group exercise for 0≤t 1
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Group exercise
For 0≤t1<…<tn real and 0≤r1≤…≤rn
integers, define X(t) by X(0)=0 and
(a) Find P(X(t)=0)
(b) Determine P(X(t)=k)
P(X(t1) =r1, ...,X(tn) =rn) =
λrne−λtnt1r1 (t2 −t1)
r2 −r1L (tn −tn−1)rn−rn−1
r1!(r2 −r1)!L (rn −rn−1)!
P(X(t1) =r1) =
λr1e−λt1t1r1
r1!: Po(λt1)
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(c) Show that X(t1) and X(t2)-X(t1) are independent
(d) Show that Kolmogorov’s consistency condition is satisfied (try this at home, xc)
P(X(t1) =r1, ...,X(tn) =rn) =
λrne−λtnt1r1 (t2 −t1)
r2 −r1L (tn −tn−1)rn−rn−1
r1!(r2 −r1)!L (rn −rn−1)!
P(X(t1) =r1,X(t2 ) =r2 ) =P(X(t1) =r1,X(t2 ) −X(t1) =r2 −r1)
λr2e−λt2 t1r1 (t2 −t1)
r2 −r1
r1!(r2 −r1)!
=P(X(t1) =r1)[λ(t2 −t1)]
r2 −r1e−λ(t2 −t1 )
(r2 −r1)!
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Markov chains
Chapters 5 and 6
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Precipitation data
January data from Snoqualmie Falls, Washington, 1948-1983
325 dry and 791 wet days
Rt=1(rain day t) ~ Bern(p) independently
E(#WW days) = N x 30 x p2
36 x 30 x (791/1116)2 = 543Today wet Today dry
Yesterday wet 643 (543) 128 (223) 771
Yesterday dry 123 (223) 186 (91) 309
766 314 1080
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A conditional model
P(Rt = 1 | Rt-1 = 1) = p11
P(Rt = 1 | Rt-1 = 0) = p01
Special case of
ik=0 or 1
Transition matrix
pd = P(rain following dry day)
P(R t =1R t−1 =i1, ...,R0 =it )
P =p00 p10
p01 p11
⎛
⎝⎜⎞
⎠⎟=
1−pd pd
1−pw pw
⎛
⎝⎜⎞
⎠⎟
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More generally
Consider a stochastic process (Xn, n≥0) taking values in a discrete state space S. It is a Markov chain if
The quantities are called the transition probabilities. The chain is homogeneous if the transition probabilities do not depend on n.
We will usually assume this.
P(Xn =k Xn−1 =k1 , ...,X1 =kn−1,X0 =kn) =
P(Xn =k Xn−1 =k1) ≡pk1,k(n)
pij (n)
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The transition matrix
The matrix P=(pij) is called the transition matrix.
Theorem: P has nonnegative entries and all row sums are one.
Such matrices are called stochastic.
Proof: pij =j∈S∑ P(Xn =j Xn−1 =i)
j∈S∑
=P(Xn ∈S Xn−1 = i) = 1
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n-step transitions
are called n-step transition probabilities. The matrix of them is denoted P(n).
Theorem (Chapman-Kolmogorov):
P(n+m) = P(n)P(m)
Proof:
using the Markov property.
pij(n) =P(Xm+n =j Xm =i)
P(Xm+n =j X0 =i) = P(Xm+n =j,Xm =k X0 =i)k∈S∑
= P(Xm+n = j Xm = k,X0 = i)P(Xm = k | X0 = j)k∈S∑
= P(Xm+n = j Xm = k)P(Xm = k | X0 = i)k∈S∑
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Consequences
1. P(n) = Pn
2. Let k(n) = P(Xn = k) and (n) = (k
(n)). Then (m+n)= (n)Pm
3. In particular, (n)= (0)Pn
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Back to precipitation
Let 1(0)=p1. Then
1(n) = P(Xn = 1)
= P(Xn = 1,Xn−1 = 0) + P(Xn = 1,Xn−1 = 1)
=0(n−1)p01 + μ1
(n−1)p11
=1(n−1) (p11 − p01) + p01
1(1) = p1(p11 − p01) + p01
1(2) = μ1
(1) (p11 − p01) + p01
= p1(p11 − p01)2 + p01(1+ (p11 − p01))
1(n) = p1(p11 − p01)
n + p01 (p11 − p01)j
j=0
n−1
∑
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p00 = p11 = 1
P(Xn=1) = p1
p01 ≠ p11
If it rains on Jan 1, what is the chance that it rains on Jan 6?
1(n) = p1(p11 − p01)
n + p01 (p11 − p01)j
j=0
n−1
∑
1(n) =
p01
1− (p11 − p01)+ (p1 −
p01
1− (p11 − p01)(p11 − p01)
n
P =.602 .398.166 .834⎛⎝⎜
⎞⎠⎟
P5 =.305 .695.290 .710⎛⎝⎜
⎞⎠⎟
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Generating functions
a=(a0,a1,a2,...) sequence of real numbers
is its generating function
Special case:
probability generating function (pgf)
Ga (s) = aisi
i=0
∞
∑
G(s) = pisi
i=0
∞
∑ =EsX
dk
dsk G(s) s=0 =
dk
dsk G(s)s↑1=
k!pk
E(X(X −1)L (X−k + 1))
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Examples
Bernoulli:
GX(s) = s0 (1-p)+sp = 1+p(s-1)
Geometric:
GX(s) =
Poisson:
GX(s) =
skp(1−p)kk=0
∞
∑ =p
1− (1− p)s
sk λk
k!e−λ
k=0
∞
∑ =e−λesλ = eλ(s−1)
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Convolutions
The convolution of sequences a and b is c=a*b where ci=a0bi+a1bi-1+...+aib0
Theorem: Gc(s) = Ga(s)Gb(s)
Proof: cis
i
i=0
∞
∑ = akbi−kk=0
i
∑⎛⎝⎜⎞⎠⎟i=0
∞
∑ = aksk bi−ks
i−k
i=k
∞
∑k=0
∞
∑
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Sum of iid random variables
Xi nonnegative integervaluedMarkov chain?
E(Sn) = = nE(X)
P(Sn=1) =Bernoulli case:
Random walk case:
Xi=1 w pr p, -1 w pr 1-p
GSn(s) =GX (s)n
n ′GX (1−) GX (1)[ ]n−1
n ′GX (0) GX (0)[ ]n−1
=np1p0n−1
GSn(s) =(1+p(s-1))n