1. ## 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 1-p=0.5(1-t).

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: $0.5^x*(1+t)^x*(1-t)^{(n-x)}$ Classic binomial situation.

b)Find the maximum likelihood estimator of t and show that it is unbiased.

My answer: Find log likelihood, differentiate yields
2x-n(1+t)/[(1+t)(1-t)]

setting to zero yields t=(2x-n)/n

To show unbiased, I need to find E(t). This is an integral including $0.5^x*(1+t)^x*(1-t)^{(n-x)}$ which I have no idea how to integrate.

2. ## Re: Maximum likelihood

im not sure your liklihood is correct.

The likelihood is proportional to $0.5^x(1+t)^x \times 0.5^{n-x}(1-t)^{n-x} = 0.5^n(1+t)^x (1-t)^{n-x}$

this shouldn't affect your final answer as long as you did your algebra correctly (the constant term at the front vanishes anyway).

$E(\frac{2x-n}{n}) = \frac{2E(x) - n}{n} = \frac{2E(x)}{n} - 1$

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

$\frac{2 \times 0.5n(1+t)}{n} - 1$

$=1+t-1$
$=t$

3. ## Re: Maximum likelihood

I forgot I could distribute the expectation. Thanks