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ECE 541 Probability and Stochastic Processes; Fall 2008 Assignment 6; Due date: Tuesday, Oct. 30, 2008
- Suppose that you start with an initial fortune of x 0 dollars, and you bet one dollar each time. The probability of winning each hand is p. You quit if you reach your goal of L dollars or when you go broke. Let Q(x 0 ) denote that probability that you eventually win. We need to characterize this probability. Next we give a more mathematical description of the problem.
Let Ω =^4 X∞ i=1Ωi, where for each i, Ωi = {H, T }. Let Yi = I{ωi=H} be a { 0 , 1 }-valued random variable,
and define Y on X
∞ i=1Ωi^ as^ Y^
= (^4 Y
1 , Y 2 ,.. .). Finally, we define^ P^ on Ω cylinders with measurable bases (as done in the notes) by forming the product of the probabilities of the coordinates while enforcing independence. For each (Ωi, Fi), define Pi{H} = p and q = 1 − p. Now define your fortune at time n ≥ 1 by
Xn = Xn− 1 + I{Yn=1} − I{Yn=0}
with the initial condition X 0 = x 0 ≥ 0. Let L ∈ IN be given, and define T 4 = inf{k : Xk = 0 or Xk = L}. Finally, let Q(x 0 ) =^4 P{XT = L}.
a) Use conditional expectations to prove that
Q(x) = pQ(x + 1) + (1 − p)Q(x − 1), x = 1,... , L − 1 , (1)
with Q(0) = 0 and Q(L) = 1.
Let q = 1 − p and define Y (^) n′ = Yn+1. Note that P{XT = L|X 1 } = Q′(X 1 ), where Q′(x) = P{X T′ ′ = L} (with X′(0) = x), and T ′^ and X n′ are defined similarly to T and Xn, respectively, but with the Yns replaced with Y (^) n′s. Note that Q and Q′^ are identically distributed. Thus, Q(x) = P{XT = L} = E[Q′(X 1 )] = Q(x + 1)p + Q(x − 1)q.
b) Solve the difference equation in (1) analytically.
The Characteristic equation of the above difference equation is pr^2 − r + q = 0. The roots are r 1 = 1 and r 2 = q/p. Then, Q(x) = cr 1 x + drx 2. Case (1): p 6 = 1/ 2 Using the boundary conditions Q(0) = 0 and Q(L) = 1, we get c = − (^) (q/p^1 )L− 1 , d = (^) (q/p^1 )L− 1 , and Q(x) = (q/p)
x− 1 (q/p)L− 1
Case (2): p = 1/ 2
Here, r 1 = r 2 = 1, and Q(x) = xL.
c) let P (x 0 ) =^4 P{XT = 0}. Determine P (x) directly and show that P (x) + Q(x) = 1.
Let P (x) = P{XT = 0}. Similarly to the case of Q(x), we derive P (x) = P (x+1)p+P (x−1)q, with boundary conditions P (0) = 1 and P (L) = 0. After solving for P (x), we discover that P (x) = 1 − Q(x).
d) Use the above results to describe in which sense Xn is convergent and characterize its limit X.
In class we will show that T < ∞ a.s. Let X = LIA, where A is any event with P(A) = Q(x). Since 0 and L are absorbing boundaries for the sequence Xn, {T < ∞}∩({XT = L}∪{XT = 0 }) = {Xn = L, for some finite n} ∪ {Xn = 0, for some finite n}. Since T < ∞ a.s., P({Xn = L, for some finite n} ∪ {Xn = 0, for some finite n}) = 1. Hence, Xn converges almost surely to X.
- Consider the following sequence:
Bn = ρBn− 1 + W (n), n > 0 , 0 < |ρ| < 1 ,
where the initial state B 0 is a random variable uniformly distributed in [0, 32]. Assume that W (1), W (2),... , are iid, and let the constants μW and σ W^2 denote the mean and variance of the sequence W (n), respec- tively. Finally, let μB (n) =^4 E[Bn] and σ B^2 (n) = var(^4 Bn).
a) Calculate μB (0) and σ^2 B (0)
Suppose that B 0 is uniformly distributed in the interval [0, 32] (you may assume other dis- tributions as well). Hence, μB (0) =^4 μB 0 = 322 − 0 = 16 and σ^2 B (0) =^4 σ^2 B 0 = (32−0)
2 12 = 85.^3.
b) Find an expression for μB (n) that explicitly shows the dependence on μB (0) and μW. Show that μB (n) is a constant if we select μW = (1 − ρ)μB (0). What is the constant mean μB?
