Chapter 8: Q. 8.10 (page 393)

It $X$ is a Poisson random variable with a mean $λ$ , showing that for $i < λ$ ,
$P {X \leq i} \leq \frac{e^{- λ} (e λ)^{i}}{i^{i}}$

Short Answer

Expert verified

Therefore,

\frac{d}{d t} λ (e^{t} - 1) - i t = 0 \Rightarrow e^{t} = \frac{i}{λ}

Hence proved.

Step by step solution

Given Information.

$X$ is a Poisson random variable with a mean $λ$ ,

Explanation.

We are going to be using Chernoff bounds. Specifically, it $X$ is a random variable with $M G F M (t)$ and $i > 0, t < 0$

$\Rightarrow P (X \leq i) \leq e^{- i t} M (t)$

Explanation.

$X$ $p o i s s$ $(λ)$ then the $M G F$ is $M (t) = e^{λ (e^{t} - 1)}$ . Then using Chernoff bound:

$i < λ, P (X \leq i) \leq e^{- i t} e^{λ (e^{t} - 1)} = e^{λ (e^{t} - 1) - i t}$

We want $t$ to minimize this. Since $e^{x}$ is a strictly increasing function, the minimizing argument is given by the first-order conditions of

$λ (e^{t} - 1) - i t$

So we derive:

$\frac{d}{d t} λ (e^{t} - 1) - i t = 0 \Rightarrow e^{t} = \frac{i}{λ}$

Since $i \leq λ$ this means that $t < 0$ and so the Chernoff bound is valid.

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It $X$ is a Poisson random variable with a mean $λ$ , showing that for $i < λ$ ,
$P {X \leq i} \leq \frac{e^{- λ} (e λ)^{i}}{i^{i}}$

Short Answer

Step by step solution

Given Information.

Explanation.

Explanation.

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It $X$ is a Poisson random variable with a mean $λ$ , showing that for $i < λ$ ,
$P {X \leq i} \leq \frac{e^{- λ} (e λ)^{i}}{i^{i}}$