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]=
- unanswered
2. Using the answer to part 1, find E[R].
E[R]=
- unanswered
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]=
- unanswered
4. If k∈{1,2,…,n−1}, then
E[IkIk+1]=
- unanswered
5. If k≥1, ℓ≥2, and k+ℓ≤n, then
E[IkIk+ℓ]=
- unanswered
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
To find the expected value E[R] and variance var(R), we need to break down the problem into smaller sub-problems.
1. Find E[Ik]:
The reward Ik is given at time k if the toss at time k resulted in Tails and the toss at time k-1 resulted in Heads. This means that the only way to get a reward at time k is if the previous toss was a Head and the current toss is a Tail. So, the probability of getting a reward at time k is (1-p) * p. Therefore, E[Ik] = (1-p) * p.
2. Find E[R]:
To find E[R], we sum up the expected rewards for each time k from 1 to n:
E[R] = E[I1] + E[I2] + ... + E[In]
= (1-p) * p + (1-p) * p + ... + (1-p) * p (summing from 1 to n times)
= n * (1-p) * p
3. Find E[I2k]:
The reward I2k is given at time 2k if the toss at time 2k resulted in Tails and the toss at time 2k-1 resulted in Heads. This means that the only way to get a reward at time 2k is if the toss at time 2k-1 resulted in Heads, and the toss at time 2k resulted in Tails. So, the probability of getting a reward at time 2k is (1-p) * p. Therefore, E[I2k] = (1-p) * p.
4. Find E[IkIk+1]:
The reward IkIk+1 is given at time k if the toss at time k resulted in Tails and the toss at time k+1 resulted in Tails. Since the tosses are independent, the probability of getting a reward at time k is (1-p) * (1-p) = (1-p)^2. Therefore, E[IkIk+1] = (1-p)^2.
5. Find E[IkIk+ℓ]:
The reward IkIk+ℓ is given at time k if the toss at time k resulted in Tails and the toss at time k+ℓ resulted in Tails. Since the tosses are independent, the probability of getting a reward at time k is (1-p) * (1-p) = (1-p)^2. Therefore, E[IkIk+ℓ] = (1-p)^2.
6. Calculate var(R) when p = 3/4 and n = 10:
First, calculate E[R]:
E[R] = n * (1-p) * p
= 10 * (1 - 3/4) * 3/4
= 30/4 - 9/4
= 21/4
To calculate var(R), we need the following values:
- E[Ik]
- E[I2k]
- E[IkIk+1]
- E[IkIk+ℓ]
Using the formulas from above:
E[Ik] = (1-p) * p
= (1 - 3/4) * 3/4
= 3/16
E[I2k] = (1-p) * p
= (1 - 3/4) * 3/4
= 3/16
E[IkIk+1] = (1-p)^2
= (1 - 3/4)^2
= 1/16
E[IkIk+ℓ] = (1-p)^2
= (1 - 3/4)^2
= 1/16
Now, we can calculate var(R):
var(R) = n * (E[IkIk+1] - E[Ik]^2) + (n-1) * E[IkIk+ℓ] - 2 * (n-1) * E[Ik] * E[I2k]
= 10 * (1/16 - (3/16)^2) + 9 * (1/16) - 2 * 9 * (3/16) * (3/16)
= 10 * (1/16 - 9/256) + 9/16 - 18/256
= 160/256 - 81/256 + 9/16 - 18/256
= (160 - 81 + 144 - 18) / 256
= 205/256
Therefore, var(R) is equal to 205/256 when p = 3/4 and n = 10.