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 X_t defined as

X_t = ∑_i=1^N_tY_i

where jumps sizes Y_i are independent and identically distributed with distribution f and (N_t) is a Poisson process with intensity λ, independent from (Y_i)_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 (T_i)_i≥1 have the same law as the jump times of the Poisson process N_t: they can be expressed as partial sums of independent exponential random variables with parameter λ.
3. The jump sizes (Y_i)_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 Y_i ≡ 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 Y_i. The compound Poisson process X_t can be obtained by changing the time of R with an independent Poisson process N_t: X_t = R(N_t). X_t thus describes the position of a random walk after a random number of time steps, given by N_t. Compound Poisson processes are Lévy processes (and part 2) and they are the only Lévy processes with piecewise constant sample paths.

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

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

N_t =# {0 < s ≤ t: X_s− ≠ X_s} —– (1)

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

N_t − N_h = #{h < s ≤ t: X_s− ≠ X_s} = #{h < s ≤ t: X_s− − X_h ≠ X_s − X_h}

Hence, N_t − N_h depends only on (X_s − X_h), h ≤ s ≤ t. Therefore, from the independence and stationarity of increments of (X_t) it follows that (N_t) also has independent and stationary increments. Using the process N, we can compute the jump sizes of X: Y_n = X_Sn − X_Sn− where S_n = inf{t: N_t ≥ 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(X_s)

A₂ ∈ σ(X_t − X_s)

B₁ ∈ σ(N_r, r ≤ s)

B₂ ∈ σ(N_r − N_s, r > s)

such that P(B₁) > 0 and P(B₂) > 0. The independence of increments of X implies that processes (X_r − X_s, r > s) and (X_r, 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 X_t − X_s and X_s are independent conditionally on the trajectory of N. In particular, choosing B₁ = {N_s = 1} and B₂ = {N_t − N_s = 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 (Y_i)_i≥1 are independent. The jump sizes have the same law, in that the two-dimensional process (X_t, N_t) has stationary increments. Therefore, for every n ≥ 0 and for every s > h > 0,

E[ƒ(X_h) | N_h = 1, N_h – N_s = n] = E[ƒ(X_s+h – X_s) | N_s+h – N_s = 1, N_s – N_h = n],

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

Let (X_t)_t≥0 be a compound Poisson process.

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

E[ƒ(X_r)g(X_s − X_r)] = E[ƒ(X_r)]E[g(X_s − X_r)]

From the representation X_r = ∑_i=1^N_r and X_s − X_r = ∑_i=Nr+1^Ns Y_i the following observations are made:

– Conditionally on the trajectory of N_t for t ∈ [0, s], X_r and X_s − X_r are independent because the first expression only depends on Y_i for i ≤ N_r and the second expression only depends on Y_i for i > N_r.
– The expectation E[ƒ(Xr) N_t, t ≤ s] depends only on N_r and the expectation E[g(X_s − X_r) N_t, t ≤ s] depends only on N_s − N_r.

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

E[ƒ(X_r)g(X_s – X_r)] = E[E[ƒ(X_r)g(X_s – X_r) | N_t, t ≤ s]]

= E[E[ƒ(X_r) | N_t, t ≤ s] E[g(X_s – X_r) |  N_t, t ≤ s]]

= E[E[ƒ(X_r) | N_t, t ≤ s]] E[E[g(X_s – X_r) |  N_t, t ≤ s]]

= E[ƒ(X_r)] E[g(X_s – X_r)]

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

E[ƒ(X_s – X_r)] = E[E[ ∑_i=Nr+1^N_s Y_i | N_t, t ≤ s]]

= E[E[∑_i=1^N_s-N_r Y_i | N_t, t ≤ s]] = E[E[∑_i=1^N_s-r Y_i | N_t, t ≤ s]] = E[ƒ(X_s-r)]

Stochastic continuity. X_t only jumps if N_t does.

P(N_s →^s<t_s→t N_t) = 1

Hence, for every t > 0,

P(X_s →^s<t_s→t X_t) = 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.