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

Solutions to Homework Set No. 4 – Probability Theory (235A), Fall 2011
1. A function ϕ : (a, b) → R is called convex if for any x, y ∈ (a, b) and α ∈ [0, 1] we
have
ϕ(αx + (1 − α)y) ≤ αϕ(x) + (1 − α)ϕ(y).
(a) Prove that an equivalent condition for ϕ to be convex is that for any x < z < y in (a, b)
we have
ϕ(z) − ϕ(x) ϕ(y) − ϕ(z)
≤ .
z − x y − z
(1)
Deduce using the mean value theorem that if ϕ is twice continuously differentiable and
satisfies ϕ ′′ ≥ 0 then it is convex.
Solution. Let x ≤ z ≤ y and α = (y − z)/(y − x) so that z = αx + (1 − α)y and
α ∈ [0, 1]. If ϕ is convex, then
ϕ(z) ≤
y − z
ϕ(x) +
y − x
subtracting ϕ(z)(z − x)/(y − x) from both sides,
ϕ(z) − ϕ(x)
z − x
≤
ϕ(y) − ϕ(z)
y − z
z − x
ϕ(y),
y − x
for any x ≤ z ≤ y
Reverse process gives the original condition. For the second part of question, use the mean
value theorem for [x, z] and [z, y] and ϕ ′′ ≥ 0 to give (1).
(b) Prove Jensen’s inequality, which says that if X is a random variable such that
P(X ∈ (a, b)) = 1 and ϕ : (a, b) → R is convex, then
ϕ(EX) ≤ E(ϕ(X)).
Hint. Start by proving the following property of a convex function: If ϕ is convex then at
any point x0 ∈ (a, b), ϕ has a supporting line, that is, a linear function y(x) = ax + b
such that y(x0) = ϕ(x0) and such that ϕ(x) ≥ y(x) for all x ∈ (a, b) (to prove its existence,
use the characterization of convexity from part (a) to show that the left-sided derivative of
ϕ at x0 is less than or equal to the right-sided derivative at x0; the supporting line is a line
1

passing through the point (x0, ϕ(x0)) whose slope lies between these two numbers). Now
take the supporting line function at x0 = EX and see what happens.
Solution. If you can show that ϕ has a supporting line y(x) = ax + b, it is easy to
prove the inequality. As mentioned in Hint, put x0 = EX, and then
ϕ(X) ≥ y(X) = aX + b, which implies E(ϕ(X)) ≥ aEX + b = y(EX) = ϕ(EX).
Existence of supporting line
Choose and fix z ∈ R in (1). Then from (1),
where A := infw≥z (ϕ(w)−ϕ(z))
(w−z)
where B := sup w≤z
(ϕ(z)−ϕ(w))
(z−w)
ϕ(x) ≥ ϕ(z) + A(x − z) for any x ≤ z
that depends only on z. Similarly,
ϕ(y) ≥ ϕ(z) + B(y − z) for any y ≥ z
that depends only on z. By defining α = (A + B)/2, we have
ϕ(x) ≥ ϕ(z) + α(x − z) := L(x) for any x ∈ R
because A ≤ B. So, for each fixed z, L(x) is a linear function, and ϕ(z) = L(z) and
ϕ(x) ≥ L(x) for all x.
2. If X is a random variable satisfying a ≤ X ≤ b, prove that
and identify when equality holds.
V(X) ≤
(b − a)2
, (2)
4
Solution. Suppose that a = b. Since (X − E(X)) 2 is a quadratic function of X, the line
intersecting (a, (a − E(X)) 2 ) and (b, (b − E(X)) 2 ) is above the (X − E(X)) 2 for a ≤ X ≤ b.
That is,
(X − E(X)) 2 ≤ (b − EX)2 − (a − EX) 2
(X − a) + (a − EX)
b − a
2
2

