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Phone calls arrive at a telephone exchange at an average rate of $1$ hit per minute ($\lambda=1$). Let $T_4$ be the time in minutes when the fourth hit occurs after the counter is switched on.

Find $\mathbb{P}(3\le T_4\le 5)$.

So to do this we calculate $\mathbb{P}(T_4\le 5)-\mathbb{P}(T_4<3)$. I calculate $\mathbb{P}(T_4\le 5)=1-\mathrm{e}^{-5(118/3)}$, using the Poisson process $1-\mathbb{P}(\mathrm{Pois}(5)\le 3)$. But the problem is I'm not sure how to compute $\mathbb{P}(T_4<3)$, namely that the fourth arrives in less than 3 minutes. I know that it will be another Poisson with mean $3$, but I'm confused on the sign of the inequality and number of terms.

Thanks!!

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Use $\backslash$le instead of <= and Poisson instead of poison. –  Did Oct 28 '11 at 8:29

2 Answers 2

The answer given by @Sasha is probably the fastest way to obtain the result, but your approach is fine as well (and the only possible one in case you don't know yet that inter-arrival times are exponentially distributed).

As you have said, $\mathbb{P}(3\leq T_4\leq5) = \mathbb P(T_4\leq5)-\mathbb P(T_4<3)$. From there, we have (with $N(t)$ the number of Poisson events that have happened before $t$): \begin{align*} \mathbb P(T_4\leq 5) &= \mathbb P(N(5)\geq 4)\\ &= 1 - \mathbb P(N(5)\leq 3)\\ &= 1 - \left(1 + \frac{5}{1} + \frac{5^2}{2} + \frac{5^3}{6}\right)e^{-5}\\ &= 1 - \frac{118}{3}e^{-5}, \end{align*} and similarly \begin{align*} \mathbb P(T_4<3) &= \mathbb P(N(3)\geq 4)\\ &= 1 - \mathbb P(N(3)\leq 3)\\ &= 1 - \left(1 + \frac{3}{1} + \frac{3^2}{2} + \frac{3^3}{6}\right)e^{-3}\\ &= 1 - 13e^{-3}. \end{align*} So this approach does yield the same result $$\mathbb{P}(3\leq T_4\leq5) = 13e^{-3} - \frac{118}{3}e^{-5}.$$

I now realize that what might be troubling you is that in one case the inequality is strict, and in the other it isn't, yet the formula for both cases is the same. This is essentially because the events have probability zero of happening at a given time. This is why the strict probability and the other are equal: \begin{align*} \mathbb P(T_4\leq3) &= \mathbb P(T_4<3) + \underbrace{\mathbb P(T_4=3)}_{=0} \end{align*}

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Since inter-arrival times of the Poisson process with rate $\lambda$ are exponentially distributed with parameter $\lambda$, we have $T_4 = \tau_1 + \tau_2 + \tau_3 + \tau_4$, with $\tau_i$ independent identically distributed following $\mathcal{E}(\lambda)$.

The sum of 4 exponentials is equal in distribution to $\Gamma(4, \frac{1}{\lambda})$, whence $$ \mathbb{P}( 3 \le T_4 \le 5) = \int_3^5 \frac{1}{6} \mathrm{e}^{- \lambda x} x^3 \lambda^4 \mathrm{d} x = 13 \mathrm{e}^{-3} - \frac{118}{3} \mathrm{e}^{-5} $$

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