# Maximum likelihood estimator

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• April 29th 2009, 04:17 PM
chella182
Maximum likelihood estimator
Similar to another thread I've posted, only that one's asking how to sketch the functions, this time I'm stumped on the calculations.

So the full Question (apart from part (i) because I've done that, and it doesn't relate) is...

A manufacturer of watches claims that the weekly error, in seconds, of their watches follow a $N(0,1)$ distribution. An inspection of the performance of 5 randomly chosen watches gave the following weekly errors $-0.68$, $0.25$, $0.29$, $-1.41$ and $1.59$.

(ii) Assume that the weekly errors follow a $N(0,\sigma^{2})$ distribution. Sketch the likelihood function and log-likelihood for $\sigma$ and calculate the most likely value of $\sigma$.

All I've got so far is
$f_{X}(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\{{-\frac{1}{2}(\frac{x}{\sigma})^{2}}\}$

But then anything could be wrong after that (Worried) I got this for the likelihood function:
$\sigma^{-n}(2\pi)^{-n/2}\exp\{-\frac{1}{2}\frac{\sum_{i=1}^{n}x_{i}^{2}}{\sigma^{ 2}}\}$

Which gave me the log-likelihood function of:
$K-n\ln{\sigma}-\frac{\sum_{i=1}^{n}x_{i}^{2}}{2\sigma^{2}}$
And now I'm stumped. I don't think what I've done is right at all. Can anyone help?
• April 29th 2009, 05:17 PM
matheagle
Rewrite as sigma squared and diff wrt sigma^2
Quote:

Originally Posted by chella182
(ii) Assume that the weekly errors follow a $N(0,\sigma^{2})$ distribution. Sketch the likelihood function and log-likelihood for $\sigma$ and calculate the most likely value of $\sigma$.

All I've got so far is
$f_{X}(x)=\frac{1}{\sigma\sqrt{2\pi}}\exp\{{-\frac{1}{2}(\frac{x}{\sigma})^{2}}\}$

But then anything could be wrong after that (Worried) I got this for the likelihood function:
$\sigma^{-n}(2\pi)^{-n/2}\exp\{-\frac{1}{2}\frac{\sum_{i=1}^{n}x_{i}^{2}}{\sigma^{ 2}}\}$

Which gave me the log-likelihood function of:
$K-{n\over 2}\ln{\sigma^2}-\frac{\sum_{i=1}^{n}x_{i}^{2}}{2\sigma^{2}}$

And now I'm stumped. I don't think what I've done is right at all. Can anyone help?

You should get ${\sum_{i=1}^{n}x_{i}^{2}\over n}$ as the MLE of sigma^2.
Hence the MLE of sigma is the square root of this.
• April 30th 2009, 03:11 AM
chella182
So is what I have right then? (Worried)
• April 30th 2009, 06:35 AM
matheagle
and what do you have now?
• April 30th 2009, 06:46 AM
chella182
I have what's in my original post, which I don't think is right. I don't like that $\sum^{n}_{i=1}x_{i}^{2}$ that I have in my log-likelihood function much... I'm gonna have another good look at it later.
• April 30th 2009, 06:50 AM
matheagle
I edited your post.
The MLE is what you have divided by n.
• April 30th 2009, 10:29 AM
chella182
I'm totally lost, sorry. I don't know what you mean.
• April 30th 2009, 04:56 PM
hametceq
please see attached.
EK(Hi)
• May 1st 2009, 01:04 AM
chella182
Thankyou. That's what I got. I think the $\sum_{i=1}^{n}x_{i}^{2}$ just threw me a bit.