Use independent colour marks, also independent of arrival times, as in the standard marked-process model. Let

K

be the index of the first white arrival. Then

P (K = k) = p (1 - p)^{k - 1}

and

K

is independent of the interarrival times. Thus

T = \sum_{j = 1}^{K} X_{j}

. Conditional on

K

M_{X}

, and summing the geometric series gives

M_{T} (θ) = \sum_{k = 1}^{\infty} p (1 - p)^{k - 1} (\frac{λ}{λ - θ})^{k} = \frac{p λ}{p λ - θ}, θ < p λ .

p λ

, so

T \sim Exp (p λ)

. The calculation assumes

p > 0

; if

p = 0

, there is no white arrival and

T = + \infty

ii

Solution

Write

Z_{t} = \sum_{j = 1}^{N_{t}} Y_{j}

, with an empty sum equal to zero. Conditioning on the independent Poisson random variable

N_{t}

E e^{θ Z_{t}} = \sum_{n = 0}^{\infty} e^{- λ t} \frac{( λ t ) ^{n}}{n !} M_{Y} (θ)^{n} = exp {λ t (M_{Y} (θ) - 1)} .

This is the compound Poisson distribution. Differentiating its log moment-generating function at zero, when the first two moments are finite, gives

E Z_{t} = λ t E Y, Var (Z_{t}) = λ t E (Y^{2}) .

The variance includes variability in the number of coaches, and so uses the second moment, not merely

Var (Y)

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