Question

2. Let X be a continuous random variable with pdf
f(x) = cx^2 , |x| ≤ 1,
0, otherwise,
where the parameter c is constant (with respect to x).
(a) Find the constant c.
(b) Compute the cumulative distribution function F(x) of X.
(c) Use F(x) (from b) to determine P(X ≥ 1/2).
(d) Find E(X) and V (X).

Answer 1

a) We know that for $f(x)$ to be a probability distribution

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

We integrate $f(x)$ with respect to $x,$ set the result equal to 1 and solve for $c.$

$$\displaystyle\int_{-\infin}^\infin f(x)dx=\displaystyle\int_{-1}^{1} cx^2dx=c\big[{x^3 \over 3}\big]\begin{array}{c} 1 \\ -1 \end{array}=$$

$$=c\big({1^3 \over 3}-{(-1)^3 \over 3}\big)={2 \over 3}c=1$$

$$c={3 \over 2}$$

b) Cumulative distribution function (cdf):

$$F(x)=P(X\leq x)=\displaystyle\int_{-\infin}^x f(x)dx$$

$$\displaystyle\int_{-1}^x {3 \over 2}x^2 dx ={3 \over 2}\big[{x^3 \over 3}\big]\begin{array}{c} x \\ -1 \end{array}={x^3 \over 2}+{1 \over 2}$$

$$F(x) = \begin{cases} 0 &\text{if } x<-1 \\ {x^3 \over 2}+{1 \over 2} &\text{if }-1\leq x<1 \\ 1 &\text{if } x\geq 1 \end{cases}$$

c)

$$P(X\geq {1 \over 2})=1-P(X< {1 \over 2})=1-P(X\leq {1 \over 2})=$$

$$=1-F({1 \over 2})=1-({({1 \over 2})^3 \over 2}+{1 \over 2})={7 \over 16}$$

d)

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

$$={3 \over 2}\big[{x^4 \over 4}\big]\begin{array}{c} 1 \\ -1 \end{array}=0$$

$$Var(X)=\sigma^2=E[(X-\mu)^2]=\displaystyle\int_{-\infin}^\infin (x-0)^2f(x)dx=$$

$$=\displaystyle\int_{-1}^{1} {3 \over 2}x^4dx={3 \over 2}\big[{x^5 \over 5}\big]\begin{array}{c} 1 \\ -1 \end{array}={3 \over 5}$$

$$E(X)=0$$ $$Var(X)={3 \over 5}$$