Taking the expectation and simplifying,
V(X) ≤ (EX − a)(b − EX) ≤
(b − a)2
4
Equalities are attained in the first inequality when X = a or X = b (when the parabola and
the line intersect) and in the second inequality when EX − a = b − EX, that is, EX = a+b
2 .
Hence, X must be a random variable on {a, b} with EX = a+b.
If a = b, the inequality (in
2
fact, equality) is obvious.
3. Let X1, X2, . . . be a sequence of i.i.d. (independent and identically distributed) random
variables with distribution U(0, 1). Define events A1, A2, . . . by
An = {Xn = max(X1, X2, . . . , Xn)}
(if An occurred, we say that n is a record time).
(a) Prove that A1, A2, . . . are independent events. Hint: For each n ≥ 1, let πn be the
random permutation of (1, 2, . . . , n) obtained by forgetting the values of (X1, . . . , Xn) and
only retaining their respective order. In other words, define
πn(k) = #{1 ≤ j ≤ n : Xj ≤ Xk}.
By considering the joint density fX1,...,Xn (a uniform density on the n-dimensional unit
cube), show that πn is a uniformly random permutation of n elements, i.e. P(πn = σ) =
1/n! for any permutation σ ∈ Sn. Deduce that the event An = {πn(n) = n} is independent
of πn−1 and therefore is independent of the previous events (A1, . . . , An−1), which are all
determined by πn−1.
Solution. Let σ = (i1, · · · , in) ∈ Sn. Since Xi ∼ U(0, 1) and Xi’s are independent,
fX1,··· ,Xn = 1 only when Xi ∈ (0, 1) for all i. So,
P(πn = σ) = P(Xi1 ≤ · · · ≤ Xin)
1 xin xi2 = · · · fX1,··· ,Xndxi1 · · · dxin
0 0 0
1 xin xi2 = · · · 1dxi1 · · · dxin
0
= 1
n!
0
3
0

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.
4

4. Compute E(X) and V(X) when X is a random variable having each of the following
distributions:
1. X ∼ Binomial(n, p) ; E(X) = np, V(X) = np(1 − p)
2. X ∼ Poisson(λ) ; E(X) = λ, V(X) = λ
3. X ∼ Geom(p) ; E(X) = 1/p, V(X) = (1 − p)p −2
4. X ∼ U{1, 2, . . . , n} ; E(X) = n+1
(n+1)(n−1)
, V(X) = 2 12
5. X ∼ U(a, b) ; E(X) = a+b
(b−a)2
, V(X) = 2 12
6. X ∼ Exp(λ) ; E(X) = 1
1 , V(X) = λ λ2 6. (a) If X, Y are independent r.v.’s taking values in Z, show that
P(X + Y = n) =
∞
k=−∞
P(X = k)P(Y = n − k) (n ∈ Z)
(compare this formula with the convolution formula in the case of r.v.’s with density).
Solution. {X + Y = n} =
{X = k, Y = n − k}. Since the events in the union
are disjoint,
P(X + Y = n) =
∞
k=−∞
k∈Z
P(X = k, Y = n − k) =
∞
k=−∞
P(X = k)P(Y = n − k)
(b) Use this to show that if X ∼ Poisson(λ) and Y ∼ Poisson(µ) are independent then
X + Y ∼ Poisson(λ + µ).
Solution. By part (a),
P(X + Y = n) =
n
e
k=0
−λ λk µn−k
e−µ
k!
−(λ+µ) 1
= e
n!
n
k=0
5
(n − k)!
n
k
= e−(λ+µ)
λ k µ n−k = e
n
k=0
λ k µ −(n−k)
k!(n − k)!
−(λ+µ) 1
(λ + µ)n
n!

This implies that X + Y ∼ Poisson(λ + µ).
(c) Use the same “discrete convolution” formula to prove directly that if X ∼ Bin(n, p)
and Y ∼ Bin(m, p) are independent then X + Y ∼ Bin(n + m, p). You may make use
of the combinatorial identity (known as the Vandermonde identity or Chu-Vandermonde
identity)
Solution.
k
j=0
P(X + Y = k) =
n m
=
j k − j
n + m
k
P(X = j, Y = k − j)
k
, (n, m ≥ 0, 0 ≤ k ≤ n + m).
j=0
k
n
= p
j
j=0
j (1 − p) n−j
m
p
k − j
k−j (1 − p) m−(k−j)
= p k (1 − p) n+m−k
k
n m
= p
j k − j
j=0
k (1 − p) n+m−k
n + m
6. Prove that if X is a random variable that is independent of itself, then X is a.s.
constant, i.e., there is a constant c ∈ R such that P(X = c) = 1.
Solution. Let FX be the c.d.f. of X. Since X is independent of itself,
P(X ≤ x) = P(X ≤ x, X ≤ x) = P(X ≤ x)P(X ≤ x),
which implies that FX(x) = 0 or 1 for each x ∈ R. Recall that FX(x) is nondecreasing,
limx→−∞ FX(x) = 0 and limx→∞ FX(x) = 1. Thus, there exists c ∈ R such that
lim
x→c− FX(x) = 0 and lim
x→c + FX(x) = 1
Since FX is right continuous, P(X = c) = FX(c) − FX(c−) = 1.
6
k
.