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}^{Nt}Y_{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

- The sample paths of X are cadlag piecewise constant functions.
- 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. - 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=0}^{n} 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

*) and they are the only Lévy processes with piecewise constant sample paths.*

**part 2**(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:

A_{1} ∈ σ(X_{s})

A_{2} ∈ σ(X_{t} − X_{s})

B_{1} ∈ σ(N_{r}, r ≤ s)

B_{2} ∈ σ(N_{r} − N_{s}, r > s)

such that P(B_{1}) > 0 and P(B_{2}) > 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[A_{1} ∩ B_{1} ∩ A_{2} ∩ B_{2}] = P[A_{1} ∩ B_{1}]P[A_{2} ∩ B_{2}]

Moreover,

– A_{1} and B_{1} are independent from B_{2}.

– A_{2} and B_{2} are independent from B_{1}.

– B_{1} and B_{2} are independent from each other.

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

P[A_{1} ∩ A_{2} | B_{1} ∩ B_{2}] = (P[A_{1} ∩ B_{1}]P[A_{2} ∩ B_{2}])/P[B_{1}]P[B_{2}]

= (P[A_{1} ∩ B_{1} ∩ B_{2}]P[A_{2} ∩ B_{1} ∩ B_{2}])/P[B_{1}]^{2}P[B_{2}]^{2} = P[A_{1} | B_{1} ∩ B_{2}]P[A_{2} | B_{1} ∩ B_{2}]

This proves that X_{t} − X_{s} and X_{s} are independent conditionally on the trajectory of N. In particular, choosing B_{1} = {N_{s} = 1} and B_{2} = {N_{t} − N_{s} = 1}

we obtain that Y_{1} and Y_{2} 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, Y

_{1}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}^{Nr} and X_{s} − X_{r} = ∑_{i=Nr+1}^{N}s 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}^{Ns} Y_{i} | N_{t}, t ≤ s]]

= E[E[∑_{i=1}^{Ns-Nr} Y_{i} | N_{t}, t ≤ s]] = E[E[∑_{i=1}^{Ns-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.