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
8 answers
Can someone please answer?
These questions are from online courses, and you need to study and know the answers to get a credit.
3. second choice
q = k/sum(k, i=1) t_i
q = k/sum(k, i=1) t_i
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
2. part II: use the mathematical definition of expectation (integration) to solve for the estimator
Can anonymous be more specific (please!)
can anyone provide the answers??????
1) 1/(t*(t+1))
2) 2/(t+2)
2) 2/(t+2)
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).
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).
0 answers