For a Poisson process N of rate lambda, show that M_t = N_t - lambda t is a martingale, give Var(M_t), and distinguish the quadratic variation [M]_t from the predictable compensator.

For a Poisson process N of rate lambda, show that M_t = N_t - lambda t is a martingale, give Var(M_t), and distinguish the quadratic variation [M]_t from the predictable compensator.

Approach: Use stationary independent increments for the martingale property, then count the contribution of each jump to the sum of squared increments and compare it with the predictable bracket.

M is a martingale with Var(M_t) = lambda t, [M]_t = N_t and the predictable compensator is lambda t. The Poisson process has independent increments with E[N_t - N_s] = lambda(t-s), so E[M_t | F_s] = M_s + E[N_t - N_s] - lambda(t-s) = M_s, which is the martingale property of the compensated Poisson process. Its variance is the variance of N_t, which is lambda t. For the quadratic variation, the path is piecewise constant with jumps of size exactly 1, so the sum of squared increments over a fine partition converges to the number of jumps, giving [M]_t = N_t, a random quantity. The predictable compensator is the unique predictable increasing process making M^2 minus it a martingale, and here it is lambda t, a deterministic one. They agree in expectation since E[N_t] = lambda t, which is why variance calculations rarely expose the difference.

Follow-up: What is the quadratic variation of a compensated compound Poisson process with jump sizes Y, and when does it agree with its compensator in expectation?

Key concepts: compensated Poisson process, predictable compensator, quadratic variation, independent increments.