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By Harkins W.D., Ewing D.T.

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Example text

So we can apply the Tauberian theorem with γ = 1. 23) can be derived for the case where X1 has a 2 density in some interval. Assume that σX = Var(X1 ) and σXY = Cov(X1 , Y1 ) are finite. Let fX and fXY denote the density function of X1 and the joint density of X1 and Y1 respectively. Let M (t) = E[R(t)]. 25) 0 t ∞ where K(t) = 0 0 yfXY (x, y)dydx. Define for t ≥ 0 the function Z(t) = M (t) − µY t. 25), we find that Z(t) satisfies the following integral equation t Z(t) = a(t) + Z(t − x)fX (x)dx, 0 where ∞ µY (x − t)fX (x)dx.

1 E(e−αR(t) ) e−αR(t)(µ) Pν (dµ) = Mp (E) = exp −α Mp (E) E = Mp (E) E 1[0,x) (t − AX (s, µ))AY (s, µ)µ(dsdxdy) Pν (dµ) 1[0,x) (t − AX (s, µ)) exp − αAY (s, µ) µ(dsdxdy)Pν (dµ). Applying the Palm formula for Poisson point processes we obtain ∞ E(e−αR(t) ) = ∞ 0 0 exp ∞ 0 Mp (E) 1[0,x) (t − AX (s, µ)) − αAY (s, µ) Pν (dµ)dH(x, y)ds. Using Fubini’s theorem we obtain ∞ E(e−αR(t) )e−βt dt 0 = 1 [1 − F ∗ (β)] β ∞ exp 0 − βAX (s, µ) + αAY (s, µ) Pν (dµ)ds. Mp (E) Using the Laplace functional of Poisson point processes we obtain exp − βAX (s, µ) + αAY (s, µ) Pν (dµ) Mp (E) = exp − Mp (E) ∞ = exp − 1−e 0 = exp 1[0,s) (r)(βu + αv)µ(drdudv) E ∞ ∞ −1[0,s) (r)[βu+αv] 0 Pν (dµ) dH(u, v)dr 0 − s[1 − H ∗ (β, α)] .

N=1 Renewal reward processes 27 Then for α, β > 0 ∞ E(e−αR(t) )e−βt dt = 0 1 − F ∗ (β) . 1 E(e−αR(t) ) e−αR(t)(µ) Pν (dµ) = Mp (E) = exp −α Mp (E) E = Mp (E) E 1[0,x) (t − AX (s, µ))AY (s, µ)µ(dsdxdy) Pν (dµ) 1[0,x) (t − AX (s, µ)) exp − αAY (s, µ) µ(dsdxdy)Pν (dµ). Applying the Palm formula for Poisson point processes we obtain ∞ E(e−αR(t) ) = ∞ 0 0 exp ∞ 0 Mp (E) 1[0,x) (t − AX (s, µ)) − αAY (s, µ) Pν (dµ)dH(x, y)ds. Using Fubini’s theorem we obtain ∞ E(e−αR(t) )e−βt dt 0 = 1 [1 − F ∗ (β)] β ∞ exp 0 − βAX (s, µ) + αAY (s, µ) Pν (dµ)ds.

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