Calculating Poisson CDF I have a quite basic question about Poisson CDF. I have the following function $P(x;\mu)$ referring to the Poisson CDF.
Hadley and Whitin (1963) defined the Poisson density as
$p(x;\mu)= \frac{\mu^x}{x!}e^{-\mu}$
and complementary cumulative distribution function as
$P(x;\mu)=\sum_{j=x}^{\infty}\frac{\mu^j}{j!}e^{-\mu}$.
To calculate $P(x;\mu)$, can I use the following:
$P(x;\mu)=1-\sum_{j=0}^{x-1}\frac{\mu^j}{j!}e^{-\mu}$
knowing that
$\sum_{j=0}^{\infty}\frac{\mu^j}{j!}e^{-\mu}=1$?
I have seen several applications of CDF calculation online and they made me confused. Can you help to verify?
 A: Yes, you simply sum the probabilities up to $P(X=x-1)$ and subtract it from $1$, to obtain $P(X \geq x)$ which is correct. However, CDF normally is defined as $P(X\leq x)$, however note that, in your book (or post), it is defined as $P(X\geq x)$. Check the definition out from here. Your book may be consistent about its definition, which is perfectly OK if you stay within the book, but trying to compare what you do in your exercises and the outside world, you may find little discrepancies.
A: A cumulative distribution function is defined as $$F(x)=\mathbb{P}(X\le x)$$or more rarely as$$F(x)=\mathbb{P}(X< x)$$in older (e.g., French or Hungarian) texts, but never as the complement$$F(x)=\mathbb{P}(X\ge x)$$which is the complementary cumulative distribution or the tail distribution.
A: Another option to calculate the Poisson CDF is to use its mathematical correspondence with the Chi-square distribution. Let $X \sim Poisson(\mu)$. Then $P(X \leq x_0) = P(Y > 2\mu)$ where $Y \sim \chi^2_{2(x_0+1)}$. 
For example, in R both of the following equal .61596.  
ppois(5,5)

1-pchisq(10,12)

