Page 48 - Elements of Distribution Theory

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052184472Xc01 CUNY148/Severini May 24, 2005 17:52

34 Properties of Probability Distributions

Proof. Consider the function f (x) =− log(x), x > 0. Then
1

f (x) = 2 > 0, x > 0
x
so that f is convex. It follows that
f (αa + βb) ≤ αf (a) + β f (b);

that is,
− log(αa + βb) ≤−[α log(a) + β log(b)].

Taking the exponential function of both sides of this inequality yields
α β
αa + βb ≥ a b ,
proving part (i).
Consider part (ii). Let α = 1/p and β = 1/q. Then, by part (i) of the theorem applied
p
q
to a and b ,
pα qβ
p
q
ab = a b ≤ αa + βb ,
proving part (ii).
p q
Proof of Theorem 1.15. The result is clearly true if E(|g 1 (X)| ) =∞ or E(|g 2 (X)| ) =∞
p
q
so we may assume that E(|g 1 (X)| ) < ∞ and E(|g 2 (X)| ) < ∞.
p
If E(|g 1 (X)| ) = 0 then |g 1 (X)|= 0 with probability 1, so that E(|g 1 (X)g 2 (X)|) = 0
q
and the result holds; similarly, the result holds if E(|g 1 (X)| ) = 0. Hence, assume that
p
q
E(|g 1 (X)| ) > 0 and E(|g 1 (X)| ) > 0.
Applying part (ii) of Lemma 1.1 to
|g 1 (X)| |g 2 (X)|
and ,
1 1
q
p
E(|g 1 (X)| ) p E(|g 2 (X)| ) q
it follows that
p q
|g 1 (X)||g 2 (X)| |g 1 (X)| |g 2 (X)|
≤ + .
1 1 p q
p
q
E(|g 1 (X)| ) E(|g 2 (X)| ) q pE(|g 1 (X)| ) qE(|g 2 (X) )
p
Taking expectations of both sides of this inequality shows that
E(|g 1 (X)g 2 (X)|) 1 1
≤ + = 1;
1 1
q
p
E(|g 1 (X)| ) E(|g 2 (X)| ) q p q
p
the result follows.
1.9 Exercises
In problems 1 through 6, let and P denote the sample space and probability function, respectively,
of an experiment and let A 1 , A 2 , and A 3 denote events.
1.1 Show that
P(A 1 ∪ A 2 ) = P(A 1 ) + P(A 2 ) − P(A 1 ∩ A 2 ).