Solutions to Homework Set No. 4 – Probability Theory (235A), Fall ...

Now, the probability that Xi ≤ Xn for all 1 ≤ i ≤ n is
by fixing the largest, Xn, and permuting the rest.
P(An) = P({πn(n) = n}) =
(n − 1)!
, where (n − 1)! is obtained
n!
(n − 1)!
n!
For any σ ∈ Sn, P(πn = σ) = 1
n! and for any σn−1 ∈ Sn−1, P(πn−1 = σn−1) = 1 . Hence
(n−1)!
= 1
n
P(An, πn−1 = σn−1) = 1
n! = P(An)P(πn−1 = σn−1),
which shows the independence. Thus, Ai’s are independent since πn−1 are determined by
A1, · · · , An−1.
(b) Define
Rn =
n
k=1
1Ak
= #{1 ≤ k ≤ n : k is a record time}, (n = 1, 2, . . .).
Compute E(Rn) and V(Rn). Deduce that if (mn) ∞ n=1 is a sequence of positive numbers such
that mn ↑ ∞, however slowly, then the number Rn of record times up to time n satisfies
P |Rn − log n| > mn log n −−−→
n→∞ 0.
Solution.
E(Rn) =
n
k=1
V(Rn) = V n
k=1
E(1Ak ) =
1Ak
=
n
k=1
n
P(Ak) =
k=1
V(1Ak ) =
n
k=1
n
k=1
1
k
1
k
= log n + O(1)
1
− = log n + O(1)
k2 √
The second equality follows from independence. The claim about P(|Rn−log n| > mn log n)
follows directly from Chebyshev’s inequality.
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