This is a question from Exercises 8.3 from Introduction to Mathematical Statistics by Hogg, Craig, McKean.

Question: Let $X_1, X_2, \cdots ,X_n$ and $Y_1, Y_2, \cdots ,Y_n$ be independent random samples from two normal distributions $\mathcal{N}(μ_1, σ^2)$ and $\mathcal{N}(μ_2, σ^2)$, respectively, where $σ^2$ is the common but unknown variance. Find the likelihood ratio $\Lambda$ for testing $H_0 : μ_1 = μ_2 = 0$ against all alternatives. Rewrite $\Lambda$ so that it is a function of a statistic $Z$ which has a well-known distribution. Give the distribution of $Z$ under both null and alternative hypotheses.

I have viewed a similar question here. It had the case of known variance, and so I wasn't able to check my answer.

My attempt is as follows. I would really appreciate if someone could check it and give some advice.

Solution attempt: I started by writing the likelihood function. To do this, I use that the samples are independent, and that within each sample, the observations are iid. I write $\theta = (\mu_1, \mu_2, \sigma^2).$ We have $$L(\theta; \mathbf{x}, \mathbf{y}) = \dfrac{1}{(2\pi \sigma^2)^n} \exp \left({-\dfrac{1}{2 \sigma^2} (\sum (x_i - \mu_1)^2 + \sum(y_i - \mu_2)^2)}\right).$$

Under $H_0$, the MLEs are $\mu_1 = \mu_2 = 0$, $\hat{\sigma_0}^2 = \dfrac{1}{2n} \left(\sum x_i^2 + \sum y_i^2\right)$. I denote the MLE as $\theta_0$. This gives the likelihood as $$L(\hat{\theta_0})= \dfrac{1}{(2 \pi \sigma_0^2)^{n}} e^{-n}.$$

The unrestricted MLEs are $\hat{\mu_1} = \bar{x}, \hat{\mu_2} = \bar{y}$ and $\hat{\sigma_1^2} = \dfrac{1}{2n} (\sum (x_i-\bar{x})^2 - \sum(y_i-\bar{y})^2)$. I denote this MLE by $\hat{\theta}$. I evaluate the likelihood and obtain $$L(\hat{\theta})= \dfrac{1}{(2 \pi \sigma_1^2)^{n}} e^{-n}.$$

Thus $\Lambda = \left(\dfrac{\hat{\sigma_1}^2}{\hat{\sigma_0}^2}\right)^n.$ I then obtain $$\Lambda^{1/n} = \dfrac{ \sum(x_i - \bar{x})^2 + \sum(y_i - \bar{y})^2}{\sum x_i^2 + \sum y_i^2}.$$

I'm not really sure what to do now. I know that $\sum(x_i-\bar{x})^2 = \sum x_i^2 - n \bar{x}^2$ so I could write $$\Lambda^{1/n} = \dfrac{\sum x_i^2 - n \bar{x}^2 + \sum y_i^2 - n\bar{y}^2}{\sum x_i^2 + \sum y_i^2}.$$

At this point I am stuck as I do not know what statistic this is a function of. Could someone please help me? Thank you!


1 Answer 1


I am leaving a general sketch here.

$\bullet$ The likelihood function is

\begin{align}\mathcal L &=\frac{1}{(2\pi \sigma^2)^n} \exp \left[{-\frac{1}{2 \sigma^2} \left\{\sum_{i=1}^n (x_i - \mu_1)^2 + \sum_{j=1}^n(y_j- \mu_2)^2\right\}}\right]. \tag 1\label 1\end{align}

$\bullet$ From $\eqref 1,$ it is easy to check \begin{align}\hat \mu_1 &= \bar x, \\ \hat \mu_2 &= \bar y, \\ \hat \sigma^2 &= \frac1{2n}\left[\sum_{i=1}^n (x_i - \bar x)^2 + \sum_{j=1}^n(y_j- \bar y)^2\right] \\ &= \frac1{2n}\left[ns_1^2 +ns_2^2\right].\tag 2\label 2 \end{align}

$\bullet$ In $\Theta_0,$ $$\begin{align}\hat\mu &= \frac{n\bar x+ n\bar y }{2n}, \\\hat \sigma^2 &=\frac1{2n}\left[\sum_{i=1}^n (x_i - \hat \mu)^2 + \sum_{j=1}^n(y_j- \hat \mu)^2\right] \\ &= \frac{1}{2n}\left[ns_1^2+ ns_2^2 + \frac{n^2}{2n}(\bar x-\bar y)^2\right].\end{align}\tag 3 \label 3 $$

Therefore \begin{align}\Lambda = \frac{\mathcal L\left(\hat \Theta_0\right)}{\mathcal L\left(\hat \Theta\right)} &= \left[\frac{ns_1^2 +ns_2^2}{ns_1^2+ ns_2^2 + \frac{n^2}{2n}(\bar x-\bar y)^2}\right]^n;\tag 4\end{align}

defining \begin{align}S^2 &:= \frac1{2n-2}\left[ns_1^2 + ns_2^2\right]\\ t &:= \frac{\bar x-\bar y}{S\sqrt{\frac1n+\frac1n}}, \tag 5\label 5 \end{align} $\Lambda$ can be expressed as

$$\Lambda = \frac{1}{\left[1 + \frac{t^2}{2n-2}\right]^{n}}.\tag 6$$

$\bullet$ $t$ can be shown to follow $\mathsf t_{2n-2}.$ The LR test is then based on $t.$

$\bullet$ In general, if $\delta := \mu_1-\mu_2,$ then it is also an easy exercise to check the distribution of $\frac{(\bar x-\bar y) - \delta}{S\sqrt{\frac1n+\frac1n}}.$

  • $\begingroup$ Thank you! I have studied your post and managed to understand. But I think we are testing for $(\mu_1, \mu_2) = (0,0)$ rather than $\mu_1 = mu_2$ and that may make things a bit different? Could you let me know your view on this? Thanks again $\endgroup$
    – Balkys
    Commented Dec 21, 2022 at 21:35
  • 1
    $\begingroup$ I see. My bad. I would update this post holidays. $\endgroup$ Commented Dec 22, 2022 at 8:39

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