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Let $(X_{1},...,X_{n})$ a $n$ sample of the law $N( \mu, \sigma^{2})$. We assumed we don't know $\mu$ and $\sigma^{2}$.

Let $\mu_{0} \in \mathbb{R}$.

Show that the Likelihood-ratio test for $\mu = \mu_{0}$ against $\mu \ne \mu_{0}$ is function of $$ 1 + \frac{(\bar{X_{n}} - \mu_{0} )^{2}}{\sigma_{n}^{2}} $$ with $\sigma_{n}^{2} = \sum_{i=1}^{n} \frac{(X_{i} - \bar{X_{n}})^{2}}{n}$

EDIT : I showed that the Likelihood-ratio test is
$$ \exp\left( \frac{n}{2 s_{n}^{2}(X)} (\bar{X_{n}} - \mu_{0} ) \right) $$ with $s_{n}^{2}(X) = \sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n}$ But I don't know how to conclude.

Thanks and regards.

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    $\begingroup$ I think the equation for the variance is wrong. What does $>$ do there? $\endgroup$ Feb 19, 2020 at 11:35
  • $\begingroup$ Oh the > is just to make the text highlighted. Let me edit. Oh and yes thank you it was a copy error. $\endgroup$
    – user347910
    Feb 19, 2020 at 11:38
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    $\begingroup$ also the value of the test is $exp(-\frac{n}{2 \sigma ^2}(\bar{X}-\mu_0)^2)$ $\endgroup$ Feb 19, 2020 at 11:40
  • $\begingroup$ I found this computing $ \frac{\sup_{H_{1}}}{\sup_{H_{0}}}$. Oh okay. But it doesn't change my question. $\endgroup$
    – user347910
    Feb 19, 2020 at 11:42
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    $\begingroup$ It is $ \frac{\sup_{H_0}}{\sup_{H_1}}$ $\endgroup$ Feb 19, 2020 at 11:44

1 Answer 1

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$L_{H_0} = (2\pi\sigma^2)^{-\frac{n}{2}}exp(-\frac{1}{2\sigma^2}\sum_{i=1}^n(X_i-\mu_0)^2)$

$supL_{H_1} = (2\pi\sigma^2)^{-\frac{n}{2}}exp(-\frac{1}{2\sigma^2}\sum_{i=1}^n(X_i-\bar{X})^2)$ ($\bar{X}$ is the maximum likelihood estimate for $\mu$ if $\mu$ is unknown).

The problem is that you are treating $\sigma^2$ as given parameter instead of unknown. If you maximize the likelihood wrt $\sigma^2$ for ${H_0}$ you will obtain that the maximum likelihood estimate for $\sigma^2$ is $\sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n}$ (analogously for ${H_1}$ it is $\sum_{i=1}^n\frac{(X_i-\bar{X})^2}{n}$).

After plugging in the estimate for $\sigma^2$ you will obtain:

$\sup L_{H_0} = (2\pi\sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n})^{-\frac{n}{2}}exp\Bigg(-\frac{\sum_{i=1}^n(X_i-\mu_0)^2}{2{\sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n}}}\Bigg)=\big(\frac{2\pi}{n}\sum_{i=1}^n(X_i-\mu_0)^2)^{-\frac{n}{2}}exp(-\frac{n}{2})$

and $\sup L_{H_1} = (2\pi\sum_{i=1}^n\frac{(X_i-\bar{X})^2}{n})^{-\frac{n}{2}}exp\Bigg(-\frac{\sum_{i=1}^n(X_i-\bar{X}^2)}{2{\sum_{i=1}^n\frac{(X_i-\bar{X})^2}{n}}}\Bigg) =\big(\frac{2\pi}{n}\sum_{i=1}^n(X_i-\bar{X})^2)^{-\frac{n}{2}}exp(-\frac{n}{2})$.

I.e. $\Lambda = \Bigg(\frac{\sum_{i=1}^n(X_i-\mu_0)^2}{\sum_{i=1}^n(X_i-\bar{X})^2}\Bigg)^{-\frac{n}{2}}$.

$\sum_{i=1}^n(X_i-\mu_0)^2 = \sum_{i=1}^n((X_i-\bar{X})+(\bar{X}-\mu_0))^2 = \sum_{i=1}^n(X_i-\bar{X})^2+n(\bar{X}-\mu_0)^2$.

Plugging back into the equation for $\Lambda$ we obtain: $\Lambda = \Bigg(1+\frac{n(\bar{X}-\mu_0)^2}{\sum_{i=1}^n(X_i-\bar{X})^2}\Bigg)^{-\frac{n}{2}}$.

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  • $\begingroup$ why $exp(-\frac{n}{2})$ ? I though of $\exp\left( \frac{-\| X-\mu_{0} \|)^{2}}{2(\sum_{i=1}^n\frac{(X_i-\mu_0)^2}{n})} \right)$ for $H_{0}$ $\endgroup$
    – user347910
    Feb 19, 2020 at 12:43
  • $\begingroup$ @CechMS I hav edited my post to make it clearer. $\endgroup$ Feb 19, 2020 at 12:51
  • $\begingroup$ @ I think $L_{H_{0}}$ still need a update. $\endgroup$
    – user347910
    Feb 19, 2020 at 12:52
  • $\begingroup$ My course just tell me that $\frac{\sup_{\Theta_{1}} L(\theta)}{\sup_{\Theta_{0}}L(\theta)}$. So I just tried to apply the formula for $\Theta_{0} = \{ \mu_{0} \}$. But it doesn't really match with the set of parameters $ \mathbb{R} \times ]0;+\infty[$. So if I try now to consider $\Theta_{0} = \{ \mu_{0} \} \times ]0;+\infty[$ and $\Theta_{1} = \mathbb{R}- \{ \mu_{0} \} \times ]0;+\infty[$. It could be more coherent. $\endgroup$
    – user347910
    Feb 19, 2020 at 12:54
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    $\begingroup$ @CechMS The density of the normal distributions is a smooth, infinitely differentiable function over the entire $\mathbb{R \times R_+}$, I took the first order conditions for both parameters. This is perfectly legal in this case. $\endgroup$ Feb 19, 2020 at 14:34

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