Hi guys,
Parameter Estimation Question
I tried using [IMG][/IMG] tags, but they would not work
Any help is greatly appreciated!
Hi guys,
Parameter Estimation Question
I tried using [IMG][/IMG] tags, but they would not work
Any help is greatly appreciated!
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.
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:
$\displaystyle var[X] = E[X^2] - \left( E[X] \right)^2 =\theta^2$
I equate:
$\displaystyle m_{1} = a + \theta$
$\displaystyle a = m_{1} - \theta$
Similarly for $\displaystyle m_{2}$ and then re-written in terms of $\displaystyle \theta$:
$\displaystyle m_{2} = \theta^2$
$\displaystyle \theta = \sqrt{m_{2}} = \sqrt{\sigma^2} = \sigma$
Substituting this into the expression for $\displaystyle a$ the estimates are as follows?:
$\displaystyle \hat{a} = \bar{X} - \sigma $
$\displaystyle \hat{\theta} = \sigma$
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:
$\displaystyle var[X] = \theta^2$
$\displaystyle var[X]=E[X^2] - (E[X])^2$
$\displaystyle E[X^2] = var[X] +(E[X])^2 $
Now equate $\displaystyle m_{2}=E[X^2]$, NOT $\displaystyle m_{2}=var[X]$ as I did before?
Doing this I should get:
$\displaystyle m_{2}=\theta^2 + m_{1}^2$
and so rearranging will give me:
$\displaystyle \theta = \sqrt{m_{2} - m_{1}^2}$
Instead of
$\displaystyle \theta = \hat{\sigma} $
You start with a pair of simultaneous equations:
$\displaystyle \overline{X}=\mu(\theta)$
$\displaystyle \overline{X^2}=m_2(\theta)$
Then subtract the square of the first from the second to get the equations:
$\displaystyle \overline{X}=\mu(\theta)$
$\displaystyle \overline{X^2}-\overline{X}^2=m_2(\theta)-(\mu(\theta))^2$
The left hand side of the second equation above is the sample variance (that is with $\displaystyle n$ rather than $\displaystyle (n-1)$ 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