Download Math 472 Homework Assignment 5 Problem 5.1.1. Let {a n} be a

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript
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.