Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 34 / 6

For $0<\delta <1$, define the compensated small-jump Poisson integralThis is an ordinary compensated finite-intensity sum. Independent Poisson random variables give mean zero and the L2 construction of compensated Poisson integrals gives, for $0<\eta <\delta$,Thus $Y_t^{\delta }$ is Cauchy in L2 space as $\delta \downarrow 0$, defining the first integral. It is not the difference of two separately convergent integrals. On a compact time interval $[0,L]$, the Doob L2 maximal inequality bounds the squared supremum of the same difference by $8cL(\delta -\eta )$. Hence the truncated martingales are Cauchy in the expected squared supremum norm. A sufficiently rapidly decreasing sequence of cutoffs converges uniformly almost surely; its limit has càdlàg paths, so the construction also supplies a consistent process.

For large jumps, $K(\{|y|>1\})=2c<\infty$. There are only finitely many such atoms before any fixed time, almost surely, and each atom has a finite mark. The uncompensated large-jump integral is therefore a finite random sum, even though its absolute first moment is infinite. The two terms are well-defined for these different reasons.

There is also a literal signed-integral defect in the PDF: $\int yK(dy)$ on either symmetric region is not $+\infty$. Its positive and negative parts both have infinite mass, so the ordinary signed Lebesgue integral is undefined. The correct divergence statement isA symmetric principal value may be zero, but is not an ordinary signed integral. Square-integrable compensation defines the small jumps; finite activity defines the large jumps.

For a finite-intensity Poisson random measure, independence of counts gives the exponential characteristic function formula. Apply it to the cutoff construction from part (a), including the deterministic compensation, and pass to the L2 space limit of the small jumps. The compensated integrand is $O(y^2)$ at zero and bounded on the large-jump region, so its integral converges absolutely:Symmetry makes the absolutely integrable imaginary part zero. Substituting $z=|u|y$ in the real part and using the specified normalization givesThus $\mathbb{E}{e}^{iu{X}_1}=e^{-|u|}$, and this construction is the standard Cauchy process. To calculate the density, its integrable characteristic function permits Fourier inversion:This is the standard Cauchy distribution. The normalization also gives $c=1/\pi$: integration by parts turns ${\int }_0^{\infty }(1-\cos z)z^{-2}dz$ into ${\int }_0^{\infty }\sin zdz/z=\pi /2$. For completeness, with exponential damping the latter integral equals $\arctan (1/\eta )$, because its derivative in frequency $b$ is ${\int }_0^{\infty }{e}^{-\eta z}\cos (bz)dz=\eta /({\eta }^2+b^2)$ and its value at $b=0$ is zero. Let $\eta \downarrow 0$; integration by parts bounds the undamped and damped tails uniformly by a constant divided by the lower cutoff, justifying the limit.

Disjoint time strips of the Poisson random measure are independent, and its intensity is translation invariant in time. The construction and its limits therefore give independent increments and stationary increments, withFor the scaling actually printed in the original PDF,For example, $\alpha =2$, $t=1$, $u=1$ gives $e^{-4}$ rather than $e^{-1}$. Thus the requested equality of process laws is false unless $\alpha =1$.

The intended self-similarity of a Cauchy process rescales space and time in opposite directions:Indeed, for $0\le s<t$,The transformed process retains independent increments. For any ordered times, the joint characteristic function of its successive increments is therefore the same product as for $X$; cumulative summation proves equality of all finite-dimensional distributions. Both constructions have càdlàg paths, so these distributions also determine their laws on the usual path space. This derives the corrected invariance and explicitly diagnoses the false printed assertion.