Asked by A

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. Choose the correct expression for q^.


q^=k−1∑ki=1tiq^=k∑ki=1tiq^=k+1∑ki=1tinone of the above

Answers

Answered by Anonymous
Can someone please answer?
Answered by sist pist
These questions are from online courses, and you need to study and know the answers to get a credit.
Answered by RVE
3. second choice

q = k/sum(k, i=1) t_i

Answered by Anonymous
1. part I: use conditional probability to solve the PMF

2. part II: use the mathematical definition of expectation (integration) to solve for the estimator
Answered by mary
Can anonymous be more specific (please!)
Answered by mary
can anyone provide the answers??????
Answered by Anonymous
1) 1/(t*(t+1))
2) 2/(t+2)
Answered by knobby
to explain previous Anonymous answer:

1. based on the "useful integral" in the question you can substitute in 1 for "alpha" (since there is only 1 head in the sequence of t tosses), and "beta" is t - 1 (since there were t tosses in total, one of which was a head, so there were t - 1 tails) i.e. a simple geometric distribution. This gives (t - 1)!/(t + 1)!

Splitting out some of the terms:
(t - 1)! / (t - 1)! * t * (t + 1)
cancelling out the (t - 1)! terms gives the answer

2. The LMS can be shown to be just the Estimated Value, so k + 1 / t + 2 (where k is the number of successes, and t is the number of tosses). k = 1 (again geometric distribution). So 2 / (t + 2).
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