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Math 472 Homework Assignment 5 Problem 5.1.1. Let {an } be a sequence of real numbers. Hence, we can also say that {an } is a sequence of constant (degenerate) random variables. P → a. Let a be a real number. Show that an → a is equivalent to an − Solution 5.1.1. Recall that an → a means that for each > 0 there is a positive integer N such that n > N implies |an − a| < or, equivalently, |an − a| ≥ implies n ≤ N . Said another way, for each > 0 only finitely many terms of the sequence {an } satisfy the condition |an − a| ≥ . Let > 0 be given. Then for each positive integer n, P (|an − a| ≥ ) = 1 if and only if |an − a| ≥ and P (|an − a| ≥ ) = 0 if and only if |an − a| < . P By definition, an − → a if and only if limn→∞ P (|an − a| ≥ ) = 0. Since the only possible values of P (|an − a| ≥ ) are 0 and 1, the limit is 0 if and only if at most finitely many of these probabilities are equal to 1, and this is equivalent to the condition that only finitely many terms of the sequence {an } satisfy |an − a| ≥ . By definition, this means that an → a. Thus, P an − → a is equivalent to an → a. Problem 5.1.2. Let the random variable Yn have a distribution that is b(n, p). (a) Prove that Yn /n converges in probability to p. This result is one form of the weak law of large numbers. (b) Prove that 1 − Yn /n converges in probability to 1 − p. (c) Prove that (Yn /n)(1 − Yn /n) converges in probability to p(1 − p). Solution 5.1.2. (a) Let X1 , . . . , Xn be iid random variables where the common distribution is a Bernoulli distribution with parameter p. We know that the expected value of the Bernoulli distribution is p and the variance of a Bernoulli distribution is p(1 − p), which is finite. Therefore, by the weak law of large P P numbers, X n − → p. Since ni=1 Xi has a b(n, p) distribution, which is the same distribution as Yn , we see that Yn /n has the same distribution as P P X n = ni=1 Xi /n. Therefore Yn /n − → p. P (b) Let g(x) = 1−x. Since g is a continuous function and since Yn /n − → p, P Theorem 5.1.4 shows that 1 − Yn /n = g(Yn /n) − → g(p) = 1 − p. P P (c) Since Yn /n − → p and 1 − Yn /n − → 1 − p, Theorem 5.1.5 shows that P (Yn /n)(1 − Yn /n) − → p(1 − p). Problem 5.1.3. Let Wn denote a random variable with mean µ and variance b/np , where p > 0, µ, and b are constants (not functions of n). Prove that Wn converges in probability to µ. Hint: Use Chebyshev’s inequality. 1 2 Solution 5.1.3. Let > 0 be given. By Chebyshev’s inequality P (|Wn − µ| ≥ ) = P (|Wn − E[Wn ]| ≥ ) 1 b ≤ 2 Var [Wn ] = 2 p . n 2 p Since p > 0, limn→∞ b/( n ) = 0 and therefore limn→∞ P (|Wn − µ| ≥ ) = P → µ. 0, which shows that Wn − Problem 5.1.5. Let X1 , . . . , Xn be iid random variables with common pdf ( e−(x−θ) x > θ, −∞ < θ < ∞ f (x) = 0 elsewhere. This pdf is called the shifted exponential. Let Yn = min{X1 , . . . , Xn }. Prove that Yn → θ in probability, by first obtaining the cdf of Yn . Solution 5.1.5. Let X by a random variable whose R ∞pdf is the above shifted exponential. Then for any a > θ, P (X ≥ a) = a e−(x−θ) dx = e−(a−θ) . From this we see that P (Yn ≥ a) = P (X1 ≥ a, . . . , Xn ≥ a) = P (X1 ≥ a) · · · P (Xn ≥ a) in h = e−(a−θ) = e−n(a−θ) . Let > 0 be given. Then since the support of Yn is the interval (θ, ∞) we know Yn > θ and P (|Yn − θ| ≥ ) = P (Yn − θ ≥ ) = P (Yn ≥ + θ) = e−n . Therefore limn→∞ P (|Yn − θ| ≥ ) = limn→∞ e−n = 0, which shows that P Yn − → θ. Problem 5.1.6. Using the assumptions behind the confidence interval given in expression (4.2.9), show that s ,s S12 S22 σ12 σ22 P + + − → 1. n1 n2 n1 n2 Solution 5.1.6. Example 5.1.1 on page 292 shows that the sample variance S 2 converges in probability to the variance σ 2 . Applying this result to each of the random variables X and Y from page 217 shows that S12 converges in probability to σ12 and S22 converges in probability to σ22 . Applying Theorems 5.1.2 and 5.1.3 we conclude that S12 S22 P σ12 σ22 (1) + − → + . n1 n2 n1 n2 3 Finally, applying Theorem 5.1.4 to (1), with g(x) = proves s ,s 2 2 S1 S σ12 σ22 P + 2 + − → 1. n1 n2 n1 n2 √ p x/ σ12 /n1 + σ22 /n2 , Problem 5.1.7. For Exercise 5.1.5, obtain the mean of Yn . Is Yn an unbiased estimator of θ. Obtain an unbiased estimator of θ based on Yn . Solution 5.1.7. From problem 5.1.5 we see that Fn (y) = P (Yn ≤ y) = 1 − e−n(y−θ) for y > θ, zero elsewhere. A little calculation shows that Z ∞ Z ∞ 1 0 ny e−n(y−θ) dy = θ + . y Fn (y) dy = E[Yn ] = n θ θ From this we see that Yn is a biased estimator for θ, but Yn − 1/n is an unbiased estimator for θ since E[Yn − 1/n] = θ + 1/n − 1/n = θ.