Q.24

Prove the Cauchy-Schwarz inequality, namely,

$(E [X Y])^{2} \leq E [X^{2}] E [Y^{2}]$

Hint: Unless $Y = - t X$ for some constant, in which case the inequality holds with equality, it follows that for all $t$ ,

$0 < E [(t X + Y)^{2}] = E [X^{2}] t^{2} + 2 E [X Y] t + E [Y^{2}]$

Hence, the roots of the quadratic equation

$E [X^{2}] t^{2} + 2 E [X Y] t + E [Y^{2}] = 0$

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

Verified

Answer

We proved the Cauchy-Schwarz inequality

$(E [X Y])^{2} \leq E [X^{2}] E [Y^{2}]$

1Step 1: Given information

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

$(E [X Y])^{2} \leq E [X^{2}] E [Y^{2}]$

2Step 2: Explanation

Let us assume that $E [Y^{2}] \neq 0$ , otherwise, we have $Y = 0$ with probability and hence $E [X Y] = 0$ , so the inequality holds.

We have,

$= E [X^{2}] - 2 \frac{E [X Y]}{E [Y^{2}]} E [X Y] + \frac{(E [X Y])^{2}}{E {[Y^{2}]}^{2}} E [Y]^{2}$

$= E [X^{2}] - \frac{E [X Y]}{E [Y^{2}]} \Rightarrow$

$(E [X Y])^{2} \leq E [X^{2}] E [Y^{2}]$

3Step 3: Final answer

We proved the Cauchy-Schwarz inequality

$(E [X Y])^{2} \leq E [X^{2}] E [Y^{2}]$