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Chapter 7: Q.24 (page 360)

Prove the Cauchy-Schwarz inequality, namely,

(E[XY])2≤EX2EY2

Hint: UnlessY=-tXfor some constant, in which case the inequality holds with equality, it follows that for all t,

0<E(tX+Y)2=EX2t2+2E[XY]t+EY2

Hence, the roots of the quadratic equation

EX2t2+2E[XY]t+EY2=0

must be imaginary, which implies that the discriminant of this quadratic equation must be negative.

Short Answer

Expert verified

We proved the Cauchy-Schwarz inequality

(E[XY])2≤EX2EY2

Step by step solution

01

Given information

Given in the question that, We need to prove the Cauchy-Schwarz inequality

(E[XY])2≤EX2EY2

02

Explanation

Let us assume that EY2≠0, otherwise, we have Y=0with probability and hence E[XY]=0, so the inequality holds.

We have,

=EX2−2E[XY]EY2E[XY]+(E[XY])2EY22E[Y]2

=EX2−E[XY]EY2⇒

(E[XY])2≤EX2EY2

03

Final answer

We proved the Cauchy-Schwarz inequality

(E[XY])2≤EX2EY2

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