Answer to Question #236605 in Statistics and Probability for john k

Question

Let X be a r.v with pdf f(x) = 3x^2, 0 < x < 1. Then

(a) Calculate the quantities E(X), E(X^2) and Var(X).

(b) If the r.v.Y is denoted by Y = 3X − 2, calculate the E(Y ) and Var(Y).

Answer 1

(a)

$$E(X)=\displaystyle\int_{-\infin}^{\infin}xf(x)dx=\displaystyle\int_{0}^{1}x(3x^2)dx$$

$$=\big[\dfrac{3x^4}{4}\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{3}{4}$$

$$E(X^2)=\displaystyle\int_{-\infin}^{\infin}x^2f(x)dx=\displaystyle\int_{0}^{1}x^2(3x^2)dx$$

$$=\big[\dfrac{3x^5}{5}\big]\begin{matrix} 1 \\ 0 \end{matrix}=\dfrac{3}{5}$$

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

$$=\dfrac{3}{5}-(\dfrac{3}{4})^2=\dfrac{3}{80}=0.0375$$

(b)

$$E(Y)=E(3X-2)=3E(X)-2$$

$$=3(\dfrac{3}{5})-2=-\dfrac{1}{5}=-0.2$$

$$Var(Y)=Var(3X-2)=(3)^2Var(X)$$

$$=9(\dfrac{3}{80})=\dfrac{27}{80}=0.3375$$