Solution to selected problems.

where Z t ′ = ∫ R xN t(·, dx), β = α− ∫ |x|≥1 xν(dx). By theorem 43, E(eiuZ′ t ) =
∫R (1−eiux )ν(dx). Z t ′ is
a compound Poisson process (See problem 11). Arrival rate (intensity) is λ since E( ∫ R N t(·, dx)) =
t ∫ R ν(dx) = λt. Since Z t is a martingale, EZ t ′ = −βt. Since EZ t ′ = E( ∫ R xN t(·, dx)) = t ∫ R xν(dx),
β = − ∫ R xν(dx). It follows that Z t = Z t ′ − λt ∫ R x ν λ
(dx) is a compensated compound Poisson
process. Then problem 12 shows that EU 1 = ∫ R x ν λ
(dx). It follows that the distribution of jumps
µ = (1/λ)ν.
14. Suppose EN t < ∞. At first we show that Z t ∈ L 1 .
(
Nt
)
)
∑
∞∑
E|Z t | ≤ E |U i | = E |U i | P (N t = n) =
Then
≤ sup
i
i=1
E|U i |
∞∑
n=0
E(Z t |F s ) = E(Z s +
i=1
n=0
( n∑
i=1
nP (N t = n) = EN t sup E|U i | < ∞.
i
∞∑
n=0 i=1
∞∑
∞∑
U i 1 {s