# Convergence in Probability Implying Stochastic Continuity. Part 3.

A compound Poisson process with intensity λ > 0 and jump size distribution f is a stochastic process Xt defined as

Xt = ∑i=1NtYi

where jumps sizes Yi are independent and identically distributed with distribution f and (Nt) is a Poisson process with intensity λ, independent from (Yi)i≥1.

The following properties of a compound Poisson process are now deduced

1. The sample paths of X are cadlag piecewise constant functions.
2. The jump times (Ti)i≥1 have the same law as the jump times of the Poisson process Nt: they can be expressed as partial sums of independent exponential random variables with parameter λ.
3. The jump sizes (Yi)i≥1 are independent and identically distributed with law f.

The Poisson process itself can be seen as a compound Poisson process on R such that Yi ≡ 1. This explains the origin of term “compound Poisson” in the definition.

Let R(n),n ≥ 0 be a random walk with step size distribution f: R(n) = ∑i=0n Yi. The compound Poisson process Xt can be obtained by changing the time of R with an independent Poisson process Nt: Xt = R(Nt). Xt thus describes the position of a random walk after a random number of time steps, given by Nt. Compound Poisson processes are Lévy processes (and part 2) and they are the only Lévy processes with piecewise constant sample paths.

(Xt)t≥0 is compound Poisson process if and only if it is a Lévy process and its sample paths are piecewise constant functions.

Let (Xt)t≥0 be a Lévy process with piecewise constant paths. We can construct, path by path, a process (Nt, t ≥ 0) which counts the jumps of X:

Nt =# {0 < s ≤ t: Xs− ≠ Xs} —– (1)

Since the trajectories of X are piecewise constant, X has a finite number of jumps in any finite interval which entails that Nt is finite ∀ finite t. Hence, it is a counting process. Let h < t. Then

Nt − Nh = #{h < s ≤ t: Xs− ≠ Xs} = #{h < s ≤ t: Xs− − Xh ≠ Xs − Xh}

Hence, Nt − Nh depends only on (Xs − Xh), h ≤ s ≤ t. Therefore, from the independence and stationarity of increments of (Xt) it follows that (Nt) also has independent and stationary increments. Using the process N, we can compute the jump sizes of X: Yn = XSn − XSn− where Sn = inf{t: Nt ≥ n}. First lets see how the increments of X conditionally on the trajectory of N are independent? Let t > s and consider the following four sets:

A1 ∈ σ(Xs)

A2 ∈ σ(Xt − Xs)

B1 ∈ σ(Nr, r ≤ s)

B2 ∈ σ(Nr − Ns, r > s)

such that P(B1) > 0 and P(B2) > 0. The independence of increments of X implies that processes (Xr − Xs, r > s) and (Xr, r ≤ s) are independent. Hence,

P[A1 ∩ B1 ∩ A2 ∩ B2] = P[A1 ∩ B1]P[A2 ∩ B2]

Moreover,

– A1 and B1 are independent from B2.

– A2 and B2 are independent from B1.

– B1 and B2 are independent from each other.

Therefore, the conditional probability of interest can be expressed as:

P[A1 ∩ A2 | B1 ∩ B2] = (P[A1 ∩ B1]P[A2 ∩ B2])/P[B1]P[B2]

= (P[A1 ∩ B1 ∩ B2]P[A2 ∩ B1 ∩ B2])/P[B1]2P[B2]2 = P[A1 | B1 ∩ B2]P[A2 | B1 ∩ B2]

This proves that Xt − Xs and Xs are independent conditionally on the trajectory of N. In particular, choosing B1 = {Ns = 1} and B2 = {Nt − Ns = 1}

we obtain that Y1 and Y2 are independent. Since we could have taken any number of increments of X and not just two of them, this proves that (Yi)i≥1 are independent. The jump sizes have the same law, in that the two-dimensional process (Xt, Nt) has stationary increments. Therefore, for every n ≥ 0 and for every s > h > 0,

E[ƒ(Xh) | Nh = 1, Nh – Ns = n] = E[ƒ(Xs+h – Xs) | Ns+h – Ns = 1, Ns – Nh = n],

Where ƒ is any bounded Borel function. This entails that for every n ≥ 0, Y1 and Yn+2 have the same law.

Let (Xt)t≥0 be a compound Poisson process.

Independence of increments. Let 0 < r < s and let ƒ and g be bounded Borel functions on Rd. To ease the notation, we prove only that Xr is independent from Xs − Xr, but the same reasoning applies to any finite number of increments. We must show that

E[ƒ(Xr)g(Xs − Xr)] = E[ƒ(Xr)]E[g(Xs − Xr)]

From the representation Xr = ∑i=1Nr and Xs − Xr = ∑i=Nr+1Ns Yi the following observations are made:

– Conditionally on the trajectory of Nt for t ∈ [0, s], Xr and Xs − Xr are independent because the first expression only depends on Yi for i ≤ Nr and the second expression only depends on Yi for i > Nr.
– The expectation E[ƒ(Xr) Nt, t ≤ s] depends only on Nr and the expectation E[g(Xs − Xr) Nt, t ≤ s] depends only on Ns − Nr.

On using the independence of increments of the Poisson process, we can write:

E[ƒ(Xr)g(Xs – Xr)] = E[E[ƒ(Xr)g(Xs – Xr) | Nt, t ≤ s]]

= E[E[ƒ(Xr) | Nt, t ≤ s] E[g(Xs – Xr) |  Nt, t ≤ s]]

= E[E[ƒ(Xr) | Nt, t ≤ s]] E[E[g(Xs – Xr) |  Nt, t ≤ s]]

= E[ƒ(Xr)] E[g(Xs – Xr)]

Stationarity of increments. Let 0 < r < s and let ƒ be a bounded Borel function.

E[ƒ(Xs – Xr)] = E[E[ ∑i=Nr+1Ns Yi | Nt, t ≤ s]]

= E[E[∑i=1Ns-Nr Yi | Nt, t ≤ s]] = E[E[∑i=1Ns-r Yi | Nt, t ≤ s]] = E[ƒ(Xs-r)]

Stochastic continuity. Xt only jumps if Nt does.

P(Ns →s<ts→t Nt) = 1

Hence, for every t > 0,

P(Xs →s<ts→t Xt) = 1

Since almost sure convergence entails convergence in probability, this implies stochastic continuity. Also, since any cadlag function may be approximated by a piecewise constant function, one may expect that general Lévy processes can be well approximated by compound Poisson ones and that by studying compound Poisson processes one can gain an insight into the properties of Lévy processes.