# Likelihood ratio for normal distribution

Let $X_1, \dots , X_n \sim \mathrm{N}(\mu,\sigma^2)$. $\sigma^2$ is known. We want to test $\mathrm{H}_0: \mu = 0$ versus $\mathrm{H}_1: \mu > 0$.

For the likelihood ratio I got: $\Lambda_1 = \exp(\frac{n(\mu_0^2 - \mu_1^2)}{2 \sigma^2}) \cdot \exp(\frac{\mu_1 - \mu_0}{\sigma^2} \cdot \sum x_i)$. Where the first term is a constant. Hope this is correct.

Now we know that for expected value of normal randowm sample $T(X_{1:n} = \sum X_i)$ is a sufficient statistic. I have to rewrite $\Lambda_1$ as a function of $T$, which gives me $\Lambda_2$. Can I now just exchange $\sum x_i$ in $\Lambda_1$ with $T$?

Another question is what I can say about the rejection region of $\Lambda_1$ and $\Lambda_2$, keeping in mind that $\mu$ is zero or bigger. I do not know what is meant here...

For your first question: Yes, you can substitute sufficient statistic as it is just a re-writing of $\sum x_i$
• Thanks for the hints! Is it true, that $\Lambda_1$ and $\Lambda_2$ are basically the same and have the same rejection regions? I would say so, because the substitution above did not change anything, right? – Michael Nov 5 '13 at 13:59
• Not exactly...$\Lambda_1$ is in $\mathbb{R}^N$ while $\Lambda_2$ is in $\mathbb{R}$...which is one of the nice features of a sufficient statistic...dimension reduction. The two are related in that if we let $R_T$ be the rejection region for T, then the rejection region for $\Lambda_1 = \{x_i: \sum x_i \in R_T\}$ I hope this clarifies it...the two problems are in different dimensions (literally), but are equivalent in the sense I just pointed out. – user31668 Nov 5 '13 at 14:29