Discussion
Let \(X\) and \(Y\) be independently distributed central chi-squared random variables with degrees of freedom \(m\ (\ge 3)\) and \(n\ (\ge 3)\), respectively. If \(E\left(\frac{X}{Y}\right) = 3\) and \(m + n = 14\), then \(E\left(\frac{Y}{X}\right)\) is equal to
Answer: D
\(E\left(\frac{Y}{X}\right)\) is equal to \(\frac{5}{7}\).
For a chi-squared variable with \(k\) degrees of freedom, \(E(\chi_{k}^{2}) = k\) and, for \(k \gt 2\), \(E\left(\frac{1}{\chi_{k}^{2}}\right) = \frac{1}{k - 2}\). By independence:
\[ \begin{aligned} E\left(\tfrac{X}{Y}\right) &= \tfrac{m}{n - 2} = 3, \quad m + n = 14 \\ 14 - n &= 3n - 6 \Rightarrow n = 5,\ m = 9 \\ E\left(\tfrac{Y}{X}\right) &= \tfrac{n}{m - 2} = \tfrac{5}{7} \end{aligned} \]
Taking \(E(Y/X)\) as the reciprocal \(\frac{1}{3}\) is the common mistake: the expectation of a reciprocal is not the reciprocal of the expectation, and \(\frac{1}{3}\) is not offered.
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