Let X and Y be rv wiyh correlation rho Show that Var y x

Let X and Y be r.v wiyh correlation rho . Show that Var [ y / x ]

Solution

We have var(y)=E([y-E(y)]²).

Therefore,

var(y|x)=E([y-E(y)]²|x)

=E([y²-2yE(y)+E(y)²]|x)

=E(y²|x)-2E(y|x)E(y|x)+E(y|x)²

=E(y²|x)-E(y|x)²

Since var(y|x) is a random variable, we can talk about its expected value.

E(var(y|x))=E(E(y²|x)-E(y|x)²)

=E(E(y²|x))-E(E(y|x)²)

The expected value of the conditional expectation of a random variable is the expected value of the original random variable, so applying this to Y² gives

E(var(y|x))=E(y²)-E(E(y|x)²) ....(1)

Consider the variation of the condtional expected value (using the defination of variance).

var(E(y|x))=E(E(y|x))²-E(E(y|x)²)

=E(E(y|x)²)-E(y)² ......(2)

Adding (1) and (2),

E(var(y|x))+var(E(y|x))=E(y²)-E(y)²=var(y)

=>E(var(y|x))=var(y)-var(E(y|x)) ......(3)

For a pair (X, Y) of random variables which follows a bivariate normal distribution, the conditional mean E(Y|X) is a linear function of X given by,

E(y|x)=E(y)+p*sqrt(var(y)/var(x))*(x-E(x))

Taking the variance oon both sides and using the rule of multiplication and addition,

var(E(y|x))=var(E(y))+p*sqrt(var(y)/var(x))*(var(x)-var(E(x))

=0+p*sqrt(var(y)/var(x))*(var(x)-0)

=p*sqrt(var(y)*var(x))

Therefore, from eq(3)

E(var(y|x))=var(y)-p*sqrt(var(y)*var(x))

=> var(y)-E(var(y|x))=p*sqrt(var(y)*var(x))

Now, p²<p or p²*var(y)<p*var(y) or p²*var(y)<p*sqrt(var(y)*var(x))

Therefore, var(y)-E(var(y|x))>p²*var(y)

=>E(var(y|x))<var(y)-p²*var(y)

=>E(var(y|x))<(1-p²)*var(y)