Answer to Question #261931 in Statistics and Probability for piewalt

Question

Let X be continuous with pdf(x) = 3x^2 if 0 < x < 1, and zero otherwise,

a) Find E(X).

b)Find Var(X).

c)Find E(X').

d)Find E(3X - 5X^2 + 1).

Answer 1

a)

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

$$=\dfrac{3}{4}$$

b)

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

$$=\dfrac{3}{5}$$

$$Var(X)=E(X^2)-(E(X))^2=\dfrac{3}{5}-(\dfrac{3}{4})^2$$

$$=\dfrac{3}{80}$$

c)

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

$$=\dfrac{3}{r+3}, r\not=-3$$

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

$$=[3\ln(|x|)]\begin{matrix} 1 \\ 0 \end{matrix}=does \ not\ exist$$

d)

$$E(3X-5X^2+1)=3E(X)-5E(X^2)+1$$

$$=3(\dfrac{3}{4})-5(\dfrac{3}{5})+1=\dfrac{1}{4}$$