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

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