86
S. Goto and H. Hino
Finally, to obtain the solution to (4.26) for ℘ θ , one rewrites
℘ θ (s, t) with p θ (s, 0) = 1
as
℘ θ (s, t) = e
γ (s−1) t
℘ θ (s, 0) =
∞
j=0
(γ s t)
j
j !
e
−γ t
=
∞
j=0
(γ t)
j
j !
e
−γ t
s
j
.
Since the term [· · · ] of the equation above should be equal to ℘ θ ( j, t), one has
℘ θ ( j, t) =
(γ t)
j
j !
e
−γ t
,
j = 0, . . . , ∞,
where this distribution function is known as the Poisson distribution with the parameter γ t.
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