Question 7.12: The time between two successive emails arriving at Julia’s e......
The time between two successive emails arriving at Julia’s email account follows the exponential distribution with a mean value of 90 minutes.
(i) What is the probability that the time between two successive emails exceeds one hour?
(ii) Given that 90 minutes have passed since Julia received her last email, what is the probability that she has to wait for at least 30 minutes until the next one?
Let us choose one hour as the time unit.³ Then, if X denotes the time between successive emails arriving, we are given that E(X) = 3/2. Since from Proposition 7.8 the mean value of an exponential random variable is the reciprocal of the parameter of its distribution, we see that X ∼ ε(𝜆) with 𝜆 = 2/3.
(i) We need the probability P(X > 1), which is given by
P(X\gt 1)=1-F(1)=\mathrm{e}^{-2/3}.(ii) If no email arrives within the next 30 minutes, this means that the time between the last email and the next one that arrives will be at least 90 + 30 = 120 minutes, i.e. two hours. Moreover, due to the lack of memory property, we have
P\left(X\gt 2|X\gt {\frac{3}{2}}\right)=P\left(X\gt {\frac{3}{2}}+{\frac{1}{2}}|X\gt {\frac{3}{2}}\right)=P\left(X\gt {\frac{1}{2}}\right).Hence,
P(X\gt 2|X\gt 3/2)=1-F(1/2)=\mathrm{e}^{-1/3},which is the desired probability.
In Example 6.3 of the previous chapter, we have obtained the distribution for a linear transformation of an exponential random variable. In the following example, we use the result of Proposition 6.2 to obtain the distribution of a nonlinear transformation for such a variable.
3A different choice would obviously have no effect on the result. Bear in mind, however, that we have to be consistent while dealing with a particular problem, in that we should use the same unit throughout