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The normal distribution of MLE is start $({x_1\ldots x_n})$ random sample from $N(u,\sigma^2)$

This family of distributions has two parameters: $\theta = (\mu, \sigma^2)$, so we maximize the likelihood, over both parameters simultaneously using logarithm...

I know this. but what if the normal distribution's $({x_1\ldots x_n})$ random sample from $N(u,\sigma^2)$

That is the variance is heteroskedasticity...

My questions is that What is the MLE if variance change $\sigma^2$ to $\sigma^2_i$.

$\sigma^2$ - homoscedasticity,

$\sigma^2_i$- variance is heteroskedasticity.

So $f\left(x ; u,\sigma _i^2\right)=\frac{e^{\frac{-(x-u)^2}{2 \sigma _i^2}}}{\sqrt{2 \pi } \sigma _i}$ I don't know.... How can I solved...

In the homoscedasticity, u hat is xbar and sigma^2 hat is sigma^2/n.

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Can you clafiry the difference between $\mu$ and $u$, that you both used for mean. –  Caran-d'Ache Mar 22 '13 at 9:36
    
Shouldn't there be a negative sign in the exponent term? And what do you anderstand under $x$: a vector $(x1 \ldots x_n)$ or single sample $x_i$? –  Caran-d'Ache Mar 22 '13 at 9:42
    
yes... you are right. exponent term is negative sign...but i don't using latex... –  Woo Suk Lee Mar 22 '13 at 9:55
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