Download STAT 724/ECO 761 Spring 2007 Problem Set Four Solutions

STAT 724/ECO 761
Problem Set Four Solutions
Spring 2007
1. Consider the symmetric random walk on the integers: S0 = 0 and if
Sn = i then the probability is p = 1/2 that Sn+1 = i + 1 and the probability
is q = 1/2 that Sn+1 = i − 1.
We showed in class that this symmetric random walk will eventually return to 0 with probability 1. Compute the expected time it takes to return
to 0, and show that it is infinity. I.e. the symmetric random walk, starting
at 0, will return to 0 with probability 1, but it will take infinitely long to do
so, on average!
(Hint: Use the theorem we proved in class that says
F0 (s) = 1 − (1 − 4pqs2 )1/2
P
n
where F0 (s) = ∞
n=1 f0 (n)s is the generating function of the sequence defined by f0 (n) = P (S1 6= 0, . . . , Sn−1 6= 0, Sn = 0).)
Solution: The quantity f0 (n) is the probability that the walk returns to
0 on the nth step, for P
the first time. Thus the expected lenth of time to
0
return to 0 is given by ∞
n=1 nf0 (n). But this is exactly F0 (1) since we can
differentiate a power series term by term with its radius of convergence. Since
we have a formula for F0 (s) we can easily compute its derivative. The result
1
s
is F00 (s) = − (1 − 4pqs2 )−1/2 (−8pqs) = √
since p = 1/2 and q = 1/2.
2
1 − s2
s
goes to infinity as s goes to 1.
Now, notice √
1 − s2
2. The probability density function of a certain continuous random variable
X is given by
(
c(4x − 2x2 ) if 0 < x < 2,
f (x) =
0
otherwise.
a) what is the value of c?
Solution:
Z 2
11
8
c(4x − 2x2 ) dx = c[2x2 − 2/3x3 ]20 = c(8 −
) = c
= 1
3
3
0
1
3
so c = .
8
b) Compute P (1/2 < X < 3/2).
Solution: Compute
Z 3/2
3
23
(4x − 2x2 ) dx = .
32
1/2 8
3. a) Calculate E[X], the expected value, for the random variable X of the
preceding problem.
Solution:
Z
Z
3 2
3 2 2
2
E(X) =
x(4x − 2x ) dx =
4x − 2x3 dx =
8 0
8 0
3 4 3 1 4 2 3 32
x − x 0 = ( − 8) = 1
8 3
2
8 3
b) Calculate V ar(X) , the variance of X.
Solution:
Z
3 2 3
3
2 2
V ar(X) =
4x − 2x4 dx − 1 = x4 − x5 0 − 1 =
8 0
8
5
3
64
3 16
1
(16 − ) − 1 = ( ) − 1 =
8
5
8 5
5
4. Finish the calculation we started in class to show that if X has a normal
distribution N (0, 1), i.e. if the density function of X looks like:
1
2
√ e−x /2 dx
2π
then the mean of X is 0 and the variance of X is 1.
2
(HINT: to show E(X) = 0 use the fact that xe−x is an odd function,
2
and to show that V ar(X) = 1, use integration by parts to integrate x2 e−x .)
1 R ∞ 2 −x2 /2
Solution: We need to compute √
xe
dx. We use integration
2π −∞
2
by parts with u = x and dv = xe−x /2 dx. Note that with this choice of dv
2
we get v = −e−x
1
√
2π
2 /2
. We get
√ Z ∞
√
Z ∞
2
2
2
2
2
−x2 /2 ∞
2 −x /2
xe
dx = √
−e−x /2 dx
|0 −
xe
dx = √ −xe
π 0
π
−∞
0
√ Z ∞
√ Z ∞
√
2
21
1
2
2
=√
e−x /2 dx = √
e−x /2 dx = √
2π = 1.
