Solution

X

have rate

λ

, let

\overline{F} (t) = P (ξ_{1} > t)

be the survival function of an inter-renewal time of

N

, and let

\overline{H}

be the survival function of the first event time of

Y

. Independence gives

\overline{H} (t) = e^{- λ t} \overline{F} (t) .

Put

μ_{F} = \int_{0}^{\infty} \overline{F} (t) d t, μ_{H} = \int_{0}^{\infty} \overline{H} (t) d t .

At a large time, the excess of

Y

is the minimum of the excesses of

X

and

N

. The Poisson excess is an independent rate-

λ

exponential random variable. Applying the renewal excess limit theorem to

N

and to the assumed renewal process

Y

gives

\frac{1}{μ _{H}} \int_{x}^{\infty} \overline{H} (y) d y = \frac{e ^{- λ x}}{μ _{F}} \int_{x}^{\infty} \overline{F} (y) d y .

Substituting

\overline{F} (y) = e^{λ y} \overline{H} (y)

yields the integral equation

\frac{1}{μ _{H}} \int_{x}^{\infty} \overline{H} (y) d y = \frac{e ^{- λ x}}{μ _{F}} \int_{x}^{\infty} e^{λ y} \overline{H} (y) d y .

Set

A (x) = \int_{x}^{\infty} \overline{H} (y) d y

and

c = μ_{H} / μ_{F} < 1

. Differentiating the equation almost everywhere gives

(1 - c) \overline{H} (x) = λ A (x) .

Since

A^{'} = - \overline{H}

, it follows that

A^{'} (x) = - \frac{λ}{1 - c} A (x) .

Using

\overline{H} (0) = 1

shows that

\overline{H} (x) = e^{- x / μ_{H}}

. Thus the first event time of

Y

Solved by gpt-5.6-sol high.

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