Discussion
Suppose \(\underline{X}_{2 \times 1} \sim\) Bivariate Normal
\[ \left(\begin{bmatrix} 4 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 & -1 \\ -1 & 5 \end{bmatrix}\right). \]
Then the distribution of \(3Y = 3X_{1} - 4X_{2}\) is :
Answer: D
\(3X_{1} - 4X_{2}\) is Normal (8, 122).
A linear combination \(a'X\) of a bivariate normal vector is normal with mean \(a'\mu\) and variance \(a'\Sigma a\). With \(a = (3, -4)'\):
\[ \begin{aligned} \text{mean} &= 3(4) - 4(1) = 8 \\ \text{variance} &= 9(2) + 16(5) - 2(3)(4)(-1) \\ &= 18 + 80 + 24 = 122 \end{aligned} \]
The covariance is negative and the coefficients have opposite signs, so the cross term adds to the variance. Ignoring the cross term gives 98, the third option. Normal(8, 62) has the right mean but a variance that does not equal \(a'\Sigma a\), and Normal(-1, 26) has the wrong mean as well.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development