We have k coins. The probability of Heads is the same for each coin and is the realized value q of a random variable Q that is uniformly distributed on [0,1]. We assume that conditioned on Q=q, all coin tosses are independent. Let Ti be the number of tosses of the ith coin until that coin results in Heads for the first time, for i=1,2,…,k. (Ti includes the toss that results in the first Heads.)

You may find the following integral useful: For any non-negative integers k and m,

∫10qk(1−q)mdq=k!m!(k+m+1)!.
Find the PMF of T1. (Express your answer in terms of t using standard notation.)

For t=1,…, pT1(t)=- unanswered
Find the least mean squares (LMS) estimate of Q based on the observed value, t, of T1. (Express your answer in terms of t using standard notation.)

E[Q∣T1=t]=- unanswered
We flip each of the k coins until they result in Heads for the first time. Compute the maximum a posteriori (MAP) estimate q^ of Q given the number of tosses needed, T1=t1,…,Tk=tk, for each coin.

The question is not in correct order ; so the first thing we are required to do is to put them together in the right form to make it easier to proof; having said that. let's get started!.

From the second part " You may find the following integral useful: For any non-negative integers k and m,"

The next equation goes thus : [tex]\int\limits^1_0 \ \ q^k (1-q)^m dq = \frac{k!m!}{(k+m+1)!}[/tex]

a. Find the PMF of [tex]T_1[/tex] . (Express your answer in terms of t using standard notation.)

For t=1,2…,

[tex]p_T_1[/tex](t)= [tex]\frac{1}{(t*(t*1))}[/tex] by using conditional probability to solve PMF

b) Find the least mean squares (LMS) estimate of Q based on the observed value, t, of T1. (Express your answer in terms of t using standard notation.)

By using the mathematical definition of integration to solve the estimate of Q:

[tex]E|Q|T_1=t|= \frac{2}{t+2}[/tex]

c) We flip each of the k coins until they result in Heads for the first time. Compute the maximum a posteriori (MAP) estimate q^ of Q given the number of tosses needed, T1=t1,…,Tk=tk, for each coin.

The MAP estimate of q is :

[tex]\bar q = \frac{k}{\sum^k_{i=1}}t_i[/tex]