传有草 由条件(2)~(4)，得： 1951 Po(t+h)=PIN(t+h)=0}=PIN(t)=0,N(t+h)-N(t)=0} =P{N(t)=0}P{N(t+h)-N(t)=O} =P(t)[1-λh+o(h)] Po(t+h)-Po( → h 2=-An(e)+o h 令h→0，得 d@=-Po(t)入 11/100 dt P(0)=1, 条件(1)N(0)=0. 解得 P(t)=e-t,t≥0. 2.当n>1,根据全概率公式有 Pn(t+h)=Pn(t)Po(h)+Pn-1(t)Pi(h)+o(h) Pn(t+h)=(1-λh)Pn(t)+λhPn-1(t)+o(h) Pn(t+h)-Pn(t 2=-P.(0+P-10+ o(h) GoBack h h FullScreen Close Quit11/100 kJ Ik J I GoBack FullScreen Close Quit d^á(2)∼(4),µ P0(t + h) = P{N(t + h) = 0} = P{N(t) = 0, N(t + h) − N(t) = 0} = P{N(t) = 0}P{N(t + h) − N(t) = 0} = P0(t)[1 − λh + o(h)] → P0(t + h) − P0(t) h = −λP0(t) + o(h) h -h → 0, (dP0(t) dt = −P0(t)λ P0(0) = 1, ^á£1§N(0) = 0. ) P0(t) = e −λt, t ≥ 0. 2. n ≥ 1,ä‚V«˙™k Pn(t + h) = Pn(t)P0(h) + Pn−1(t)P1(h) + o(h) Pn(t + h) = (1 − λh)Pn(t) + λhPn−1(t) + o(h) → Pn(t + h) − Pn(t) h = −λPn(t) + λPn−1(t) + o(h) h