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I am fairly new to statistics. I am working on a list of items in stochastic vehicle routing problem with Poisson distribution and I need to do a normal approximation. I read a paper with the following formulation (image attached). I am not sure what I am missing in my algebraic skills but I can't get to the final equation. $$\begin{align}&\operatorname{Prob}\left\{z\ge\frac{c-\bar\mu}{\sqrt{\bar\mu}}\right\}=\alpha\\\implies&(c-\bar\mu)=z_{1-\alpha}\sqrt{\bar\mu}\\\implies&(c-\bar\mu)^2=\bar\mu z_{1-\alpha}^2\quad\text{which, after some algebra, yields}\end{align}\\\bar\mu=\frac{2c+z_{1-\alpha}^2-\sqrt{z_{1-\alpha}^4+4cz_{1-\alpha}^2}}2.$$

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    $\begingroup$ I know nothing about the context, but the last equation seems to be just derived with quadratic formula $\endgroup$
    – xd y
    Dec 1, 2022 at 2:03
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    $\begingroup$ Please don't upload images of text, use MathJax. $\endgroup$
    – Rob
    Dec 1, 2022 at 2:18

1 Answer 1

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Context Free Algebra.
As pointed out by xd y in the comments, one way to derive this without context (i.e., understanding the Gaussian approximation to the Poisson) is just to apply the quadratic formula to the expression you provide.

For notation simplicity, let $z_{1-\alpha} \equiv z$.

Expanding and factoring permit immediately give the target form: $\begin{align} (c-\bar \mu)^2 &=\bar\mu z^2 \\ \bar\mu^2 -2c\bar\mu -\mu z^2 +c^2 &= 0 \\ \bar\mu^2 + \bar\mu(-2c-z^2) + c^2 &=0 \tag{Target Form} \\ \end{align}$

All that remains is to apply the quadratic formula and simplify:

$$ \bar \mu = \frac{-(-2c-z^2) \pm \sqrt{(-2c-z^2)^2-4c^2}}{2} \\ \bar \mu= \frac{2c+z^2 \pm \sqrt{4c^2+4cz^2+z^4-4c^2}}{2} \\ \bar \mu= \frac{2c+z^2 \pm \sqrt{z^4 +4cz^2}}{2} \quad \square\\ $$

With Context: Normal (Gaussian) approximation to the Poisson distribution.
The ideas behind the equations in your question come from this approximation.

If $X \sim \text{Poisson}(\lambda)$ then $$X \approx \text{Normal}(\mu = \lambda,\; \sigma = \sqrt{\lambda}) \tag{1}$$ for a sufficiently large $\lambda$ value. As $\lambda$ increases, the approximation gets better and better.

One way to understand this is a $\text{Poisson}(20)$ random variable is equivalent to adding up the results of 20 independent and identically distributed (i.i.d.) $\text{Poisson}(1)$ random variables. This permits invoking the hallowed Central Limit Theorem which immediately results in Equation $(1)$.

Related:
On CrossValidated (Stats.SE): Normal approximation to the Poisson distribution

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