Question

Consider two independent random variables: $X ∼ N(5, 4)$ and $Y~N(3, 2)$. If $(2X + 3Y)~N(\mu, \sigma ^2)$, then the values of mean ($\mu$) and variance ($\sigma ^2$) are

• A. $\mu$ = 19 and $\sigma ^2$ = 34
• B. $\mu$ = 8 and $\sigma ^2$ = 14
• C. $\mu$ = 19 and $\sigma ^2$ = 14
• D. $\mu$ = 8 and $\sigma ^2$ = 34

Answer 1

The Correct Option is A

Mean and Variance of a Linear Combination of Independent Random Variables

For two independent random variables X and Y, the mean and variance of the linear combination 2X + 3Y are calculated as follows:

Step 1: Calculating the Mean

The expected value (mean) follows the linearity property: 

E(2X + 3Y) = 2E(X) + 3E(Y)

Given:

• E(X) = 5
• E(Y) = 3

Substituting values:

E(2X + 3Y) = 2 × 5 + 3 × 3 = 10 + 9 = 19

Step 2: Calculating the Variance

For independent random variables, the variance of a linear combination is given by:

Var(aX + bY) = a² Var(X) + b² Var(Y)

Given:

• Var(X) = 4
• Var(Y) = 2

Applying the formula:

Var(2X + 3Y) = 2² × Var(X) + 3² × Var(Y)

= 4 × 4 + 9 × 2

= 16 + 18 = 34

Final Answer:

• Mean (E(2X + 3Y)) = 19
• Variance (Var(2X + 3Y)) = 34