Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 201 / 6 / d

Solution

Let

N (d s, d y)

(0, \infty)^{2}

with intensity

d s K (d y)

X_{t} = \int_{(0, t] \times (0, \infty)} y N (d s, d y) .

The assumption

\int yK (d y) < \infty

makes this integral finite on compact time intervals. The exponential formula for a Poisson random measure gives

E e^{i u X_{t}} = exp {t \int_{(0, \infty)} (e^{i u y} - 1) K (d y)},

X

is a Lévy process with exponent

ψ

q_{n} \geq 0

which vanish off

[1/ n, n]

and satisfy

\int ∣ q_{n} (y) - y ∣ K (d y) ⟶ 0.

This is possible by truncation followed by approximation by simple functions. Put

X_{t}^{n} = \int_{(0, t] \times (0, \infty)} q_{n} (y) N (d s, d y) .

The measure of the support of

q_{n}

is finite, and

q_{n}

takes finitely many values, so

X^{n}

is a simple pure-jump Lévy process. Under this common coupling,

E sup_{s \leq t} ∣ X_{s}^{n} - X_{s} ∣ \leq E \int_{(0, t] \times (0, \infty)} ∣ q_{n} (y) - y ∣ N (d s, d y) = t \int ∣ q_{n} - y ∣ d K ⟶ 0.

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