Question

Let X be a random variable with the following probability distribution:
x : 3 − 6 9
P(X=x) : 1/6 1/2 1/3
Find E(X) and E(X^2)and using the laws of expectation, evaluate E . (2X+1)^2.

Answer 1

$\def\arraystretch{1.5} \begin{array}{c:c:c:c} x & 3 & -6 & 9 \\ \hline P(X=x) & 1/6 & 1/2 & 1/3 \\ \end{array}$

$E(X), E(X^2), E((2X+1)^2) \ - ?$

Solution:

$E(X)=3\times 1/6+(-6)\times 1/2+9\times1/3=0.5$

$E(X^2)=3^2\times 1/6+(-6)^2\times 1/2+9^2\times1/3=46.5$

$Var(X)=E(X^2)-(E(X))^2=46.5-0.5^2=46.25$

$Var(2X+1)=E((2X+1)^2)-(E(2X+1))^2= E((2X+1)^2)-(2E(X)+1)^2$

$Var(2X+1)=Var (2X)=4Var(X)$

$E((2X+1)^2) =4Var(X)+(2E(x)+1)^2=4\times 46.25+(2\times 0.5+1)^2=189$

Answer: $E(X)=0.5,\ E(X^2)=46.5, \ E((2X+1)^2)=189.$