consider a sequence of independent tosses of a biased coin at times k=0,1,2,…,n. On each toss, the probability of Heads is p, and the probability of Tails is 1−p.
A reward of one unit is given at time k, for k∈{1,2,…,n}, if the toss at time k resulted in Tails and the toss at time k−1 resulted in Heads. Otherwise, no reward is given at time k.
Let R be the sum of the rewards collected at times 1,2,…,n.
We will find E[R] and var(R) by carrying out a sequence of steps. Express your answers below in terms of p and/or n using standard notation. Remember to write '*' for all multiplications and to include parentheses where necessary.
We first work towards finding E[R].
1. Let Ik denote the reward (possibly 0) given at time k, for k∈{1,2,…,n}. Find E[Ik].
E[Ik]=
2. Using the answer to part 1, find E[R].
E[R]=
The variance calculation is more involved because the random variables I1,I2,…,In are not independent. We begin by computing the following values.
3. If k∈{1,2,…,n}, then
E[I2k]=
4. If k∈{1,2,…,n−1}, then
E[IkIk+1]=
5. If k≥1, ℓ≥2, and k+ℓ≤n, then
E[IkIk+ℓ]=
6. Using the results above, calculate the numerical value of var(R) assuming that p=3/4, n=10.
var(R)=
1. p*(1-p)
2. n*p*(1-p)
3. p*(1-p)
4. 0
5. p^2*(1-p)^2
6. 57/64
Oh sorry, misunderstood.
1. Let Ik denote the reward (possibly 0) given at time k, for k∈{1,2,…,n}. Find E[Ik].
E[Ik] = P(Tails at time k and Heads at time k-1) = p(1-p)
2. Using the answer to part 1, find E[R].
E[R] = E[I1] + E[I2] + ... + E[In] = (p(1-p)) + (p(1-p)) + ... + (p(1-p)) = np(1-p)
The variance calculation is more involved because the random variables I1, I2, ..., In are not independent. We begin by computing the following values.
3. If k∈{1,2,…,n}, then
E[I2k] = P(Tails at time 2k and Heads at time 2k-1) = p(1-p)
4. If k∈{1,2,…,n−1}, then
E[IkIk+1] = P(Tails at time k and Heads at time k-1, and Tails at time k+1 and Heads at time k) = p(1-p) * p(1-p) = p^2(1-p)^2
5. If k≥1, ℓ≥2, and k+ℓ≤n, then
E[IkIk+ℓ] = P(Tails at time k and Heads at time k-1, and Tails at time k+ℓ and Heads at time k+ℓ-1) = p(1-p) * p(1-p) = p^2(1-p)^2
Now, we can calculate the numerical value of var(R) assuming that p=3/4, n=10.
var(R) = E[I1^2] + E[I2^2] + ... + E[In^2] - (E[I1] + E[I2] + ... + E[In])^2
= (p^2(1-p)^2 + p^2(1-p)^2 + ... + p^2(1-p)^2) - (np(1-p))^2
= (10)(3/4)^2(1-3/4)^2 - (10)(3/4)(1-3/4))^2
= (10)(9/16)(1/4) - (10)(3/4)(1/16)
= 90/64 - 30/64
= 60/64
= 15/16
Therefore, var(R) = 15/16.
1. Let Ik denote the reward (possibly 0) given at time k, for k∈{1,2,…,n}. Find E[Ik].
E[Ik] is the expected value of the reward at time k. In order to receive a reward at time k, the toss at time k must be Tails and the toss at time k-1 must be Heads. The probability of this happening is (1-p) * p, as the probability of Tails at time k is (1-p) and the probability of Heads at time k-1 is p. Therefore,
E[Ik] = (1-p) * p
2. Using the answer to part 1, find E[R].
E[R] is the expected sum of rewards collected at times 1,2,…,n. Since each reward Ik is independent, we can calculate the expected value of the sum by summing the expected values of each reward:
E[R] = E[I1] + E[I2] + ... + E[In]
= (1-p) * p + (1-p) * p + ... + (1-p) * p (n times)
= n * (1-p) * p
3. If k∈{1,2,…,n}, then
E[I2k] is the expected value of the reward at time 2k. In order to receive a reward at time 2k, the toss at time 2k must be Tails and the toss at time 2k-1 must be Heads. The probability of this happening is (1-p) * p, as before. Therefore,
E[I2k] = (1-p) * p
4. If k∈{1,2,…,n−1}, then
E[IkIk+1] is the expected value of the product of the rewards at times k and k+1. In order to receive a reward at time k and k+1, the toss at time k must be Tails, the toss at time k-1 must be Heads, the toss at time k+1 must be Tails, and the toss at time k must be Heads. The probability of this happening is (1-p) * p * (1-p) * p, as the probability of Tails at times k and k+1 is (1-p) and the probability of Heads at times k-1 and k is p. Therefore,
E[IkIk+1] = (1-p) * p * (1-p) * p
5. If k≥1, ℓ≥2, and k+ℓ≤n, then
E[IkIk+ℓ] is the expected value of the product of the rewards at times k and k+ℓ. In order to receive a reward at time k and k+ℓ, the tosses at times k, k-1, k+ℓ, and k+ℓ-1 must follow the pattern Tails, Heads, Tails, Heads. The probability of this happening is (1-p) * p * (1-p) * p, as before. Therefore,
