1. LetNY =X,.i=1Q2(i.) Let øy (t) = E[etY] be the moment generating function of the random variable Y. Using conditioningargument (i.e., E[X] = E[E[X[Y]) show thatØy (t) = E[(¢x(t))^],where ox...

Extracted text: Q2 Let X1, X2, ·. be independent and identically distributed random variables. Let N be a non-negative, integer valued random variable that is independent of the sequence X;, i>1. Let N Y =X,. i=1 Q2(i.) Let øy (t) = E[etY] be the moment generating function of the random variable Y. Using conditioning argument (i.e., E[X] = E[E[X[Y]) show that Øy (t) = E[(¢x(t))^], where ox (t) is the m.g.f. of X. Q2 (ii.) Using the moment generating function from part (i) show that E[Y] = E[N]E[X] Q2 (iii.) Using the moment generating function from part (i) and the results from part (ii) show that Var[Y] = E[N]Var[X] + (E[X])²Var[N]. Q2 (iv.) If X1, X2, · . - are independent and identically distributed exponential random variables with parameter A and N is a geometric random variable with parameter p independent of the sequence X1, X2, ... (i.e P(N = n) = p(1 – p)"-1), then show that N Y = X; i=1 is exponentially distributed with parameter Ap using the moment generating function from part (i). Hint: The m.g.f. of an exponential random variable with parameter A is X/(A – t). Q(3.) (Continuation of Q2) If X1, X2, - .... are independent and identically distributed exponential random variables with parameter A and N is a geometric random variable with parameter p independent of the sequence X1, X2, ... (i.e P(N = n) = p(1 – p)"¯), then show that N Y = ) X; i=1 is exponentially distributed with parameter Xp using conditioning arguments directly with the random sum, instead of using the m.g.f. method. Hint: Write the C.D.F of the Gamma random variable in terms of the Poisson process.