μB (n) = E[B(n)] = E[ρBn− 1 + W (n)] =... = E[ρnB(0) +
∑n i=1 ρ (n−i)W (i)] = ρnE[B(0)] + ∑n i=1 ρ(n−i)E[W^ (i)] =^ ρnμB 0 +^ 1 −ρn 1 −ρ μW^.^ If^ μW^ = (1^ −^ ρ)μB^0 then^ μB^ (n) =^ ρ nμB 0 + (1 − ρn)μB 0 = μB 0.
c) Find an expression for σ B^2 (n), for the case when μW = (1 − ρ)μB (0). Show that σ^2 B (n) is a constant if we select σ W^2 = (1 − ρ^2 )σ B^2 (0). What is the constant variance σ^2 B.
σ^2 B (n) = var(B(n)) = E[B^2 (n)] − E[B(n)]^2 = E[(ρBn− 1 + W (n))^2 ] − μ^2 B (n) = E[(ρnB(0) +
∑n i=1 ρ (n−i)W (i)) (^2) ] − μ 2 B (n) = E[(ρnB(0))^2 + ( ∑n i=1 ρ(n−i)W^ (i))^2 + 2ρnB(0)^
∑n i=1 ρ(n−i)W^ (i)]^ −^ μ^2 B (n) = ρ^2 nE[B^2 (0)]+
∑n i=1 ρ (2n− 2 i)E[W 2 (i)]+[(∑n i=1 ρ (n−i)) (^2) −∑n i=1 ρ (2n− 2 i)]E[W (i)] (^2) +2ρnE[B(0)] ∑n i=1 ρ (n−i)E[W ( μ^2 B = ρ^2 n(σ^2 B 0 + μ^2 B 0 ) + 1 −ρ
2 n 1 −ρ^2 (σ
2 W + (μW^ ) (^2) ) + [( 1 −ρn 1 −ρ )
(^2) − 1 −ρ^2 n 1 −ρ^2 ](μW^ )
(^2) + 2ρnμB 0
1 −ρn 1 −ρ μW^ −^ μ
2 B = ρ^2 n(σ^2 B 0 + μ^2 B 0 ) + 1 −ρ 2 n 1 −ρ^2 σ
2 W + (^
1 −ρn 1 −ρ )
(^2) (μW ) (^2) + 2ρnμB 0
1 −ρn 1 −ρ μW^ −^ μ
2 B. Since μW = (1 − ρ)μB 0 σ^2 B (n) = ρ^2 n(σ B^20 + μ^2 B 0 ) + 1 −ρ 2 n 1 −ρ^2 σ
2 W + (1^ −^ ρ n) (^2) μ 2 B 0 + 2ρ n(1 − ρn)μ 2 B 0 −^ μ 2 B 0 = ρ^2 nσ^2 B 0 + 1 −ρ 2 n 1 −ρ^2 σ
2 W Now if we select σ W^2 = (1 − ρ^2 )σ B^20 , we obtain σ^2 B (n) = σ^2 B 0.
d) Use Matlab to generate 100 realizations, each of length 1000, of the process Bn using the conditions μW = (1 − ρ)μB (0) and σ^2 W = (1 − ρ^2 )σ^2 B (0). Find the sample mean for the 1000-point realizations by averaging the 100 realizations pointwise (i.e., for each n) and plot five of the 1000-point realizations as functions of n. Assume that ρ = 0.8.
11/4/08 11:41 PM C:\Documents and Settings\hayat\My Documents\My St...\Program3.m 1 of 2
% Random Process Simulation % ECE 541, HW6, Fall 2008 % Created by Prof. Hayat, Nov. 2006, revised Nov. 2008
close all m n % Parameter setting % Range of uniform distribution a = b = % rho r B=zeros(m,n) % mean and variance of B var_B_0=(32) mean_B_ % mean and variance of W mean_W=(1-r)mean_B_ var_W=(1-r^2)var_B_ sdev_W=sqrt(var_W) % Generate B0, uniform distribution B(:,1)=a+(b-a)rand(m,1) for i=2:n % Generate W % Assume W~Gaussian for convenience % Because Gaussian has the mean and var in the pdf equation. W=mean_W+(sdev_Wrandn(100,1)) % Generate B B(:,i)=r*B(:,i-1)+W end
figure(1) N=[1:n] K=[1 10 20 75 100] plot(N,B(1,:),'b.',N,B(20,:),'m.',N,B(75,:),'g.') hold on mean_B_n=mean(B) plot(N,mean_B_n,'k') xlabel('TIME {\it n}') ylabel('Sample mean of B(n) and five sample realizations') legend(['m = ' num2str(K(1))],['m = ' num2str(K(3))],['m = ' num2str(K(4))],'average over 100 realizations') mean(mean_B_n) hold off
figure(2) var_B_n=var(B) plot(N,var_B_n,'k-') xlabel('TIME {\it n}') ylabel('Sample variance of B(n)') mean(var_B_n)
Program
Current plot held
Current plot released
Time average of sample mean = 16.
Time average of sample variance = 86.
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Sample mean of B(n) and five sample realizations
m = 1 m = 20 m = 75 average over 100 realizations
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