Question

need answer for q4,5,6.

MEe ORS O e GaR RRKR O 360 nvestme nctions I Additional theo a Course 14.310x I edX x et 6.431x Progress I edx 5. Indicator variables Bookmark this page Problem 5. Indicator variables 3/6 points (graded) Consider a sequence of n +1 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 E {1,2.....n), if the toss at time k resulted in Tails and the toss at time &- 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,, We will find ER and Var (R) by carrying out a sequence of steps. Express your answers below in terms of p and/or n using standard notation (available through the "STANDARD NOTATION" button below.) Remember to write " for all multiplications and to include parentheses where necessary. We first work towards finding ER. 1. Let I denote the reward (possibly 0) given at time k, for k e {1,2, ...,n}. Find E Il E[ p*(1-p) p (1-p) 2. Using the answer to part 1, find E R. ER p) n° n.p.(1-p) The variance calculation is more involved because the random variables I,1..I are not independent. We begin by computing the following values Oan o Ole G RE O60RP'lnvestme Functions Additional thec x a Course 14310x edx x 6.431x Progress edX EE p*(1-p) p-(1-p) 4. If k E (1,2,...,7 1}, then ,n- E[ 2 p*(1-p) 2 p (1-p) 5, If k>1,> 2, and k +lsn, then E p (1-p) x p (1-p) 6. Using the results above, calculate the numerical value of Var (R), assuming that p3/4, n-10. Var (R) 1.64 STANDARD NOTATION Submit You have used 3 of S attempts Save Partially correct (3/6 points) Hide Discussion Discussion Topic: Unt 4 Dscrete random variablesProblem Ser 4/5 Indcaor vanables Add a Post

Answer 1

4)

If k∈{1,2,…,n−1}, then

E[$I_kI_k_+_1$] = 0

5)

If k ≥ 1, ℓ ≥ 2, and k+ℓ ≤ n, then

E[
k+1
] = $p^2(1-p)^2$

6)

Using the results above, calculate the numerical value of var(R) assuming that p=3/4, n=10.

Var(R) = E[R}^2 - E[R^2] = 57/64 = 0.8906

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