We toss a biased coin n times. The probability of heads, denoted by y, is the...

  

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We toss a biased coin n times. The probability of heads, denoted by y, is the value of a random variable Y with a given mean and variance o2. Let Xį be a Bernoulli random variable that models the outcome of the i-th toss (i.e., X = 1 if the i-th toss is a head). In other words, for each 1 ≤ i ≤n, Xi = ={₁ 0 where Y€ [0, 1] is a random variable with [ 1 w.p. Y w.p. 1-Y : EY] =, and Var(Y) = ². We assume that X₁, X2..... Xn are conditionally independent given the event {Y = y} for each y € [0, 1]. let Sn = X₁ + X₂+...+ Xn be the total number of heads in the n tosses. (a) (5 points) Use the law of iterated expectations to find E[Xi] and E[Sn.]. (b) (3 points) Using the fact that X? = X₁, show that Var(X₂) = μ - ². (c) (5 points) Using the law of iterated expectations, find Cov(X₁, Xj) = E[X₁X₁] - E[X]E[X], for ij. Are X, and X, independent? (d) (7 points) By writing Var (S₂) = E[S2]- (E[S])2, show that Var (S₂) = E[Var (SnlY)] + Var(E[Sn|Y]). (1) where Var (S,Y) is the random variable that takes on the value Var(S, Y = y) with probability Pr(Y = y). (e) (5 points) Calculate the variance of Sn by using the formula (1) in part (d) above. We toss a biased coin n times. The probability of heads, denoted by y, is the value of a random variable Y with a given mean and variance o2. Let Xį be a Bernoulli random variable that models the outcome of the i-th toss (i.e., X = 1 if the i-th toss is a head). In other words, for each 1 ≤ i ≤n, Xi = ={₁ 0 where Y€ [0, 1] is a random variable with [ 1 w.p. Y w.p. 1-Y : EY] =, and Var(Y) = ². We assume that X₁, X2..... Xn are conditionally independent given the event {Y = y} for each y € [0, 1]. let Sn = X₁ + X₂+...+ Xn be the total number of heads in the n tosses. (a) (5 points) Use the law of iterated expectations to find E[Xi] and E[Sn.]. (b) (3 points) Using the fact that X? = X₁, show that Var(X₂) = μ - ². (c) (5 points) Using the law of iterated expectations, find Cov(X₁, Xj) = E[X₁X₁] - E[X]E[X], for ij. Are X, and X, independent? (d) (7 points) By writing Var (S₂) = E[S2]- (E[S])2, show that Var (S₂) = E[Var (SnlY)] + Var(E[Sn|Y]). (1) where Var (S,Y) is the random variable that takes on the value Var(S, Y = y) with probability Pr(Y = y). (e) (5 points) Calculate the variance of Sn by using the formula (1) in part (d) above.