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Let $X$ be a random variable with a Poisson distribution, such that $P(X>2) = 0.3$. Find $P(X<2)$.

This is bunched in with some easy Poisson questions in the textbook. I don't see how to answer this easily though.

$ e^{-\lambda} + e^{-\lambda}\lambda + \frac{e^{-\lambda}\lambda^2}{2} = 0.7,\ $ but I don't see where to go from here without going into numerical methods, which is definitely not what they want...

Adam Rubinson
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    I don't see a clever way to dodge the numerical work. It yields $\lambda = 1.91378$ which gives $P(X<2)=0.429845777$ and those numbers don't suggest anything obvious. – lulu Nov 24 '20 at 01:15
  • Well that does match the answer in the back of the book: $0.430$. But if they want us to do the numerical work then it is out of place in the entire book, let alone the exercise. So perhaps we are both missing something. – Adam Rubinson Nov 24 '20 at 01:17
  • Can't help you there. The numerical work is not terribly difficult...Wolfram Alpha can do it, for instance. – lulu Nov 24 '20 at 01:19
  • Sure, but I know that they don't want us to use numerical methods. So I'm more interested in what method they want us to use. – Adam Rubinson Nov 24 '20 at 01:21

2 Answers2

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A crude but surprisingly effective way to solve for the rate parameter $\lambda$ is to set up a recursion. Given $$\Pr[X \le 2] = e^{-\lambda}(1 + \lambda + \lambda^2/2) = 0.7,$$ we may write this as $$\lambda = \log \frac{1 + \lambda + \lambda^2/2}{0.7}$$ which induces a recursion relation $$\lambda_{n+1} = \log \frac{1 + \lambda_n + \lambda^2_n/2}{0.7}.$$ Then with an initial guess, say $\lambda_0 = 1$, it is relatively simple to implement this recursion in a typical pocket calculator; e.g. first entering 1, then entering ln((1 + Ans + Ans^2/2)/0.7), then repeating this entry by pressing =. By the thirtieth iterate, we get at least six digits of precision, sufficient to compute $\Pr[X < 2]$.

We can improve upon the initial guess by noting that for $\lambda = 1$, $\Pr[X \le 2] \approx 0.91$, which is too large; but $\lambda = 2$ gives $\Pr[X \le 2] \approx 0.676$, much closer but on the other side of $0.7$; so we know that the precise value must be slightly less than $2$, say $\lambda_0 = 1.9$. This saves us about $15$ iterations.

heropup
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  • This counts as a "numerical method" though. I've got a feeling they wrote something in the question wrong for what they intended it to be. – Adam Rubinson Nov 25 '20 at 11:56
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    @AdamRubinson As others have already pointed out, there is no elementary closed form solution for the equation $\Pr[X \le 2] = 0.7$. My answer is about the simplest way you could solve for $\lambda$ without using a computer. If your question is whether the author made a mistake, I can't speak to that, since all one can do is speculate. – heropup Nov 25 '20 at 19:04
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I suppose that graphing you noticed that the solution is close to $\lambda=2$.

On the other hand, you can rewrite the equation as $$e^\lambda=\frac{5}{7} \left(\lambda ^2+2 \lambda +2\right)$$ Make $\lambda=x+2$ to face $$e^2\, e^x=\frac{5}{7} \left(x^2+6 x+10\right)$$ Since $x$ is small rewrite it as $$e^2 \left(1+x+\frac{x^2}{2}+O\left(x^3\right)\right)=\frac{5}{7} \left(x^2+6 x+10\right)$$ Ignoring the higher order terms, you face a quadratic equation in $x$ $$\left(14 e^2-100\right)+\left(14 e^2-60\right) x+\left(7 e^2-10\right) x^2=0$$ and the approximate solution is then $$x=\frac{60-14 e^2+\sqrt{-400+1680 e^2-196 e^4}}{14 e^2-20}\approx -0.0865227$$ which implies $\lambda \approx 1.913473$

Claude Leibovici
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