Math 171 Homework #8
8.4
a.
P{X2>2}=Z∞
2
1
√4πe−x2/4dx
≈0.07865
b.
P{X2> X1}=P{X1>0}=1
2
d. Let Yt=1
√2X2t. Then by the scaling rules, Ytis a standard Brownian motion, and
P{Xt= 0,2≤t≤3}=P{Yt= 0,1≤t≤3
2}
= 1 −2
πarctan √2
≈0.392
f. Since 1-D Brownian motion is recurrent, P{Xt>0, t > 10}=P{Xt6= 0, t > 10}= 0;
8.9
P{X2>0|X1=b}=P{X1< b|X0= 0}
= 1 −P{X1≥b|X0= 0}
= 1 −Z∞
b
1
√2πe−x2/2dx
P{X2>0|X1>0}= 2 Z∞
0
p1(0, b)h1−Z∞
b
1
√2πe−x2/2dxidb
= 2 Z∞
0
p1(0, b)db −1
πZ∞
0Z∞
b
e−(b2+x2)/2dxdb
= 1 −1
πZ∞
0Zπ/2
arctan(1)
re−r2/2dθdr
= 1 −1
π(π
2−arctan(1))
= 0.75
8.15
We need 2nintervals of length 2n/5nto cover An. This gives
2n=2
5−nD
D=ln 2
ln(5/2)