1. ## MLEs

Does anyone know how to find maximum likelihood estimates? I have a sample with 15 values but am not understanding exactly what I am supposed to be looking for with the maximum likelihood estimates. Thanks for any potential help. I haven't come across an explanation that would help me tackle this problem.

2. Originally Posted by jerryc
Does anyone know how to find maximum likelihood estimates? I have a sample with 15 values but am not understanding exactly what I am supposed to be looking for with the maximum likelihood estimates. Thanks for any potential help. I haven't come across an explanation that would help me tackle this problem.
to Find MLE.. We assume a distribution first and for estimation of parameters we use MLE
Maximum likelihood - Wikipedia, the free encyclopedia

3. First you need the likelihood function, which in the continuous case id the product of the densities via independence.
In the discrete case you take the product of the distribution function.
Then, as it's name says, you maximize wrt the parameter.
In many cases that's just calculus.
But in a case where the parameter is a boundary in your support, then calculus fails, but common sense prevails.

4. I am to find the MLE estimates for $\theta_{1}=\mu$and $\theta_{2}=\sigma^{2}$ if a random sample of 15 from $N(\mu,\sigma^{2})$ has the values 31.5, 36.9, 33.8, 30.1, 33.9, 35.2, 29.6, 34.4, 30.5, 34.2, 31.6, 36.7, 35.8, 34.5, 32.7

All I am coming up with is, for the mean,
$f(x_{k},\mu)=1/\sqrt{2\pi\sigma^{2}}*e^{-\frac{x_{k}-\mu)^{2}}{2\sigma^{2}}}$

$L=f(x_{1},\mu)$... $f(x_{n},\mu)=(2\pi\sigma^{2})^{-\frac{n}{2}}e^{-\Sigma\frac{(x_{k}-\mu)^{2}}{2\sigma^{2}}}$

$lnL=-\frac{n}{2};n(2\pi\sigma^{2})-\frac{1}{2\sigma^{2}}\Sigma(x_{k}-\mu)^{2}$

Partil derivative= $\frac{1}{\sigma^{2}}\Sigma(x_{k}-\mu)$

0= $\frac{1}{\sigma^{2}}\Sigma(x_{k}-\mu)$

$\mu=\frac{\Sigma x_{k}}{n}$

And similar equations for the variance leading me to $\sigma^{2}=\frac{\Sigma(x_{k}-\mu)^{2}}{n}$

The mean calculated from the sample is 33.4267 and the variance is 5.0979, but all these equations are straight out of a book from the library, so I'm not sure if they had to be tweaked or are standard. Also, I could find what the k stood for. It wasn't explained anywhere near that section. Thanks

5. The MLE of $\mu$ is $\bar X$

and the MLE of $\sigma^2$ is ${\sum_{k=1}^n (X_k-\bar X)^2\over n}$.

6. It is really that simple? The MLE for the mean IS the mean, which is 33.4267, and the MLE for the variance IS the variance, which is 5.0979? I though I needed a whole slew of equations. Thanks

7. Originally Posted by jerryc
It is really that simple? The MLE for the mean IS the mean, which is 33.4267, and the MLE for the variance IS the variance, which is 5.0979? I though I needed a whole slew of equations. Thanks
Not exactly... The MLE for the mean is the empirical mean... the MLE for the variance is the empirical variance

8. so...
empirical mean $=\frac{\Sigma X_{i}}{15}=\frac{501.4}{15}=33.4267$

empirical variance $=\frac{\Sigma(X_{i}-EmpiricalMean)^{2}}{15}=\frac{76.4673}{15}=5.0979$?

9. Originally Posted by jerryc
so...
empirical mean $=\frac{\Sigma X_{i}}{15}=\frac{501.4}{15}=33.4267$

empirical variance $=\frac{\Sigma(X_{i}-EmpiricalMean)^{2}}{15}=\frac{76.4673}{15}=5.0979$?
I don't have your data, but I guess you're correct

I just wanted to underline that the MLEs are the EMPIRICAL mean and variance, mean and variance issued from the data

10. Originally Posted by jerryc
I am to find the MLE estimates for $\theta_{1}=\mu$and $\theta_{2}=\sigma^{2}$ if a random sample of 15 from $N(\mu,\sigma^{2})$ has the values 31.5, 36.9, 33.8, 30.1, 33.9, 35.2, 29.6, 34.4, 30.5, 34.2, 31.6, 36.7, 35.8, 34.5, 32.7
My data is above. It seems to be the same formula as finding the sample mean and variance. That's why I wasn't sure. All the answers seem to be the same.