clear how to obtain the sum of wait times and the sum of exponential distributions.. If say the question was to find the time between the start of the shift and the arrival of the 3rd call, how could we do this. $\endgroup$ – Denson Feb 22 '18 at 0:36

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