People arrive at random times and independently at a bus stop and wait for the bus to arrive. The bus arrives at this stop once every hour. Thus, the waiting times of the people follow a uniform distribution over the unit interval, i.e., $T_1, T_2,\ldots\sim U(0,1)$. We are interested in the minimum number of people that will have a combined waiting time that exceeds $45$ minutes i.e.,
$$N=\min_{k\geq 1}\left(\{X_1+X_2+\dots+X_k > 3/4\} \cap \{X_1+X_2+\dots+X_{k-1} \leq 3/4\}\right).
$$
a) Show by induction that $$
P(N=k) = \frac{(3/4)^{k-1}}{(k-1)!}-\frac{(3/4)^k}{k!},\quad k = 1,2,\dots
$$
b) Find the mean and variance of $N$.
I got stuck on the induction part of the problem and I got the a different answer to the mean and variance of $N$ than the book's. Also I apologize for the format of the problem. I am new to this website and do not know how to use latex very well. Can someone please help? Thank you for your time and consideration!
Best Answer
Denote $X_1+X_2+..+X_k = S_k$. Next, notice that event $A_k=${$(S_{k-1}<3/4)\cap(S_{k}\geq3/4)$} is opposite to event $B=${$(S_{k-1}\geq3/4)\cup(S_k<3/4)$}. Let's find the probability of $B$: $P(B)=P({S_{k-1}\geq3/4})+P(S_k<3/4).$ The first summand we get from induction assumption because event ${S_{k-1}\geq 3/4}=A_{k-1}+A_{k-2}+...A_1$. The second summand equals to volume of pyramid in $R^k$ restricted by hyperplanes $x_1+x_2+..+x_k=3/4$; $x_i=0\ (i=1,2,..,k)$. As a result, we have: $P(B)=\sum_{i=1}^{k-1} A_i + \frac {(3/4)^k} {k!}$. And in the end we can get: $P(A)=1-P(B)$.
$Mean = \sum_{i=1}^{\infty} {kP(N=k)}$ and $Variance = \sum_{i=1}^{\infty} {(k-Mean)^2}P(N=k)$