Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For one observation $Y$ from a normal distribution with variance $1$ and mean $0$ or $2$, consider $H_{0}:\mu=0$ and $H_{1}:\mu=2$. Suppose first that we observe only $Y$. Construct a size $\alpha$ likelihood ratio test. Give explicitly the rejection region in terms of $Y$, and find the power of this test.

--I found the LRT test statistic of $$\lambda(Y)=\frac{e^{-y^{2}/2}}{e^{-y^{2}/2}+e^{-(y-2)^{2}/2}},$$ and a rejection region of $\{Y:\lambda(Y)\geq c\}$, where $\sup_{H_{0}}P(\lambda(Y)\leq c)=\alpha$.

--I am finding the calculation a bit messy for finding the power. I would have to find $\beta(\mu)=P(Y\in R)$, where $R$ is the rejection region.

Any way I can clean this up?

share|cite|improve this question
Seems related to this previous question of yours. By the way, what happened to it? – Did Jul 19 '13 at 23:08
It's a completely different question, and I solved the other one. – Kirk Fogg Jul 19 '13 at 23:10
You're missing some minus signs. – Michael Hardy Jul 19 '13 at 23:11
Oh oops, forgot to type those in. I will edit this. – Kirk Fogg Jul 19 '13 at 23:11
Compare "I'm not sure where to even start for this problem. That is why I posted it. – Kirk Fogg Jul 13 at 19:48" and "deleted by Kirk Fogg Jul 13 at 19:52". This is what I call a steep learning curve... Anyway, the usual procedure in such a case is to post your own solution instead of deleting the question. After a while, you may even accept it. – Did Jul 19 '13 at 23:16
up vote 2 down vote accepted

$$ L(\mu) \propto e^{-(y-\mu)^2/2}. $$ The value of $\mu\in\{0,2\}$ that maximizes this is $$ \widehat\mu = \begin{cases} 0 & \text{if }y<1, \\ 2 & \text{if }y>1. \end{cases} $$ Thus the maximized value is $$ L(\widehat\mu) = \begin{cases} e^{-y^2/2} & \text{if }y<1, \\ e^{-(y-2)^2/2} & \text{if }y>1. \end{cases} $$ That's the denominator in your fraction; the numerator is $e^{-y^2/2}$. Hence the value of the fraction is $$ \begin{cases} 1 & \text{if }y<1, \\[10pt] \frac{e^{-y^2/2}}{e^{-(y-2)^2/2}} & \text{if }y>1. \end{cases} $$ The second case simplifies: $$ e^{-y^2/2 + (y-2)^2/2} = e^{(-4y+4)/2}. $$ So your test statistic is $$ \begin{cases} 1 & \text{if }y<1, \\[10pt] e^{(-4y+4)/2} & \text{if }y>1. \end{cases} $$ or any monotone function of that quantity.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.