Question

If X and Y are independent Poisson variates having means 1 and 3 respectively, find the variance of 3X+Y.

• A. 6
• B. 9
• C. 12
• D. 18

Hint:

Always remember:
- For Poisson: Mean = Variance.
- For Variance of \(aX\): Pulling out the constant squares it (\(a^2\)).
- Since \(X\) and \(Y\) are independent, there is no covariance term:
\[ V(3X + Y) = 9 \cdot V(X) + V(Y) = 9(1) + 3 = 12 \]

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a Poisson distribution, a key mathematical property is that the variance is equal to the mean.
Additionally, we can calculate the variance of a linear combination of independent random variables using standard algebraic rules of variance.
Key Formula or Approach:
If \(X\) is a Poisson random variable with parameter \(\lambda_1\):
\[ V(X) = \lambda_1 \]
If \(Y\) is a Poisson random variable with parameter \(\lambda_2\):
\[ V(Y) = \lambda_2 \]
If \(X\) and \(Y\) are independent random variables, the variance of their linear combination is:
\[ V(aX + bY) = a^2 \cdot V(X) + b^2 \cdot V(Y) \]

Step 2: Detailed Explanation:
From the problem description, we have:
- \(X\) is a Poisson variable with mean \(\lambda_1 = 1 \implies V(X) = 1\)
- \(Y\) is a Poisson variable with mean \(\lambda_2 = 3 \implies V(Y) = 3\)
We want to find the variance of the linear combination \(3X + Y\).
Here, the constants are \(a = 3\) and \(b = 1\).
Substitute these values into our variance formula:
\[ V(3X + Y) = 3^2 \cdot V(X) + 1^2 \cdot V(Y) \]
\[ V(3X + Y) = 9 \cdot V(X) + 1 \cdot V(Y) \]
Now, substitute the values of the variances:
\[ V(3X + Y) = 9(1) + 3 = 9 + 3 = 12 \]
Thus, the variance of \(3X + Y\) is 12.

Step 3: Final Answer:
The correct option is (C).