E[IkIk+ℓ] = (1-p) * p * (1-p) * p
6. Using the results above, calculate the numerical value of var(R) assuming that p=3/4, n=10.
To calculate var(R), we need to calculate the variance of each individual reward, and then sum up the variances. The variance of a reward Ik is defined as:
var(Ik) = E[Ik^2] - (E[Ik])^2
Since each reward Ik is 0 or 1, we have Ik^2 = Ik for all k. Therefore,
var(Ik) = E[Ik] - (E[Ik])^2
Using the formulas from above, we can calculate the variances:
var(Ik) = (1-p) * p - ((1-p) * p)^2 for k=1,2,...,n
Finally, we can calculate var(R) by summing up the variances:
var(R) = var(I1) + var(I2) + ... + var(In)
= (1-p) * p - ((1-p) * p)^2 + (1-p) * p - ((1-p) * p)^2 + ... + (1-p) * p - ((1-p) * p)^2 (n times)
= n * [(1-p) * p - ((1-p) * p)^2]
1. To find E[Ik], we need to consider the probabilities of the outcomes that result in a reward at time k. In order for Ik to be 1 (indicating a reward), the toss at time k must result in Tails and the toss at time k-1 must result in Heads.
P(Ik = 1) = P(Tails at time k and Heads at time k-1) = (1 - p) * p
Since Ik can only be 0 or 1, we have the following probability distribution for Ik:
P(Ik = 0) = P(Tails at time k and Tails at time k-1) + P(Heads at time k) = (1 - p) * (1 - p) + p
Therefore, E[Ik] is the expected value of this probability distribution:
E[Ik] = 0 * P(Ik = 0) + 1 * P(Ik = 1) = p - p^2.
2. To find E[R], we can sum up the expected rewards at each time k:
E[R] = E[I1] + E[I2] + ... + E[In]
Since E[Ik] is the same for all k, we can express this as:
E[R] = n * E[Ik] = n * (p - p^2)
3. To find E[I2k], we look at the probability that a reward is given at time 2k. This happens when the toss at time 2k results in Tails and the toss at time 2k-1 results in Heads.
P(I2k = 1) = P(Tails at time 2k and Heads at time 2k-1) = (1 - p) * p
Therefore, E[I2k] is the expected value of this probability:
E[I2k] = p - p^2.
4. To find E[IkIk+1], we need to consider the probability that a reward is given at both time k and time k+1. This happens when the toss at time k results in Tails and the toss at time k-1 results in Heads, and the toss at time k+1 also results in Tails and the toss at time k results in Heads.
P(IkIk+1 = 1) = P(Tails at time k and Heads at time k-1 and Tails at time k+1 and Heads at time k) = (1 - p) * p * (1 - p) * p
Therefore, E[IkIk+1] is the expected value of this probability:
E[IkIk+1] = p^2 - p^3.
5. To find E[IkIk+ℓ] for k ≥ 1, ℓ ≥ 2, and k + ℓ ≤ n, we need to consider the probability that a reward is given at both time k and time k+ℓ. This happens when the toss at time k results in Tails and the toss at time k-1 results in Heads, and for each intermediate time j, the toss at time j also results in Tails and the toss at time j-1 results in Heads, and the toss at time k+ℓ also results in Tails and the toss at time k+ℓ-1 results in Heads.
P(IkIk+ℓ = 1) = P(Tails at time k and Heads at time k-1 and Tails at time k+ℓ and Heads at time k+ℓ-1) * (product of P(Tails at time j and Heads at time j-1) for each j from k+1 to k+ℓ-1)
= (1 - p) * p * (product of (1 - p) * p for each j from k+1 to k+ℓ-1)
= (1 - p) * p * ((1 - p) * p)^(ℓ-1)
Therefore, E[IkIk+ℓ] is the expected value of this probability:
E[IkIk+ℓ] = (1 - p) * p * ((1 - p) * p)^(ℓ-1).
6. To calculate var(R) with p = 3/4 and n = 10, we need to substitute these values back into the formulas for variance:
var(R) = n * (p - p^2) - (n * (p - p^2))^2 + (n - 1) * (p^2 - p^3) - 2 * (n - 1) * (p - p^2) * (p^2 - p^3) + summation (n - 1) * (p - p^2) * (1 - p) * p * ((1 - p) * p)^(ℓ-1) for each ℓ from 2 to n.
Substituting p = 3/4 and n = 10 into these formulas will give the numerical value of var(R).