Limit of series with poisson probability mass function

veneratifer

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What is the limit (if any exists) of this formula? Is it 1/2?​
limλk=λλkeλk!\lim_{\lambda\rightarrow\infty}\sum_{k=\lambda}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}
 
What is the limit (if any exists) of this formula? Is it 1/2?​
limλk=λλkeλk!\lim_{\lambda\rightarrow\infty}\sum_{k=\lambda}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}
Your summation doesn't make much sense.

1) λ\lambda is a constant parameter, you can't take the limit.
2) For the lower bound of the summation, k normally start at 0. Can you explain why is it starting at λ\lambda?
 
Your summation doesn't make much sense.

1) λ\lambda is a constant parameter, you can't take the limit.
2) For the lower bound of the summation, k normally start at 0. Can you explain why is it starting at λ\lambda?
Thank you for reply !

Probably you are right - my summation have no sense, but I don't know how to write this formally correct. My idea is as follow:
there is a poisson distribution of some event, I want to calculate probability that event occurs λ\geq\lambda times, so I sum PMF from λ\lambda to \infty (the reason why lower bound of the summation start at λ\lambda)

But when λ\lambda is large poisson distribution getting more and more similar to gaussian distribution, which is ideal symmetric.
Subsequently
1) Probability must sum to 1
2) Half of probabilities sum of ideal symmetric distribution = 1/2 ?
I search for analytical solution of this problem.
 
Thank you for reply !

Probably you are right - my summation have no sense, but I don't know how to write this formally correct. My idea is as follow:
there is a poisson distribution of some event, I want to calculate probability that event occurs λ\geq\lambda times, so I sum PMF from λ\lambda to \infty (the reason why lower bound of the summation start at λ\lambda)
Essentially you're looking for the probability of the random variable being greater than or equal to its mean i.e. λ\lambda. Typically, we simply write Pr(Xλλ=c)\Pr(X\ge \lambda | \lambda = c) , where cc is some positive real number.

Since λ\lambda can be non-integer, we can't use it as an index and thus don't use the sigma notation in the general case. However, it is perfectly ok when λ\lambda is an integer.

You're correct that Poisson is a skewed distribution since its mean is higher than its median. However, as λ\lambda becomes large, we can approximate the Poisson with the Normal distribution by the Central Limit Theorem.
 
I generated random samples with size 10,000 for various λs\lambda 's. The plots were made using Python packages. See the code below if you're interested in replicating it and testing it yourself. Final note, if λ is greater than about 10, then the Normal Distribution is a good approximation if an appropriate continuity correction is performed.
Screen Shot 2022-12-30 at 10.47.43 PM.png

Python:
import seaborn as sns
from matplotlib import pyplot as plt
import numpy as np
import math

plt.close('all')
lambdas = [.5,1,5,20,100,500] #Change different values of lambda here
f, ax = plt.subplots(2,3, figsize = (12,5))

for i,rate in enumerate(lambdas):
    plt.subplot(2,3,i+1)
    data = np.random.poisson(rate, size=10000)
    plt.hist(data, bins=50, linewidth=1,histtype = 'stepfilled')
    plt.title('\u03BB = '+ str(rate))
plt.subplots_adjust(hspace=0.5)
plt.show()
 
Last edited:
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