# Thread: How did they get this?

1. ## How did they get this?

My textbook makes a calculation and just says "with some simplications", and then gets the answer.

We have L0=(1/sqrt(2pi))^n*(e^(-1/2*sum(xi-m0)^2)...i.e. this the max likelihood function for a normal with variance 1 and mean m0. Sum is from 1 to n. We have L1 is the same except replace m0 with m1.

Then they compute L0/L1 and get

e^((n/2)*(m1^2-m0^2)+(m0-m1)*sum(xi))

I do not see where this step comes from. Any ideas?

Thank you.

2. Originally Posted by zhupolongjoe
My textbook makes a calculation and just says "with some simplications", and then gets the answer.

We have L0=(1/sqrt(2pi))^n*(e^(-1/2*sum(xi-m0)^2)...i.e. this the max likelihood function for a normal with variance 1 and mean m0. Sum is from 1 to n. We have L1 is the same except replace m0 with m1.

Then they compute L0/L1 and get

e^((n/2)*(m1^2-m0^2)+(m0-m1)*sum(xi))

I do not see where this step comes from. Any ideas?

Thank you.
This is tedious to type out, but I literally have it sitting in my notes...

What part are you stuck on?

3. Just how they get that form of Lo/L1...I understand the analysis that follows using the Neyman Pearson lemma, but I just don't understand this, and maybe it's an algebraic thing I'm missing, but I don't see how they are getting the e^((n/2)*(m1^2-m0^2)+(m0-m1)*sum(xi)) from the two likelihood functions.....Maybe there is some property of sums I am missing, but I am not sure.

4. $\frac{f_0}{f_1} = \frac{\exp{[-\frac{1}{2\sigma^2}\sum^n (X_i - \mu_0)^2]}}{\exp{[-\frac{1}{2\sigma^2}\sum^n (X_i - \mu_1)^2]}}$ Apply the property $\frac{e^a}{e^b} = e^{a-b}$ and cancel.

$= \exp\{-\frac{1}{2\sigma^2}\sum^n[(X_i - \mu_0)^2 - (X_i - \mu_1)^2)]\}$ Now expand this.

$= \exp{[2\bar X(\mu_0 - \mu_1) + \mu_0^2 - \mu_1^2]}$

Let me know if you have any specific questions.