π 0
π 2 −∞
2π
Z
∞
−x2 /2
5. Suppose that X1 , . . . , X20 are independent random variables that all have
density function f (x) = 2x, 0 ≤ x ≤ 1. Let S = X1 + · · · + X20 . Use the
central limit theorem to approximate P (S ≤ 10).
Solution:
Z 1
2
2
2x2 dx = x3 |10 =
E(Xi ) =
3
3
0
Z 1
x4
1
E(Xi2 ) =
2x3 dx = |10 =
2
2
0
9
18
1
1 4
−
=
V ar(X) = − =
2 9
18 18
18
√
1
so E(S) = 20 · 32 = 40
; V ar(X) = 20 · 18
= 10
; s.d.(S) = 310 . Also,
3
9
S = X1 + X2 + · · · + X20 is approximately normal by the Central Limit
Theorem. We have
p
S − p40/3
10
−
40/3 √
√
≤
= P (z ≤ −3.16) ∼
P (S ≤ 10) ∼
=P
= 0.
10
3
10
3
6. Suppose that you have a chance to bet $5 on a fair game, meaning your
chances are 1/2 of losing your $5 bet and 1/2 of getting your bet back plus $5
of winnings. Suppose you play this 50 times. Use the central limit theorem
to estimate the probability that you will lose more than $75.
Solution: Let xi be the r.v. which is the winnings on the ith play of the
game. So xi = 5 with probability 1/2 and xi = −5 with probability 1/2.
Clearly E(Xi ) = 0 and Var(X
p i ) = 25. Let S = X1 + . . . X50 . Then E(S) = 0
and Var(S) = 1250, and Var(S) = 35.3553. We get
P (S < −75) = P (
S
−75
<
) = P (Z < −2.1213) = .0169.
35.3553
35.3553
3
7. In this problem we will reinforce the idea that the drift rate µ is tied
in with the probability of an up or down movement in a binomial
√ model.
Consider the random walk which goes up or down by an amount σ ∆t with
probability p or 1 − p, respectively, in the time interval ∆t, where p is given
by
√ µ ∆t
p = 1/2 1 +
.
σ
Show that as ∆t → 0 the resulting limiting process is a Brownian motion
with drift rate µ and variance σ 2 .
√ Solution: Start with a random walk
√ that goes up by an amount ∆x =
σ ∆t with probability p = 1/2(1 + µ ∆t/σ) and down by the same amount
with probability 1 − p. Let
(
+1 if it goes up on ith step
Xi =
−1 if it goes down on ith step
so Xt = ∆x(X1 + X2 + . . . X[t/∆t] ). Now,
√
µ ∆t
E(Xi ) = (+1)p + (−1)(1 − p) = 2p − 1 =
σ
√
µ2 ∆t
µ ∆t 2
Var(Xi ) = E(Xi2 ) − E(Xi )2 = 1 −
=1− 2
σ
σ
√
√
√ t µ ∆t
t µ ∆t
E(Xt ) = ∆x
= σ ∆t
= µt
∆t
σ
∆t
σ
t
µ2 ∆t t
µ2 ∆t Var(Xt ) = (∆x)2
1 − 2 = σ 2 ∆t
1 − 2 → σ 2 t.
∆t
σ
∆t
σ
By the Central Limit Theorem, Xt is approximately normal, and as ∆t goes
to zero, Xt approaches a normal distribution. By the above calculation we
have
Xt = µt + σBt
where Bt stands for standard Brownian motion.
8. Use Ito’s Lemmma to show that
Z T
Z T
1 3
2
Zt dZt = ZT −
Zt dt
3
0
0
4
Here Zt is a Weiner process.
Solution: Let S = Z so that dS = dZ. Let g(S) = g(Z) =
Z3
. Then
3
∂g
∂g
∂2g
Z3
= 0,
= Z 2 , and
) = Zdt + Z 2 dZ
=
2Z.
By
Ito’s
Lemma,
d(
∂t
∂Z
∂Z 2
3
RT
Z3 R T
so
= 0 Z dt + 0 Z 2 dZ, whence the conclusion.
3
5