X_t = sum_{i=1}^{N_t} Y_i is compound Poisson with rate lambda = 3 per year, and the Y_i are independent with mean 0.02 and standard deviation 0.05. Compute Var(X_1), and give the characteristic exponent of X.

X_t = sum_{i=1}^{N_t} Y_i is compound Poisson with rate lambda = 3 per year, and the Y_i are independent with mean 0.02 and standard deviation 0.05. Compute Var(X_1), and give the characteristic exponent of X.

Approach: Condition on the number of jumps and apply the law of total variance, then write the characteristic function by conditioning the same way.

0.0087. Conditioning on N_1 and using the law of total variance, Var(X_1) = E[N_1] Var(Y) + Var(N_1) E[Y]^2 = lambda(Var(Y) + E[Y]^2) = lambda E[Y^2], since a Poisson count has equal mean and variance. The second moment of the jump size is E[Y^2] = 0.05^2 + 0.02^2 = 0.0029, so Var(X_1) = 3(0.0029) = 0.0087 and the standard deviation is 0.0933. The mean is E[X_1] = lambda E[Y] = 0.06. The characteristic function is E[exp(iuX_t)] = exp(lambda t (phi_Y(u) - 1)) where phi_Y is the characteristic function of a single jump, so the characteristic exponent is lambda(phi_Y(u) - 1), which is linear in the rate and shows that every cumulant of X_t is lambda t times the corresponding raw moment of Y.

Follow-up: If instead the jump sizes are Pareto with tail index 1.5, which of these quantities remain finite and what replaces the variance as a risk measure?

Key concepts: compound Poisson, law of total variance, second moment, characteristic function.