Discussion
Let \(X_{1}\) and \(X_{2}\) be jointly distributed with pdf \(f(x_{1}, x_{2})\) given by :
\[ \begin{aligned} f(x_{1}, x_{2}) &= \tfrac{1}{4}\{1 + x_{1}x_{2}(x_{1}^{2} - x_{2}^{2})\}; \\ &\qquad |x_{1}| \lt 1 \text{ and } |x_{2}| \lt 1 \\ &= 0; \quad \text{otherwise} \end{aligned} \]
Then, the characteristic function of \(X_{1}\) is :
Answer: A
The characteristic function of \(X_{1}\) is \(\dfrac{\sin t}{t}\).
First find the marginal density of \(X_{1}\) by integrating out \(x_{2}\) over \((-1, 1)\):
\[ \begin{aligned} f_{1}(x_{1}) &= \frac{1}{4}\int_{-1}^{1} dx_{2} \\ &\quad + \frac{x_{1}}{4}\int_{-1}^{1} x_{2}(x_{1}^{2} - x_{2}^{2})\, dx_{2} \end{aligned} \]
In the second integral, \(x_{2}(x_{1}^{2} - x_{2}^{2})\) is an odd function of \(x_{2}\), so it integrates to 0 over the symmetric interval. What remains is \(\frac{1}{4} \cdot 2 = \frac{1}{2}\) for \(|x_{1}| \lt 1\): \(X_{1}\) is uniform on \((-1, 1)\).
\[ \begin{aligned} \phi(t) &= \frac{1}{2}\int_{-1}^{1} e^{itx}\, dx \\ &= \frac{e^{it} - e^{-it}}{2it} = \frac{\sin t}{t} \end{aligned} \]
The other expressions cannot be characteristic functions: every characteristic function equals 1 at \(t = 0\), while \(\cos t/t\) and \(\sin t/t^{2}\) blow up there, and \(t/\sin t\) exceeds 1 in absolute value near \(t = \pi\).
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