Question

Conditional expectation.

Question 2 (10.0 marks) Previous 1 2 Validate Mark Unfocus Help ndependently or the Suppose the number o calls attempte per hour to a telephone exchange has Posso tribution with mean Suppose there is only 80% chance that an attempted call is connected a other calls connected. Let X be the number of calls that are connected in an hour. We ask you to find the mean and variance of X. In reality, this is mainly a theory question, and the following will guide you through it. Warning. Make sure you read the question! In particular, the 3rd part is entirely theoretical; no actual parameter values are involved. In the 4th part, you do use the parameter values 2.1 (1 mark) First, let N be the number of calls attempted per hour, to a telephone exchange. Then N Poisson(80) Your last answer, exactly as you typed it, was: Poisson(80) Your answer is correct. The mark for your last attempt was 1.00 You previously made one incorrect attempt, and were penalised 0.10, giving an overall mark of 0.90 2.2 (1 mark) Now define random variables X by | 1, ifthe ith call is connected j 0, otherwise. Xi- Then Xi-Bernoulli(0.8) Your last answer, exactly as you typed it, was Bernoulli(0.8) Your answer is correct. The mark for your last attempt was 1.00 You previously made 2 incorrect attempts, and were penalised 0.10 for each of them, giving an overall mark of 0.80. 2.3 (7 marks) Now using the following theorem from lectures, we wish you to find EX) and VarX) Theorem For random variables X and Y, (a) E(X) = E(E(X | Y)), and (b) Var(X)Var(E(XI Y)) + E(Var(XI Y)) 2.3.1 (1 mark) First find E X I N = n Do this theoretically, Le use p rather that the parameter value you entered in the last question part. Your last answer was: np Your answer is correct. The mark for your last attempt was 1.00 You previously made 6 incorrect attempts, and were penalised 0.10 for each of them, giving an overall mark of 0.40 2.3.2 (1 mark) E(X | N) = p Your last answer was: Your answer is not correct Your mark is 0.00. You have made 7 incorrect attempts 2.3.3 (1 mark) Writing X, which is entered as: lambda for the parameter of N, finally find E(X), i.e. as per part (a) of the theorem, find E(E(X | N)) = Your last answer was empty Your empty answer has been disregarded, with no penalty.The mark for your last valid attempt was 0.00 2.3.4 (1 mark) Now find Var(E(X IN)), using what you worked out for E(X I N) above, using A for the parameter of N. Var(E(X | N)) = You have not attempted this yet 2.3.5 (1 mark) Now find Var(X I N). Again, do this theoretically, i.e. use p for the relevant parameter, as before You have not attempted this yet 2.3.6 (1 mark) As before, use lambda for the parameter of N You have not attempted this yet 2.3.7 (1 mark) Still with parameters p and X, use part (b) of the theorem, to fin Var(X) Var(X)- You have not attempted this yet 2.4 (1 mark) Finally, using the actual values of the parameters given in the question what are the mean and variance of X? 2.4.1 (5 marks) E(X) You have not attempted this yet 2.4.2 (5 marks) Var(X)- You have not attempted this yet

Answer 1

x/n Y)-1

2. L-1

E NP (2.32) Var(E(xln)): Vn(Nr) : ry@): G.3.4 〉 V(x)N〉=N.FG-מ [.:x(N ~Bao..งเ»,り1 (2. 3.s)

"Van.(ヒ 2.3.7 Vm() = þ〉,0.gxso.= 64 〈2-4.2〉