P_t(n) is the probability of n events occurring during a time interval t. How will you express P_{t+h}(n) in terms of P_t and h, if P_t(n) has stationary independent increments? (Note: f(t) is the probability density function.)

GATE 1989 · Engineering Mathematics · Poisson Distribution · medium

Answer: P_{t+h}(n) = sum_{k=0}^{n} P_t(k) * P_h(n-k). For small h: P_{t+h}(n) = P_t(n)(1 - lambda*h) + P_t(n-1)*lambda*h.

1. Apply independent increments (convolution): N([0, t+h]) = N([0,t]) + N([t,t+h]). By independence and stationarity: P_{t+h}(n) = sum_{k=0}^{n} P_t(k) * P_h(n-k).
2. Small-h differential form: Keeping only O(h) terms: P_{t+h}(n) approx P_t(n)*(1 - lambda*h) + P_t(n-1)*lambda*h. This gives the Kolmogorov forward equation dP_t(n)/dt = -lambda*P_t(n) + lambda*P_t(n-1).