Question

A2. A random variable X has the cumulative distribution function given as
F(x) =



0, for x < 1
x
2 2 2x + 2
2
, for 1 ≤ x < 2
1, for x ≥ 2
Calculate the variance of X.

Answer 1

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

The magnitude of the jump in the CDF that occur at $x=1$ is $\dfrac{1}{2}.$

$$f(x)=F'(x) = \begin{cases} {\delta(x-1)\over 2}+x-1 &\text{for } 1\leq x<2 \\ 0 &\text{otherwise } \end{cases}$$

$$E(X)=\displaystyle\int_{1}^2x( {\delta(x-1)\over 2}+x-1)dx=$$

$$={x\over 2}|_{x=1}+\big[{x^3\over 3}-{x^2\over 2}\big]\begin{matrix} 2 \\ 1 \end{matrix}=$$

$$={1\over 2}+{8\over 3}-2-{1\over 3}+{1\over 2}={4\over 3}$$

$$E(X^2 )=\displaystyle\int_{1}^2x^2 ( {\delta(x-1)\over 2}+x-1)dx=$$

$$={x^2\over 2}|_{x=1}+\big[{x^4\over 4}-{x^3\over 3}\big]\begin{matrix} 2 \\ 1 \end{matrix}=$$

$$={1\over 2}+4-{8\over 3}-{1\over 4}+{1\over 3}={23\over 12}$$

$$V(X)=E(X^2)-(E(X))^2={23\over 12}-({4\over 3})^2={5\over 36}$$