Rule for indices of expected values Is it forever and always the case that for a given distribution, to use the E[X] formula as well as LOTUS, that the formula is only valid if you set the indices to the domain of the original distribution? I realize I might be using the wrong language, but as an example to illustrate:
Let's say I'm trying to figure out something like $\sum_{n=1}^{\infty}5\cdot\frac{e^{-2}2^n}{n!}$
Clearly the right hand side of the equation represents $Pois(2)$
In order for me to solve this using E[N], would it be necessary to set the beginning index to n=0?
Likewise is this true for E[X] for all probability distributions?
 A: In R, the first 101 terms of the expression for $E(X)$ are as follows:
lam = 2; i = 0:100;  pdf=exp(-lam)*lam^i/factorial(i)

The vector pdf [of probabilities $f(i)\,]$ can also be computed using the R function dpoisfor a Poisson PDF.
PDF = dpois(i, lam)

Here is a table of the first six values:
lam = 2; i = 0:100;  pdf=exp(-lam)*lam^i/factorial(i)
PDF = dpois(i, lam)
head(cbind(i, pdf, PDF))
     i        pdf        PDF
[1,] 0 0.13533528 0.13533528
[2,] 1 0.27067057 0.27067057
[3,] 2 0.27067057 0.27067057
[4,] 3 0.18044704 0.18044704
[5,] 4 0.09022352 0.09022352
[6,] 5 0.03608941 0.03608941

The exact computation of $E(X) = \sum_{i=0}^\infty if(i) = \lambda = 2,$
where $f(i)$ is the PDF. But we can get very close by summing the first 101 terms:
sum(i*pdf)
[1] 2

Even the first 10 terms would be good enough for most practical purposes.
sum((0:9)*pdf[1:10])
[1] 1.999525

You can leave off the first term because it is multiplied by $0.$  That is, $\sum_{i=0}^\infty if(i) = 0\cdot f(0)+ \sum_{i=1}^\infty if(i) = \sum_{i=1}^\infty if(i).$
However, $$P(X \le 5) = P(X=0)+P(X-1)+\cdots+P(X=5)\\ = \sum_{i=0}^{5} f(i) = 0.9834364$$ and you must add all six terms--including the first--to get the right answer.
In R, you can uxe:
sum(dpois(0:5, 2))
[1] 0.9834364
ppois(5, 2)    $ 'ppois' is a Poisson CDF
[1] 0.9834364

