Answer to Question #273953 in Statistics and Probability for Air

Question

A random variable x has the cumulative distribution function given as

f(x) = 8>x<7

0; for x<1

X^2 -2x+2

2; for 1≤x2

1; for x≥2

Calculate the variance of x.

Answer 1

$$F(x)= \begin{cases} 0 &x<1 \\ x^2-2x+1 &1\leq x<2 \\ 1 & x\geq2 \end{cases}$$

$$f(x)= \begin{cases} 2x-2 &1\leq x\leq2 \\ 0 & otherwise \end{cases}$$

Check

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

$$=[x^2-2x]\begin{matrix} 2 \\ 1 \end{matrix}=4-4-(1-2)=1$$

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

$$=[\dfrac{2x^3}{3}-x^2]\begin{matrix} 2 \\ 1 \end{matrix}=\dfrac{16}{3}-4-(\dfrac{2}{3}-1)=\dfrac{5}{3}$$

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

$$=[\dfrac{x^4}{2}-\dfrac{2x^3}{3}]\begin{matrix} 2 \\ 1 \end{matrix}=8-\dfrac{16}{3}-(\dfrac{1}{2}-\dfrac{2}{3})=\dfrac{17}{6}$$

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

$$=\dfrac{17}{6}-[\dfrac{5}{3}]^2=\dfrac{1}{18}$$