Bloody Poisson question. $X \sim Po(\lambda). P(X>2) = 0.3$. Find $P(X<2)$.

Question

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...

Answer 1

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 $\lambda \approx 1.913473$

Answer 2

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.