Paul and n other runners compete in a marathon. Their times are independent
continuous r.v.s with CDF F: Show all your supporting work.
(a) For j = 1; 2; : : : ; n, let Aj be the event that anonymous runner j completes the
race faster than Paul. Explain whether the events Aj are independent, and whether they
are conditionally independent given Paul's time to nish the race.
(b) For the rest of this problem, let N be the number of runners who nish faster
than Paul. Find E(N):
(c) Find the conditional distribution of N; given that Pauls time to nish the
marathon is t:
(d) Find V ar(N):

Any two indices will do i!=j Paul's time and the time of the ith runner are equally distributed and independent, hence according to the Symmetry, p (Ai) = p(Aj)= 1/2. Consider the intersection of Ai and Aj. Thus, Paul is outpaced by both runners I and j. There are 6 possible permutations of their times because they are i.id, with Paul being the slowest. P(Ai intersection Aj) hence equals 1/3.

But now wе have that P(Ai intersection Aj)= 1/3 != 1/4 = P(Ai)P(Aj)

thus, events They are not autonomous, Ai and Aj. However, they are only somewhat independent given Paul's timing. Paul can be removed from the narrative if such is the case. Then, only these two runners are left. Events As a result of their independent and equal distribution of times, Aj and AI are independent.

(B)

Define Ij as the random variable indication that indicates whether or not the jth runner is superior to Paul. via means of symmetry. We know p(Ij = 1)=1/2.

We determine that

E(N) = E(I1+I2 +.......In)

= nE(i1)

=nP(I1-1)

=n/2

using the linearity of expectation.

(C)

Paul's estimated time of completion is t The probability that each runner will finish ahead of Paul is P (time of rumer I t) = F. (t)

By using part (a), we can determine that other runners' times are independent of Paull's time, and that the number of runners who finish faster than Paul has a binomial distribution with the parameters n and F. To come to the conclusion that N/T= t Binom (n, F(t)) and if we choose T at random, N/T Binom (n, F(T)),

(D)

Let's figure out Var (N)

Var (N)=F(Var (N/T)) + Var(E(N/T))

= E(n F(T) (1- F(T)) + Var(on F(T))

according to Eve's Law.

Here, we can use the uniform distribution's universality to produce

F(T) ~Unif (0, 1) = U

Hence symmetry, which states that U I-U, leads to

Var(N) = nE(U2) + n2 var (U)

=n/3 + n2/12,

where E(U2)=1/3

Var(U)=1/12.

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