Estimating Parameters

**lindah** · June 1st 2011, 05:41 PM

Hi guys,

Parameter Estimation Question

I tried using [IMG][/IMG] tags, but they would not work
Any help is greatly appreciated!

**CaptainBlack** · June 1st 2011, 10:44 PM

Originally Posted by lindah

Hi guys,

Parameter Estimation Question

I tried using [IMG][/IMG] tags, but they would not work
Any help is greatly appreciated!

What do you get for the mean and variance of the distribution?

(you should also tell us that f(x)=0 for x<a and that what you give applies for x>=a)

CB

**lindah** · June 1st 2011, 11:05 PM

Hi CB,

Yes sorry I omitted the information about the bounds

The math tags don't seem to be working so my E[X] and E[X^2] are as follows:
Mean and Variance, Moments

It is at this point where I get slightly confused. I have just rearranged the parameters in each of the 2 equations to get the answer I obtained above. I've worked through the examples for a normal distribution, but its much more straightforward than this one.

**CaptainBlack** · June 2nd 2011, 01:38 AM

Originally Posted by lindah

Hi CB,

Yes sorry I omitted the information about the bounds

The math tags don't seem to be working so my E[X] and E[X^2] are as follows:
Mean and Variance, Moments

It is at this point where I get slightly confused. I have just rearranged the parameters in each of the 2 equations to get the answer I obtained above. I've worked through the examples for a normal distribution, but its much more straightforward than this one.

Your moments are wrong, how have you calculated them?

CB

**lindah** · June 2nd 2011, 01:57 AM

For m_{1}, I took the integral with respect to x of [x.f(x)]
For m_{2}, I took the integral of [x^2 f(x)]
The integral boundaries were [0,infinity)

**CaptainBlack** · June 2nd 2011, 02:55 AM

Originally Posted by lindah

For m_{1}, I took the integral with respect to x of [x.f(x)]
For m_{2}, I took the integral of [x^2 f(x)]
The integral boundaries were [0,infinity)

The limits are a to infinity.

CB

**lindah** · June 2nd 2011, 04:01 AM

Originally Posted by CaptainBlack

The limits are a to infinity.

CB

Hi CB,

In hindsight that does make sense to me now to have those limits.

I have recalculated:
1) $E[X] = a + \theta$
2) $E[X^2] = a^2 + 2a\theta + 2\theta^2$

Can I confirm if these are correct before the next step?

Thank you!

**CaptainBlack** · June 2nd 2011, 04:09 AM

Originally Posted by lindah

Hi CB,

In hindsight that does make sense to me now to have those limits.

I have recalculated:
1) $E[X] = a + \theta$
2) $E[X^2] = a^2 + 2a\theta + 2\theta^2$

Can I confirm if these are correct before the next step?

Thank you!

Yes, they are correct, but you will find the next part easier if you work with the second central moment (the variance) the effect is the same as working with the raw second moment.

CB

**lindah** · June 2nd 2011, 04:31 AM

Originally Posted by CaptainBlack

Yes, they are correct, but you will find the next part easier if you work with the second central moment (the variance) the effect is the same as working with the raw second moment.

Hi CB,

Please bear with me for this part, as its a hurdle I can't get past for these types of questions.

Using what I've derived above:
$var[X] = E[X^2] - \left( E[X] \right)^2 =\theta^2$

I equate:
$m_{1} = a + \theta$
$a = m_{1} - \theta$

Similarly for $m_{2}$ and then re-written in terms of $\theta$ :
$m_{2} = \theta^2$
$\theta = \sqrt{m_{2}} = \sqrt{\sigma^2} = \sigma$

Substituting this into the expression for $a$ the estimates are as follows?:
$\hat{a} = \bar{X} - \sigma$
$\hat{\theta} = \sigma$

**CaptainBlack** · June 2nd 2011, 04:54 AM

Originally Posted by lindah

Hi CB,

Please bear with me for this part, as its a hurdle I can't get past for these types of questions.

Using what I've derived above:
$var[X] = E[X^2] - \left( E[X] \right)^2 =\theta^2$

I equate:
$m_{1} = a + \theta$
$a = m_{1} - \theta$

Similarly for $m_{2}$ and then re-written in terms of $\theta$ :
$m_{2} = \theta^2$
$\theta = \sqrt{m_{2}} = \sqrt{\sigma^2} = \sigma$

Substituting this into the expression for $a$ the estimates are as follows?:
$\hat{a} = \bar{X} - \sigma$
$\hat{\theta} = \sigma$

That looks right (except make it clear that $\sigma$ is the sample based SD).

CB

**lindah** · June 2nd 2011, 05:13 AM

Thank you for your guidance!!!

Linda

**lindah** · June 4th 2011, 12:11 AM

Originally Posted by CaptainBlack

That looks right (except make it clear that $\sigma$ is the sample based SD).

CB

Hi CB,

Can I just confirm if I am allowed to equate the Variance as the second moment or would it have to be calculated as:

$var[X] = \theta^2$
$var[X]=E[X^2] - (E[X])^2$
$E[X^2] = var[X] +(E[X])^2$

Now equate $m_{2}=E[X^2]$ , NOT $m_{2}=var[X]$ as I did before?

Doing this I should get:
$m_{2}=\theta^2 + m_{1}^2$
and so rearranging will give me:
$\theta = \sqrt{m_{2} - m_{1}^2}$

Instead of
$\theta = \hat{\sigma}$

**CaptainBlack** · June 4th 2011, 01:10 AM

Originally Posted by lindah

Hi CB,

Can I just confirm if I am allowed to equate the Variance as the second moment or would it have to be calculated as:

$var[X] = \theta^2$
$var[X]=E[X^2] - (E[X])^2$
$E[X^2] = var[X] +(E[X])^2$

Now equate $m_{2}=E[X^2]$ , NOT $m_{2}=var[X]$ as I did before?

Doing this I should get:
$m_{2}=\theta^2 + m_{1}^2$
and so rearranging will give me:
$\theta = \sqrt{m_{2} - m_{1}^2}$

Instead of
$\theta = \hat{\sigma}$

You start with a pair of simultaneous equations:

$\overline{X}=\mu(\theta)$

$\overline{X^2}=m_2(\theta)$

Then subtract the square of the first from the second to get the equations:

$\overline{X}=\mu(\theta)$

$\overline{X^2}-\overline{X}^2=m_2(\theta)-(\mu(\theta))^2$

The left hand side of the second equation above is the sample variance (that is with $n$ rather than $(n-1)$ in it) and the right hand side is the variance of the population.

So equating the raw first two moments is the same as equating the means and variances, but in this case you will find it easier to work with the population variance than with the second moment (the algebra comes out nicer)

CB

**lindah** · June 4th 2011, 03:00 AM

Thanks for the clarification!
I see what you mean now and is definitely easier to deal with the population variance

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