Download 21-325 (Fall 2008): Homework 9 (ONE side) Due by

21-325 (Fall 2008): Homework 9 (ONE side)
Due by December 3, in class
Show FULL JUSTIFICATION for all your answers.
1. Let X1 , X2 , · · · be a sequence of independent and identically distributed random variables with
mean µ and variance σ 2 . Let X n =
1
n
Pn
i=1 Xi .
Show that E[(X n − µ)2 ] converges to 0 as n goes
to ∞.
Solution. E[(X n − µ)2 ] = E[( n1
Pn
independent, E[(
i=1 (Xi
Pn
i=1 (Xi
− µ))2 ] =
Pn
− µ))2 ] =
i=1 E[(Xi
P
1
E[( ni=1 (Xi
n2
− µ))2 ]. Since {Xi − µ}ni=1 are
− µ)2 ] = nσ 2 . Then E[(X n − µ)2 ] =
σ2
n
→ 0 as
n → ∞.
2. Let X1 , X2 , · · · be a sequence of independent and identically distributed random variables with
uniform distribution on [0, 1]. Let Yn = min{X1 , X2 , · · · , Xn } and Zn = max{X1 , X2 , · · · , Xn }.
Show that Yn converges to 0 in probability and Zn converges to 1 in probability.
Solution. Let ∈ (0, 1). P (Yn ≥ ) = P (Xi ≥ , i = 1, ..., n) = P (X1 ≥ )n = (1 − )n → 0
as n → ∞. So Yn converges in probability to 0. On the other hand, P (|Zn − 1| ≥ ) = P (Zn ≤
1 − ) = P (∩ni=1 (Xi ≤ 1 − )) = P ((X1 ≤ 1 − ))n = (1 − )n → 0 as n → ∞. So Zn converges to 1
in probability.
3. Suppose that the number of units produced daily at factory A is a random variable with mean
20 and variance 9. and the number produced at factory B is a random variable with mean 18 and
variance 36. Assuming independence, derive a (non-trivial) upper bound for the probability that
more units are produced today at factory B than at factory A.
Solution. Let X be the number of units produced daily at factory A and Y be the number of units
produced daily at factory B. X and Y are independent. We want to estimate P (Y > X). We first
try to use Markov inequality directly.
P (Y > X) = P (Y −X > 0) = P (Y −X ≥ 1) = P (Y −X −(−2) ≥ 1−(−2)) ≤ P ((Y −X +2)2 ≥ 9),
where −2 is the mean of Y − X. Then we have by the Markov inequality, P (Y > X) ≤ 91 V ar(Y −
X) =
V ar(Y )+V ar(X)
9
=
45
9
= 5. But 5 is not the non-trivial upper bound. We need to modify the
argument. Let a > 0. P (Y > X) = P (Y −X −(−2) ≥ 1−(−2)) ≤ P ((Y −X +2+a)2 ≥ (3+a)2 ) ≤
1
E[(Y
(3+a)2
− X + 2 + a)2 ] =
1
(V
(3+a)2
ar(Y − X) + a2 ) =
1
45+a2
.
(3+a)2
Since P (Y > X) ≤
45+a2
,
(3+a)2
we
then try to minimize
45+a2
.
(3+a)2
Simple Calculus yields that
45+a2
(3+a)2
is minimized at a = 15. Plug it in
to get P (Y > X) ≤ 56 , which is non-trivial.
4. The servicing of a machine requires two steps, with the time needed for the first step being an exponential random variable with mean 0.2 hour and the time for the second step being an independent
exponential random variable with mean 0.3 hour. If a repair person has 20 machines to service, approximate the probability that all the work can be completed in 8 hours. (You may use the table for
the standard normal distribution on the web at http://www.sjsu.edu/faculty/gerstman/EpiInfo/ztable.htm)
Solution. Let Xi and Yi be the time needed for the first step and the second step for the ith
machine. Then Xi is an exponential random variable with mean 0.2 and Yi is an exponential
random variable with mean 0.3. Then P (
0.26. So P (
P20
i=1 (Xi
+ Yi ) ≤ 8). Xi + Yi has mean 0.5 and variance
P20
P20
+ Yi ) ≤ 8) can be approximated by P (
i=1 (Xi
P20
P20
correction. Then P (
i=1 (Xi
i=1 (Xi
(Xi +Yi )−10
i=1
√
20∗0.26
+ Yi ) < 8.5) = P
≤
+ Yi ) < 8.5) by the continuity
√8.5−10
20∗0.26
−1.5
= Φ( √
).
5.2
5. Suppose that a fair die is rolled 100 times. Let Xi be the value obtained on the ith roll. Compute
an approximation for
P
100
Y
!
100
Xi ≤ a
,
1 < a < 6.
i=1
Solution. P
Q
100
i=1 Xi
≤ a100
= P
P
100
i=1 log(Xi )
≤ 100 log(a) . Let E[log(X1 )] =
1
6 (log(1)
+
log(2) + log(3) + log(4) + log(5) + log(6)). So
P
100
X
!
log(Xi ) ≤ 100 log(a)
P100
!
− 100E(log(X1 ))
100 ∗ log(a) − 100 ∗ E(log(X1 ))
p
p
≤
,
10 V ar(log(X1 ))
10 V ar(log(X1 ))
i=1 log(Xi )
=P
i=1
1 ))
√
which can be estimated by Φ( 100∗log(a)−100∗E(log(X
).
10
V ar(log(X1 ))
6. Let f (x) be a continuous function defined for 0 ≤ x ≤ 1. Consider the functions
n
X
!
n k
Bn (x) =
f (k/n)
x (1 − x)n−k
k
k=0
and prove that limn→∞ Bn (x) = f (x).
Solution. Let Yn be an Binomial random variable with parameters n and x. Then Bn (x) =
E[f (Yn /n)]. Since Yn has the same distribution as X1 + X2 + ... + Xn , where Xi is a Bernoulli
random variable with parameter x, i = 1, 2, ..., n. Then by the strong law of large numbers, we
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have (X1 + X2 + ... + Xn )/n → x as n → ∞. Then Bn (x) = E[f ((X1 + X2 + ... + Xn )/n)] → f (x)
as n → ∞.
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