Hi Guys, I need help on this question.
Where the conditions are shown below:
The final answer given in the paper is
Kindly enlighten me on this as I do not know how to derive to the final answer.
Thanks in advance.
Sincerely,
James
Hi Guys, I need help on this question.
Where the conditions are shown below:
The final answer given in the paper is
Kindly enlighten me on this as I do not know how to derive to the final answer.
Thanks in advance.
Sincerely,
James
First sort out the brackets in the first expression:
$\displaystyle
E\left( \frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)^2
$
This is ambiguous and could plausibly mean either:
$\displaystyle
\left[ E \left(\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)\right]^2
$
or:
$\displaystyle
E\left( \left[\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right]^2 \right)
$
Given what you are asked to prove you probably want the latter. If so expand the square and use the linearity of expectation to simplify.
CB
$\displaystyle E\left[ \frac{1}{T^2}\left( 2\sum_{s\ne t} \frac{u_yu_s}{x_t x_s} +\sum \frac{u_t^2}{x_t^2}\right)\right]$ $\displaystyle =
\frac{1}{T^2}\left[\sum_{s\ne t} E\left(\frac{u_tu_s}{x_tx_s}\right)\right]+\frac{1}{T^2}\left[\sum E\left(\frac{u_t^2}{x_t^2} \right)\right]
$
You are told that the all the terms in the first summation on the right are zero, and so:
$\displaystyle E\left[ \frac{1}{T^2}\left( 2\sum_{s\ne t} \frac{u_yu_s}{x_t x_s} +\sum \frac{u_t^2}{x_t^2}\right)\right]$ $\displaystyle =
\frac{1}{T^2}\left[\sum E\left(\frac{u_t^2}{x_t^2} \right)\right]
$ $\displaystyle
=\frac{1}{T^2}\sum \frac{E(u_t^2)}{x_t^2}
$
and you should be able to finish from there.
CB