Question 6.2: The time, in minutes, between two consecutive arrivals of a ......
The time, in minutes, between two consecutive arrivals of a train at a metro station is a continuous random variable X with distribution function
F(t)=c-\mathrm{e}^{-t/3}-\frac{t}{3}\cdot\mathrm{e}^{-t/3},\quad t\gt 0,for some constant c.
(i) Find the value of c and then obtain the density function f of X.
(ii) Suppose that Mary arrives at the platform of the metro station exactly at the time when a train departs. What is the probability that she has to wait until the arrival of the next train
(a) less than three minutes?
(b) more than one but less than four minutes?
(i) In order to find the value of c, we use that the distribution function F must satisfy
\operatorname*{lim}_{t\rightarrow\infty}F(t)=1.For the function F given in the statement, this gives
\operatorname*{lim}_{t\rightarrow\infty}\left(c-\mathrm{e}^{-t/3}-{\frac{t}{3}}\cdot\mathrm{e}^{-t/3}\right)=1.\qquad\qquad\qquad(6.4)It is clear that \textstyle{\operatorname*{lim}_{t\to\infty}\mathrm{e}^{-t/3}=0,} while for the last term inside the bracket we have
\operatorname*{lim}_{t\rightarrow\infty}{\frac{t}{3}}\cdot\mathrm{e}^{-t/3}=\operatorname*{lim}_{t\rightarrow\infty}{\frac{t}{3e^{t/3}}}=0,by applying l’Hôpital’s rule. Substituting these results into (6.4), we get immediately that c = 1. This means that the distribution function of X is
F(t)=1-\mathrm{e}^{-t/3}-{\frac{t}{3}}\cdot\mathrm{e}^{-t/3},\quad t\gt 0\qquad\qquad\qquad(6.5)and the density of X can now be obtained by differentiating this expression. This yields
f(x)=F^{\prime}(x)={\frac{1}{3}}\mathrm{e}^{-x/3}-\left({\frac{1}{3}}\mathrm{e}^{-x/3}-{\frac{x}{3}}\cdot{\frac{1}{3}}\mathrm{e}^{-x/3}\right)={\frac{x}{9}}\mathrm{e}^{-x/3}.Figure 6.3 shows the distribution function and the density function obtained above.
(ii) The probabilities we seek for this part are
(a) P(X < 3),
(b) P(1 < X < 4).
For the former, we use Proposition 6.1(b) which gives, since X is continuous, that P(X < 3) = F(3) and now putting t = 3 in (6.5) we obtain immediately
P(X\lt 3)=1-2\mathrm{e}^{-1}\cong0.264.Next, in order to find the probability P(1 < X < 4), we first note this is the same as P(1 ≤ X ≤ 4) and then use the formula in Part (d) of Proposition 6.1. In this way, we obtain by an appeal to (6.5) again,
P(1\lt X\lt 4)=F(4)-F(1)={\bigg(}1-{\mathrm{e}}^{-4/3}-{\frac{4}{3}}\cdot{\mathrm{e}}^{-4/3}{\bigg)}-{\bigg(}1-{\mathrm{e}}^{-1/3}-{\frac{1}{3}}\cdot{\mathrm{e}}^{-1/3}{\bigg)}=\frac{4}{3}\mathrm{e}^{-1/3}-\frac{7}{3}\mathrm{e}^{-4/3}\cong0.34.
