Let X be a random variable with probability mass functionx -3 6 9pX (x) 1/6 1/2 1/3Find E(X), E(X2) and E(2X + 1)2

1. Break Down the Problem

We need to find the following:

1. $E(X)$ - the expected value of $X$
2. $E(X^2)$ - the expected value of $X^2$
3. $E(2X + 1)$ - the expected value of the linear transformation of $X$

2. Relevant Concepts

• The expected value $E(X)$ can be calculated using the formula: $E(X) = \sum_{x} x \cdot p_X(x)$

• Similarly, the expected value $E(X^2)$ is calculated using: $E(X^2) = \sum_{x} x^2 \cdot p_X(x)$

• The expected value of a linear transformation can be derived from the linearity of expectation: $E(aX + b) = a \cdot E(X) + b$ For $E(2X + 1)$, $a = 2$ and $b = 1$.

3. Analysis and Detail

Step 1: Calculate $E(X)$

Given data:

• $x = -3, 6, 9$
• $p_X(-3) = \frac{1}{6}, \, p_X(6) = \frac{1}{2}, \, p_X(9) = \frac{1}{3}$

Using the formula for expected value: $E(X) = (-3) \cdot \frac{1}{6} + 6 \cdot \frac{1}{2} + 9 \cdot \frac{1}{3}$ Calculating each term: $E(X) = -\frac{3}{6} + \frac{6 \cdot 3}{6} + \frac{9 \cdot 2}{6} = -\frac{1}{2} + 9 + 6 = -\frac{1}{2} + 15 = \frac{29}{2}$

Step 2: Calculate $E(X^2)$

Using the formula: $E(X^2) = (-3)^2 \cdot \frac{1}{6} + 6^2 \cdot \frac{1}{2} + 9^2 \cdot \frac{1}{3}$ Calculating each term: $E(X^2) = 9 \cdot \frac{1}{6} + 36 \cdot \frac{1}{2} + 81 \cdot \frac{1}{3} = \frac{9}{6} + \frac{36 \cdot 3}{6} + \frac{81 \cdot 2}{6} = \frac{9}{6} + \frac{108}{6} + \frac{162}{6} = \frac{279}{6}$

Step 3: Calculate $E(2X + 1)$

Using the linear transformation formula: $E(2X + 1) = 2 \cdot E(X) + 1 = 2 \cdot \frac{29}{2} + 1 = 29 + 1 = 30$

4. Verify and Summarize

The calculations for $E(X)$, $E(X^2)$, and $E(2X + 1)$ were performed step by step, ensuring each component was accounted for accordingly.

Final Answer

• $E(X) = \frac{29}{2}$
• $E(X^2) = \frac{279}{6}$
• $E(2X + 1) = 30$