1. ## expectation proving question

Prove that E[(Z - Uz)^2] = c^2(E[X^2] - (Ux)^2) if X is a discrete random variable and b and c are constants and define a new random variable Z = b + cX

Note : Uz = mu of z = mean of z

Ux = mu of x = mean of x

2. ## this is what i get upto....

E ((Z-Uz)^2) = E(Z^2) - Uz^2 --- Eq1

Uz = cUx

subbing in z = b + cX in Eq1

=E(b^2 + 2cX + (cX)^2) - (cUx)^2
= b^2 + 2c *(E(X)) + c^2(E(X^2)) - c^2 * Ux^2
= b^2 + 2cE(X) + c^2((E(X^2)) - Ux^2)

Why do i get b^2 + 2cE(X) extra???????

3. Originally Posted by jus_a_kid
Prove that E[(Z - Uz)^2] = c^2(E[X^2] - (Ux)^2) if X is a discrete random variable and b and c are constants and define a new random variable Z = b + cX

Note : Uz = mu of z = mean of z

Ux = mu of x = mean of x
By definition: $E[(Z - \overline{z})^2] = Var[Z]$.

And $Var[Z] = Var[b + cX] = c^2 Var[X]$.

And by definition $Var[X] = E[X^2] - \overline{x}^2$.