
Maximum likelihood
An individual taken from a very large biological population is of type A with probability p=0.5(1+t) and of type B with probabiliy 1p=0.5(1t).
a) Suppose that X denotes the number of type A individuals in a random sample of size n. What is the probability that X = x?
My answer: $\displaystyle 0.5^x*(1+t)^x*(1t)^{(nx)}$ Classic binomial situation.
b)Find the maximum likelihood estimator of t and show that it is unbiased.
My answer: Find log likelihood, differentiate yields
2xn(1+t)/[(1+t)(1t)]
setting to zero yields t=(2xn)/n
To show unbiased, I need to find E(t). This is an integral including $\displaystyle 0.5^x*(1+t)^x*(1t)^{(nx)}$ which I have no idea how to integrate.
Thanks for your help.

Re: Maximum likelihood
im not sure your liklihood is correct.
The likelihood is proportional to $\displaystyle 0.5^x(1+t)^x \times 0.5^{nx}(1t)^{nx} = 0.5^n(1+t)^x (1t)^{nx}$
this shouldn't affect your final answer as long as you did your algebra correctly (the constant term at the front vanishes anyway).
To show your final answer is unbiased, evaluate:
$\displaystyle E(\frac{2xn}{n}) = \frac{2E(x)  n}{n} = \frac{2E(x)}{n}  1$
You know X is binomial and so E(X) = 0.5n(1+t).
Im assuming that you are allowed to treat the moments of a binomial distribution as standard results instead of having to derive them every time
$\displaystyle \frac{2 \times 0.5n(1+t)}{n}  1$
$\displaystyle =1+t1$
$\displaystyle =t$

Re: Maximum likelihood
I forgot I could distribute the expectation. Thanks