Page 173 - Probability and Statistical Inference

150 3. Multivariate Random Variables

–E[X X ] in the last step of (3.9.11), we get
1 2

that is since . Thus, (3.9.10) has

been proven.
Next, let us suppose that equality holds in (3.9.10), that is,
2
E [X X ] = 0. Then, from (3.9.10) we conclude that
1 2

It should be clear from (3.9.12) that with λ ≡ λ = –E[X X ]/ we have
0 1 2

and again since (X + λ X ) is a non-negative random variable w.p.1, one can
2
1
0
2
conclude that X + λ X = 0 w.p.1. Hence, X = kX w.p.1 with some real
2
1
2
0
1
number k = –λ .
0
Let us suppose that X = kX w.p.1 so that E[X X ] = k and
2
1
2
1
. Hence, one immediately has E [X X ] –
2
1 2
. Now the proof is complete. ¢
Theorem 3.9.6 (Covariance Inequality) Suppose that we have two real
valued random variables X , X , such that and E[X X ] are
1 2 1 2
all finite. Then, we have
In (3.9.13), the equality holds if and only if X = a + bX w.p.1 for some
1
2
constants a and b.
Proof Recall from (3.4.1) that Cov(X , X ) = E{(X – µ )(X – µ )} where
2
2
1
1
2
1
µ = E[X ], i = 1, 2. Let us denote U = X – µ , i = 1, 2, and then by the
i
i
i
i
i
Cauchy-Schwarz inequality (3.9.10), we have
which verifies (3.9.13). The covariance inequality will become an equality if and
only if we have equality throughout (3.9.14), that is if and only if . From the
Cauchy-Schwarz inequality it then follows that we will have equality in (3.9.13) if
and only if U = kU w.p.1 with some constant k, which will be equivalent to the
1 2
claim that X = a + bX w.p.1. where a = µ – kµ and b = k. ¢
1 2 1 2