Question

A random variable \(X\) has a cumulative distribution function \(F_X(x)\) given by 

\( F_X(x) = \begin{cases} 0 & \text{if } x < 1 \\ \frac{x^{2} - 2x + 2}{2} & \text{if } 1 \le x < 2 \\ 1 & \text{if } x \ge 2 \end{cases} \)

Then \(E(X)\) is:

• A. $\frac{3}{2}$
• B. $\frac{4}{3}$
• C. $\frac{7}{3}$
• D. $\frac{5}{6}$

Hint:

For non-negative variables, $E(X) = \int_{0}^{\infty} [1 - F(x)] dx$.
$E(X) = \int_{0}^{1} (1-0) dx + \int_{1}^{2} (1 - \frac{x^2-2x+2}{2}) dx = 1 + \int_{1}^{2} \frac{2x - x^2}{2} dx$.
$1 + \frac{1}{2} [x^2 - x^3/3]_1^2 = 1 + \frac{1}{2} [(4 - 8/3) - (1 - 1/3)] = 1 + \frac{1}{2} [4/3 - 2/3] = 1 + 1/3 = 4/3$. This "complementary CDF" method handles discrete jumps automatically and is often less prone to error.

Answer 1

The Correct Option is B

We need to find the expectation $E(X)$ from the given CDF. Notice that $F(1) = 1/2$ while $F(x)=0$ for $x<1$. This jump indicates a discrete mass at $x=1$. 

Step 1: Identify the components of the distribution
The probability at the jump point $x=1$ is $P(X=1) = F(1) - \lim_{x \to 1^{-}} F(x) = 1/2 - 0 = 1/2$. 
The continuous density $f(x)$ for $1 < x < 2$ is the derivative of the CDF: 
$f(x) = \frac{d}{dx} (\frac{x^2 - 2x + 2}{2}) = \frac{2x - 2}{2} = x - 1$. 

Step 2: Set up the expectation integral for a mixed distribution
$E(X) = \text{Discrete part} + \text{Continuous part}$. 
$E(X) = (1) \cdot P(X=1) + \int_{1}^{2} x \cdot f(x) dx$. 
$E(X) = (1) \cdot \frac{1}{2} + \int_{1}^{2} x(x - 1) dx$. 

Step 3: Evaluate the definite integral
$\int_{1}^{2} (x^{2} - x) dx = [\frac{x^{3}}{3} - \frac{x^{2}}{2}]_{1}^{2}$. 
$= (\frac{8}{3} - \frac{4}{2}) - (\frac{1}{3} - \frac{1}{2})$. 
$= (\frac{8}{3} - 2) - (\frac{2 - 3}{6}) = \frac{2}{3} - (-\frac{1}{6})$. 
$= \frac{2}{3} + \frac{1}{6} = \frac{4 + 1}{6} = \frac{5}{6}$. 

Step 4:Calculate the total expectation
$E(X) = \frac{1}{2} + \frac{5}{6}$. 
$E(X) = \frac{3}{6} + \frac{5}{6} = \frac{8}{6} = \frac{4}{3}$. 
The expected value is $\frac{4}{3}$, which corresponds to Option (2).