1. Expectation with Double Summation

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.

Sincerely,
James

2. Originally Posted by nightvision1984
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.

Sincerely,
James
First sort out the brackets in the first expression:

$
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:

$
\left[ E \left(\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)\right]^2
$

or:

$
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

3. Originally Posted by CaptainBlack
First sort out the brackets in the first expression:

$
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:

$
\left[ E \left(\frac{1}{T}\ \sum_{t=1}^T \left(\frac{u_t}{x_t}\right) \right)\right]^2
$

or:

$
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
Hi Captian Black,

I've used the latter one just as you have predicted.

But when I expand the square of the term, I got

And I'm stuck with the double summation.

Kindly help me with this. I'm quite weak in this.

Thanks,
James

4. Originally Posted by nightvision1984
Hi Captian Black,

I've used the latter one just as you have predicted.

But when I expand the square of the term, I got

And I'm stuck with the double summation.

Kindly help me with this. I'm quite weak in this.

Thanks,
James
$E\left[ \frac{1}{T^2}\left( 2\sum_{s\ne t} \frac{u_yu_s}{x_t x_s} +\sum \frac{u_t^2}{x_t^2}\right)\right]$ $=
\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:

$E\left[ \frac{1}{T^2}\left( 2\sum_{s\ne t} \frac{u_yu_s}{x_t x_s} +\sum \frac{u_t^2}{x_t^2}\right)\right]$ $=
\frac{1}{T^2}\left[\sum E\left(\frac{u_t^2}{x_t^2} \right)\right]
$
$
=\frac{1}{T^2}\sum \frac{E(u_t^2)}{x_t^2}
$

and you should be able to finish from there.

CB

5. Captain Black,

THANK YOU!!

Sincerely,
James