I have got the following likelihood:
$$l(p) = C + xlog(p) + (n-x)log(1-p)$$
I have got that $\theta_0 = 1/3$ and $\theta_1 = 1/2$
All I need to do is find the correct value of the log likelihood in its simplest terms. the problem is that I am coming to realize really fast that my algebra is pretty bad! Here is what i have got to so far:
$$l(1/2) = C + xlog(1/2) + (n-x)log(1/2)$$ $$l(1/3) = C + xlog(1/3) + (n-x)log(2/3)$$ $$log(LR) = [C + xlog(1/2) + (n-x)log(1/2)] - [C + xlog(1/3) + (n-x)log(2/3)]$$ $$log(LR) = C + xlog(1/2) + (n-x)log(1/2) - C - xlog(1/3) - (n-x)log(2/3)$$ $$log(LR) = xlog(1/2) + (n-x)log(1/2) - xlog(1/3) - (n-x)log(2/3)$$ $$log(LR) = x(log(1/2)- log(1/3)) + (n-x)(log(1/2) - log(2/3))$$ $$log(LR) = x(log(3/2)) + (n-x)(log(3/4))$$ $$log(LR) = x(log(3)-log(2)) + (n-x)(log(3)-log(4))$$
This expression can be simplified further as pert my text book multiple choice of answers but my algebraic brain has had a total meltdown! Please help!
(Copy of comment):
Maybe they want you to write it as $$x \log \frac{3}{2} - x \log \frac{3}{4} + n \log \frac{3}{4} = x \log 2 + n \log \frac{3}{4}$$