Institut für Stochastik
Prof. Dr. G. Last · Dipl.-Math. F. Gieringer
The Poisson Process SS 2016
Exercise sheet 10
Exercise 10.1:
Let f, g ∈ L1 (λ) ∩ L2 (λ) and assume moreover that the functions f g 2 , f 2 g, f 2 g 2 are all in L1 (λ).
Show that
E[I(f )2 I(g)2 ] = λ(f 2 )λ(g 2 ) + 2[λ(f g)]2 + λ(f 2 g 2 ).
In particular
E[I(f )4 ] = 3[λ(f 2 )]2 + λ(f 4 ),
provided that f ∈ L1 (λ) ∩ L4 (λ).
Solution: By definition of I, we have
E[I(f )2 I(g)2 ] = E[(η(f )2 − 2η(f )λ(f ) + λ(f )2 )(η(g)2 − 2η(g)λ(g) + λ(g)2 )]
= E[η(f )2 η(g)2 − 2η(f )2 η(g)λ(g) − 2η(g)2 η(f )λ(f ) + η(f )2 λ(g)2 + η(g)2 λ(f )2
+ 4η(f )η(g)λ(f )λ(g) − 2η(f )λ(f )λ(g)2 − 2η(g)λ(g)λ(f )2 + λ(f )2 λ(g)2 ].
Using the mutivariate Mecke equation, we get
E[η(f )2 η(g)2 ] = E[η (4) (f ⊗ f ⊗ g ⊗ g) + η (3) (f 2 ⊗ g ⊗ g + f ⊗ f ⊗ g 2 + 4f g ⊗ f ⊗ g)
+ η (2) (f 2 ⊗ g 2 + 2f g ⊗ f g) + η(f 2 g 2 )]
= λ(f )2 λ(g)2 + λ(f 2 )λ(g)2 + λ(f )2 λ(g 2 ) + 4λ(f g)λ(f )λ(g)
+ λ(f 2 )λ(g 2 ) + 2λ(f g)2 + λ(f 2 g 2 )
and similarly for the other summands:
E[η(f )2 η(g)] = E[η (3) (f ⊗ f ⊗ g) + η (2) (f 2 ⊗ g + 2f g ⊗ f ) + η(f 2 g)]
= λ(f )2 λ(g) + λ(f 2 )λ(g) + 2λ(f g)λ(f ) + λ(f 2 g),
E[η(g)2 η(f )] = λ(g)2 λ(f ) + λ(g 2 )λ(f ) + 2λ(f g)λ(g) + λ(g 2 f ),
E[η(f )2 ] = E[η (2) (f ⊗ f ) + η(f 2 )] = λ(f )2 + λ(f 2 ),
E[η(g)2 ] = E[η (2) (g ⊗ g) + η(g 2 )] = λ(g)2 + λ(g 2 ),
E[η(f g)] = E[η (2) (f ⊗ g) + η(f g)] = λ(f )λ(g) + λ(f g).
Putting the pieces together yields the result.
1
Exercise 10.2:
For f ∈ L1 (λ2 ) define
Z
I2 (f ) :=
ZZ
(2)
f (x1 , x2 ) η (d(x1 , x2 )) −
f (x1 , x2 ) η(dx1 ) λ(dx2 )
ZZ
Z
−
f (x1 , x2 ) λ(dx1 ) η(dx2 ) + f (x1 , x2 ) λ2 (d(x1 , x2 )).
Show that
(a)
E[I2 (f )] = 0
(b) and for f 0 (x1 , x2 ) := f (x2 , x1 ), x1 , x2 ∈ X, it holds
I2 (f ) = I2 (f 0 ).
Solution:
(a) The multivariate Mecke equation and Fubini’s theorem lead to
ZZ
hZ
E[I2 (f )] = E
f (x1 , x2 ) η (2) (d(x1 , x2 )) −
f (x1 , x2 ) η(dx1 ) λ(dx2 )
ZZ
Z
i
−
f (x1 , x2 ) λ(dx1 ) η(dx2 ) + f (x1 , x2 ) λ2 (d(x1 , x2 ))
Z
ZZ
= f (x1 , x2 ) λ(d(x1 , x2 )) −
f (x1 , x2 ) λ(dx1 ) λ(dx2 )
ZZ
Z
−
f (x1 , x2 ) λ(dx1 ) λ(dx2 ) + f (x1 , x2 ) λ2 (d(x1 , x2 ))
= 0.
