1. ## Maximum Likelihood.

Hello, I have a couple of exercise questions which I attempted and failed to complete, so the help I could get from you guys would be greatly appreciated!

1. It is known that the probability p of a head on a biased coin is either 1/4 or 3/4. The coin is tossed twice and a value for the number of heads Y is observed.

(a) For each possible value of Y, which of the two possible values for p maximizes the probability that Y = y.

(b) Depending on the value y actually observed, what is the maximum likelihood estimate of p (ie. 1/4 or 3/4)?

---------------------

2. Suppose that $\hat{\theta}$ is the maximum likelihood estimator of $\theta$. Let $t(\theta)$ be a function of $\theta$ that possesses a unique inverse. (That is, if $\beta = t(\theta)$ then $\theta = t^{-1}(\beta))$ . Show that $t(\hat{\theta})$ is the maximum likelihood estimator of $t(\theta)$

Thank you!

2. ## Just maximize likelihood over two possible values of p

As far as there are two trials and y is observed number of heads likelihood function looks as follows:
L(p)=C^y_2*p^y*(1-p)^(2-y)
To obtain estimate based on likelihood maximize L() over two possible values for p=1/4 or 3/4. For
y=0 you have L(p)=(1-p)^2 which is maximized by p=1/4. For y=1 we have L(p)=2*p*(1-p) is to be maximized
by both p=1/4 and p=3/4. Analogously for y=2 the p=3/4 would be maximum likelihood estimate.

Second is invariance which is one of basic properties of maximum likelihood estimates. If L(teta) is likelihood function
then L'=L( t^(-1) (beta) ) is likelihood function for beta. As far as teta' is maximum of L the beta'=t(teta') is maximum of L'.

As a consequence of invariance you have that ML estimate is generally biased.