Limit of series with poisson probability mass function

[veneratifer]

What is the limit (if any exists) of this formula? Is it 1/2?​

[math]\lim_{\lambda\rightarrow\infty}\sum_{k=\lambda}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}[/math]

 

[BigBeachBanana]

What is the limit (if any exists) of this formula? Is it 1/2?​

[math]\lim_{\lambda\rightarrow\infty}\sum_{k=\lambda}^{\infty}\frac{\lambda^{k}e^{-\lambda}}{k!}[/math]

Your summation doesn't make much sense.

1) [imath]\lambda[/imath] 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 [imath]\lambda[/imath]?

 

[veneratifer]

Your summation doesn't make much sense.

1) [imath]\lambda[/imath] 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 [imath]\lambda[/imath]?

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 [imath]\geq\lambda[/imath] times, so I sum PMF from [imath]\lambda[/imath] to [imath]\infty[/imath] (the reason why lower bound of the summation start at [imath]\lambda[/imath])

But when [imath]\lambda[/imath] 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.

 

[BigBeachBanana]

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 [imath]\geq\lambda[/imath] times, so I sum PMF from [imath]\lambda[/imath] to [imath]\infty[/imath] (the reason why lower bound of the summation start at [imath]\lambda[/imath])

Essentially you're looking for the probability of the random variable being greater than or equal to its mean i.e. [imath]\lambda[/imath]. Typically, we simply write [imath]\Pr(X\ge \lambda | \lambda = c)[/imath] , where [imath]c[/imath] is some positive real number.

Since [imath]\lambda[/imath] 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 [imath]\lambda[/imath] is an integer.

You're correct that Poisson is a skewed distribution since its mean is higher than its median. However, as [imath]\lambda[/imath] becomes large, we can approximate the Poisson with the Normal distribution by the Central Limit Theorem.

 

[BigBeachBanana]

I generated random samples with size 10,000 for various [imath]\lambda 's[/imath]. 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.

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()