(b) The assertion follows from
Z
Z
(2)
f (x1 ,x2 ) η (d(x1 , x2 )) + f (x1 , x2 ) λ2 (d(x1 , x2 ))
Z
Z
(2)
= f (x1 , x2 ) η (d(x2 , x1 )) + f (x1 , x2 ) λ2 (d(x2 , x1 ))
Z
Z
(2)
= f (x2 , x1 ) η (d(x1 , x2 )) + f (x2 , x1 ) λ2 (d(x1 , x2 ))
Z
Z
0
(2)
= f (x1 , x2 ) η (d(x1 , x2 )) + f 0 (x1 , x2 ) λ2 (d(x1 , x2 ))
and
ZZ
ZZ
f (x1 ,x2 ) η(dx1 ) λ(dx2 ) +
f (x1 , x2 ) λ(dx1 ) η(dx2 )
ZZ
ZZ
0
=
f (x2 , x1 ) η(dx1 ) λ(dx2 ) +
f 0 (x2 , x1 ) λ(dx1 ) η(dx2 )
ZZ
ZZ
0
=
f (x1 , x2 ) η(dx2 ) λ(dx1 ) +
f 0 (x1 , x2 ) λ(dx2 ) η(dx1 )
ZZ
ZZ
=
f 0 (x1 , x2 ) λ(dx1 ) η(dx2 ) +
f 0 (x1 , x2 ) η(dx1 ) λ(dx2 ).
2
Exercise 10.3:
Let f ∈ L1s (λ2 ) and g ∈ L1 (λ). Show that
2
E[I2 (f ) I(g)] = 4
Z
f (x1 , x2 )2 g(x1 ) λ2 (d(x1 , x2 ))
holds under suitable integrability assumptions on f and g.
Solution: Because of the symmetry and Fubini’s theorem, it holds
I2 (f ) = η (2) (f ) − 2η(f 0 ) + λ2 (f )
where f 0 (x) :=
R
f (x, y) λ(dy). Therefore, we have
E[I2 (f )2 I(g)] = E η (2) (f )2 + 4η(f 0 )2 + λ2 (f )2 − 4η (2) (f )η(f 0 ) + 2η (2) (f )λ2 (f ) − 4η(f 0 )λ2 (f )
× η(g) − λ(g)
It holds
E[η (2) (f )2 η(g)] = E η (5) (f (x1 , x2 )f (x3 , x4 )g(x5 ))
+ η (4) (4f (x1 , x2 )f (x3 , x4 )g(x1 ) + 4f (x1 , x2 )f (x1 , x3 )g(x4 ))
+ η (3) (4f (x1 , x2 )f (x1 , x3 )g(x1 ) + 8f (x1 , x2 )f (x2 , x3 )g(x1 ) + 2f (x1 , x2 )2 g(x3 ))
+ η (2) (4f (x1 , x2 )2 g(x1 ))
= λ2 (f )2 λ(g) + 4λ(f 0 g)λ2 (f ) + 4λ(f 02 )λ(g) + 4λ(f 02 g)
Z
+ 8 f (x1 , x2 )f 0 (x2 )g(x1 ) λ2 (d(x1 , x2 )) + 2λ2 (f 2 )λ(g)
Z
+ 4 f (x1 , x2 )2 g(x1 ) λ2 (d(x1 , x2 ))
and
E[(η (2) (f ))2 ] = E η (4) (f (x1 , x2 )f (x3 , x4 )) + 4η(3)(f (x1 , x2 )f (x1 , x3 )) + 2η (2) (f (x1 , x2 )2 )
= λ2 (f )2 + 4λ(f 02 ) + 2λ2 (f 2 )
such that
Z
(2)
2
0
2
02
E (η (f )) I(g) = 4λ(f g)λ (f ) + 4λ(f g) + 8 f (x1 , x2 )f 0 (x2 )g(x1 ) λ2 (d(x1 , x2 ))
Z
+ 4 f (x1 , x2 )2 g(x1 ) λ2 (d(x1 , x2 )),
E η(f 0 )2 η(g) = E η (3) (f 0 (x1 )f 0 (x2 )g(x3 )) + η (2) (f 0 (x1 )2 g(x2 ) + 2f 0 (x1 )g(x1 )f 0 (x2 )) + η(f 0 (x)2 g(x))
= λ(f 0 )2 λ(g) + λ(f 02 )λ(g) + 2λ(f 0 g)λ(f 0 ) + λ(f 02 g),
E η(f 0 )2 = E η (2) (f 0 (x1 )f 0 (x2 )) + η(f 02 ) = λ(f 0 )2 + λ(f 02 ),
therefore
E 4η(f 0 )2 I(g) = 8λ(f 0 g)λ(f 0 ) + 4λ(f 02 g).
Proceeding in the same way for the other summands, we get the assertion. The computations require
that all integrals are finite.
3