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
1) 1/(t*(t+1))
2) 2/(t+2)
3. second choice
q = k/sum(k, i=1) t_i
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).
Can someone please answer?
Can anonymous be more specific (please!)
can anyone provide the answers??????
To find the PMF of T1, we need to determine the probability mass function for the random variable T1, which represents the number of tosses of the first coin until it results in heads for the first time.
Given that the probability of Heads for each coin is the realized value q of a random variable Q that is uniformly distributed on [0, 1], we can calculate the PMF of T1 as follows:
P(T1 = t) = P(first coin results in heads on the t-th toss)
= P(Q > (t-1)/t) [since the t-th toss must result in heads]
= ∫[(t-1)/t, 1] q^0(1-q)^(t-1) dq [using the definition of the uniform distribution]
= [(t-1)!0!((t-1)+1)!] / [(t+1)!] [using the given integral formula]
Simplifying the expression, we get:
P(T1 = t) = 1 / (t(t+1)) for t = 1, 2, ...
Therefore, the PMF of T1 is given by:
pT1(t) = 1 / (t(t+1)) for t = 1, 2, ...
Moving on to the LMS estimate of Q based on the observed value t of T1, we want to find the expected value of Q given T1 = t.
E[Q | T1 = t] represents the conditional expectation of Q given T1 = t. To calculate this, we can use the probability density function (PDF) of Q, which is a uniform distribution on [0, 1].
Using the properties of conditional expectation, we have:
E[Q | T1 = t] = ∫0^1 q fQ|T1=q(t) dq
= ∫0^1 q fQ(t) dq
Since fQ(t) is the conditional PDF of Q given T1 = t, it is equal to 1 if 0 ≤ q ≤ 1, and 0 otherwise.
Therefore, integrating over the range [0, 1], we get:
E[Q | T1 = t] = ∫0^1 q dq
= 1/2
Hence, E[Q | T1 = t] = 1/2 for any value of t.
Finally, to compute the MAP estimate q^ of Q given the number of tosses needed for each coin (T1 = t1, ..., Tk = tk), we choose the correct expression. In this case, the correct expression is:
q^ = k/(t1 + t2 + ... + tk)
Therefore, the MAP estimate of Q is given by q^ = k/(t1 + t2 + ... + tk).
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