How can I obtain the mean of a Poisson distribution given the first improbable point of the distribution?

Question

I generated a Poisson distribution with mean equal to 3 and 10000 samples by using np.random.poisson(3,10000). The plot is the following:

from this plot I see that given the mean equal to 3, a point like 12 is very unlikely to fall into the distribution. Is there a function that given the number 12 and a probability to fall in the distribution, gives me the mean of such distribution ?

I know that from scipy.stats module there is the ppf function that returns the point such that given a mean of the distribution and the probability to have this point and the points lower than this, that is:

st.poisson.ppf(q = 0.99995,mu = 3)

It returns 12, hence I have a 99.995% probability that a Poisson with mean 3 returns a value that is 12 or less. Is there a function that does the opposite ? That given 0.95 and the number 12, gives me the right mean 3 ?

Answer 1

The cumulative distribution function (CDF) for the Poisson distribution is:

\begin{equation} CDF = exp(-\lambda)*\Sigma_{i=0}^{i=|k|}\frac{\lambda^i}{i!} \end{equation}

It sounds to me that you have particular values for $CDF$ and $k$ and you want to solve for $\lambda$.

You can do this using scipy.optimize.fsolve https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.fsolve.html?highlight=fsolve

from scipy.optimize import fsolve
from scipy.stats import poisson

def myCallable(lambda_):
    return 0.95-poisson.cdf(12, lambda_)

mu = fsolve(myCallable, x0=[2.5])

I got $\lambda = 7.68957829$ for $CDF=0.95$ and $k=12$. I double-checked this using poisson.ppf that you referenced in your question and the value is correct, so I'm not sure where the value of $3$ is coming from. Hope this helps