G (λ) = \sum_{n = 0}^{\infty} e^{λn} \frac{( μ t ) ^{n}}{n !} e^{- μ t} = exp [μ t (e^{λ} - 1)] .

Differentiating at zero gives

E N_{t} = μ t, Var (N_{t}) = μ t .

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ii

Solution

\frac{1}{t} lo g G (λ) = μ (e^{λ} - 1) .

By Cramér theorem, its Legendre transform is stationary at

e^{λ_{*}} = x / μ

for

x \geq 0

. Thus

I (x) = x lo g \frac{x}{μ} - x + μ, x \geq 0,

with

I (x) = + \infty

for

x < 0

and the continuous convention

0 lo g 0 = 0

Solved by gpt-5.6-sol high.

Assume

μ_{2} > μ_{1}

, so the upper tail is governed by its boundary. The large deviation principle gives the exponentially equivalent estimate

P (N_{t} / t \geq μ_{2}) ≍ e^{- t I (μ_{2})},

where now

I (x) = x lo g (x / μ_{1}) - x + μ_{1}

. Increasing capacity to

μ_{3} > μ_{2}

reduces this estimate by at least

1 0^{6}

when

t [I (μ_{3}) - I (μ_{2})] \geq 6 lo g 10 .

Equivalently,

μ_{3} lo g \frac{μ _{3}}{μ _{1}} - μ_{3} - μ_{2} lo g \frac{μ _{2}}{μ _{1}} + μ_{2} \geq \frac{6 lo g 10}{t} .

Solved by gpt-5.6-sol high.

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