Question

The p.m.f. of a random variable X is $P(x) = \begin{cases} \frac{2x}{n(n+1)}, & x = 1, 2, 3, ....... n \\ 0, & \text{otherwise} \end{cases}$, then $E(X)$ is

• A. $\frac{n+1}{6}$
• B. $\frac{2n+1}{6}$
• C. $\frac{n+1}{3}$
• D. $\frac{2n+1}{3}$

Hint:

The expected value of a discrete random variable is the sum of each possible value multiplied by its probability. Remember the formula for the sum of the first $n$ squares: $\sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6}$.

Answer 1

The Correct Option is D

Step 1: Write down the formula for the Expected Value. For a discrete random variable $X$ with probability mass function $P(x)$, the Expected Value $E(X)$ is given by: \[ E(X) = \sum_{x} x P(x) \] Given that $x$ ranges from $1$ to $n$, and $P(x) = \frac{2x}{n(n+1)}$ for these values, we have: \[ E(X) = \sum_{x=1}^{n} x \cdot \frac{2x}{n(n+1)} \]
Step 2: Simplify the summation expression. We can factor out the constant terms from the summation: \[ E(X) = \frac{2}{n(n+1)} \sum_{x=1}^{n} x^2 \]
Step 3: Apply the formula for the sum of the first $n$ squares. The sum of the first $n$ squares is given by the formula: \[ \sum_{x=1}^{n} x^2 = \frac{n(n+1)(2n+1)}{6} \]
Step 4: Substitute the sum back into the $E(X)$ expression and simplify. Substitute the sum of squares formula into the expression for $E(X)$: \[ E(X) = \frac{2}{n(n+1)} \cdot \frac{n(n+1)(2n+1)}{6} \] Cancel out the common terms $n(n+1)$ from the numerator and denominator: \[ E(X) = \frac{2(2n+1)}{6} \] \[ E(X) = \frac{2n+1}{3} \]