ETH Zürich, FS 2017
D-MATH
Prof. Dr. A-S Sznitman
Coordinator
Chong Liu
Applied Stochastic Processes
Solution sheet 1
Solution 1.1 Denote by Sym(n) the symmetric group of degree n. Since the Xi have a density,
we have X(1) < X(2) < . . . < X(n) P -a.s. Using that the Xi are i.i.d. and that the order of Sym(n)
is n!, we get for all B ∈ B(Rn )
P [(X(1) , . . . , X(n) ) ∈ B] = P [(X(1) , . . . , X(n) ) ∈ B, X(1) < X(2) < · · · < X(n) ]
X
=
P [(X(1) , . . . , X(n) ) ∈ B, X(1) < · · · < X(n) , X(1) = Xπ(1) , . . . , X(n) = Xπ(n) ]
π∈Sym(n)
X
=
P [(Xπ(1) , . . . , Xπ(n) ) ∈ B, Xπ(1) < Xπ(2) < · · · < Xπ(n) ]
π∈Sym(n)
= n! P [(X1 , . . . , Xn ) ∈ B, X1 < X2 < · · · < Xn ]
Z
n
Y
=
1{(x1 ,...,xn )∈B} n! 1{x1 <x2 <···<xn }
f (xi ) dx1 · · · dxn .
Rn
(1)
i=1
This establishes the claim.
Solution 1.2
(a) We have E[Sn ] = µn for all n ∈ N. As N is independent of (Xi )i∈N , it follows for all n ∈ N0
E[SN 1{N =n} ] = E[Sn 1{N =n} ] = E[Sn ]E[1{N =n} ] = µnE[1{N =n} ] = E[µN 1{N =n} ] .
Hence, as N takes value in N so that σ(N ) is generated by the sets {N = n}, we have
E[SN | N ] = µN . This implies
E[SN ] = E[E[SN |N ]] = E[µN ] = µE[N ] .
(b) We have E[Sn2 ] = σ 2 n + µ2 n2 for all n ∈ N. As N is independent of (Xi )i∈N , it follows for all
n∈N
2
E[SN
1{N =n} ] = E[Sn2 1{N =n} ]
=
E[Sn2 ]E[1{N =n} ]
=
(σ 2 n + µ2 n2 )E[1{N =n} ]
=
E[(σ 2 N + µ2 N 2 )1{N =n} ] .
2
This implies E[SN
| N ] = σ 2 N + µ2 N 2 . This leads to
V ar(SN )
2
= E[SN
] − E[SN ]2
2
= E[E[SN
| N ]] − E[SN ]2
= σ 2 E[N ] + µ2 E[N 2 ] − µ2 E[N ]2
=
σ 2 E[N ] + µ2 V ar(N ) .
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Applied Stochastic Processes, FS 2017
D-MATH
Solution sheet 1
Solution 1.3
(a) For all r ≥ 0:
P [D > r] = P [N (Br ) = 0] = exp(−λπr2 ) .
Hence, the distribution function F and the density function f of D are given by
F (r) = (1 − exp(−λπr2 ))1{r≥0} ,
f (r) = 2λπr exp(−λπr2 )1{r≥0} .
(b) For all u > r:
P [N (Bu ) = 1|N (Br ) = 1]
=
P [N (Bu \Br )=0,N (Br )=1]
P [N (Br )=1]
=
P [N (Bu \Br )=0]P [N (Br )=1]
P [N (Br )=1]
=
exp(−λπ(u2 − r2 ))
Hence, as r → 0 we obtain
P [N (Bu ) = 1|N (Br ) = 1] → exp(−λπu2 ) .
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