Question

Let \(X_1, X_2, X_3\) be three variables with means 3, 4 and 5 respectively, variances 10, 20 and 30 respectively and \(cov (X_1, X_2) = cov (X_2, X_3) = 0\) and \(cov (X_1, X_3) = 5\). If, \(Y = 2X_1 +3X_2+4X_3\) then, Var(\(Y\)) is:

• A. 700
• B. 710
• C. 690
• D. 620

Hint:

When calculating the variance of a linear combination, pay close attention to the signs of the coefficients and the covariance terms. The formula is \( \text{Var}(\sum a_i X_i) = \sum a_i^2 \text{Var}(X_i) + \sum_{i \neq j} 2 a_i a_j \text{Cov}(X_i, X_j) \). If your result isn't in the options, quickly check for potential typos, especially signs, in the problem statement.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question requires calculating the variance of a linear combination of three random variables. The means of the variables are given but are not needed for calculating the variance. The key is to use the general formula for the variance of a sum of random variables, which includes terms for their individual variances and their covariances.

Step 2: Key Formula or Approach:
For a linear combination of three random variables \(Y = aX_1 + bX_2 + cX_3\), the variance is given by: \[ \text{Var}(Y) = a^2 \text{Var}(X_1) + b^2 \text{Var}(X_2) + c^2 \text{Var}(X_3) + 2ab \text{Cov}(X_1, X_2) + 2ac \text{Cov}(X_1, X_3) + 2bc \text{Cov}(X_2, X_3) \]
Step 3: Detailed Explanation:
Let's identify the given values: - Linear combination: \(Y = 2X_1 + 3X_2 + 4X_3\), so \(a=2, b=3, c=4\). - Variances: \(\text{Var}(X_1) = 10\), \(\text{Var}(X_2) = 20\), \(\text{Var}(X_3) = 30\). - Covariances: \(\text{Cov}(X_1, X_2) = 0\), \(\text{Cov}(X_2, X_3) = 0\), \(\text{Cov}(X_1, X_3) = 5\). Substitute these values into the variance formula: \[ \text{Var}(Y) = (2^2)\text{Var}(X_1) + (3^2)\text{Var}(X_2) + (4^2)\text{Var}(X_3) + 2(2)(3)\text{Cov}(X_1, X_2) + 2(2)(4)\text{Cov}(X_1, X_3) + 2(3)(4)\text{Cov}(X_2, X_3) \] \[ \text{Var}(Y) = 4(10) + 9(20) + 16(30) + 12(0) + 16(5) + 24(0) \] \[ \text{Var}(Y) = 40 + 180 + 480 + 0 + 80 + 0 \] \[ \text{Var}(Y) = 780 \] The calculated variance is 780. This result is not among the options. This suggests a potential typo in the question. Let's consider a common typo, such as a sign error in the linear combination. For instance, if \(Y = 2X_1 + 3X_2 - 4X_3\), the variance would be: \[ \text{Var}(Y) = a^2\text{Var}(X_1) + b^2\text{Var}(X_2) + c^2\text{Var}(X_3) + 2ab\text{Cov}(X_1, X_2) - 2ac\text{Cov}(X_1, X_3) - 2bc\text{Cov}(X_2, X_3) \] \[ \text{Var}(Y) = 4(10) + 9(20) + (-4)^2(30) + 12(0) - 2(2)(4)(5) - 24(0) \] \[ \text{Var}(Y) = 40 + 180 + 16(30) - 80 \] \[ \text{Var}(Y) = 40 + 180 + 480 - 80 = 700 - 80 = 620 \] This result matches option (D). It is highly probable that the intended equation was \(Y = 2X_1 + 3X_2 - 4X_3\) or that \(\text{Cov}(X_1, X_3) = -5\).
Step 4: Final Answer:
Based on the direct calculation from the question as written, the answer is 780, which is not an option. Assuming a typo in the sign of the last term of Y, the answer is 620. We choose 620 as the intended answer.