(adsbygoogle = window.adsbygoogle || []).push({}); Suppose that X andy Y are (scalar) random variables. Show that

[cov(X,Y)]^2 ≤ var(X) var(Y). (Cauchy-Schwarz inequality)

Sow that equality holds if and only if there is a relationship of the form

m.s.

c=aX+bY (i.e. c is equal to aX+bY in "mean square").

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Proof:

E[(X+tY)^2] is quadratic in t and is ≥0. This means that the graph of it has at most one real root, i.e. discriminant ≤ 0. Setting discriminant ≤ 0 gives gives [E(XY)]^2 ≤ E(X^2) E(Y^2).

Now I am stuck with the second part. I know that equality can only possibly happen when the graph of E[(X+tY)^2] has exactly one real root, i.e. discriminant = 0, but how can I prove that c is equal to aX+bY in "mean square"?

Any help is appreciated!:)

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# Cauchy-Schwarz inequality: cov(X,Y)]^2 ≤ var(X) var(Y)

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