Poisson process, finding time t given $\lambda$ and $P$

Question

A typical Poisson process denoted by $N_t$ refers to the number of arrivals (discrete) at time $t$, with rate $\lambda$, where

$$P(N_t = k) = \frac{e^{\lambda t} (\lambda t)^k}{k!}$$

If it is given that $\lambda = 30$, $P = .3$, and $k = 1$, what is time $t$?

Substituting into the probability formula you have

$$.3 = e^{30t}30t$$

How do you solve this for $t$?

Answer 1

Solving numerically should be quite easy with newton's method to get a zero of $$ f(t)=e^{30\,t}\,30\,t-.3=0\Rightarrow t\approx 0.00789184. $$

Answer 2

You need to find the zero of function $$f(t)=30\,t \,e^{30t}-\frac 3 {10}$$ For the time being, let $x=30t$ and the problem becomes $$g(x)=xe^x-\frac 3 {10}\implies g'(x)=(x+1)e^x > 0 \quad \forall x >0$$ You could use Taylor expansion which gives $$g(x)=-\frac 3 {10}+x+x^2+O\left(x^3\right)$$ Ignoring the higher order terms, then $$x=\frac{1}{10} \left(\sqrt{55}-5\right)\implies t=\frac{1}{300} \left(\sqrt{55}-5\right)\approx 0.008054$$

You could do better using instead the $[2,2]$ Padé approximant which would make $$g(x)\sim \frac{x+\frac{x^2}{3}}{1-\frac{2 x}{3}+\frac{x^2}{6}}-\frac 3 {10}$$ Solving the quadratic equation would then give $$x=\frac{3}{17} \left(\sqrt{178}-12\right)\implies t=\frac{1}{170} \left(\sqrt{178}-12\right)\approx 0.007892$$ which does not seem to be too bad.