Question

Prove the Cauchy-Schwarz inequality, namely,

$\left(E\right[XY]{)}^2\le E\left[X^2\right]E\left[Y^2\right]$

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

$0<E\left[(tX+Y{)}^2\right]=E\left[X^2\right]t^2+2E\left[XY\right]t+E\left[Y^2\right]$

Hence, the roots of the quadratic equation

$E\left[X^2\right]t^2+2E\left[XY\right]t+E\left[Y^2\right]=0$

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

Short answer

We proved the Cauchy-Schwarz inequality

$\left(E\right[XY]{)}^2\le E\left[X^2\right]E\left[Y^2\right]$

Step-by-step solution

Given information

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

$\left(E\right[XY]{)}^2\le E\left[X^2\right]E\left[Y^2\right]$

Explanation

Let us assume that $E\left[Y^2\right]\ne 0$, otherwise, we have $Y=0$with probability and hence $E[XY]=0$, so the inequality holds.

We have,

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

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

$\left(E\right[XY]{)}^2\le E\left[X^2\right]E\left[Y^2\right]$

Final answer

We proved the Cauchy-Schwarz inequality

$\left(E\right[XY]{)}^2\le E\left[X^2\right]E\left[Y^2\right]$