2
$\begingroup$

In wikipedia about the Pascal triangle: Relation to binomial distribution

"When divided by 2n, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases."

How can I find the distribution approached by the rows of the Stirling numbers of the second kind?

$$\begin{array}{llllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 3 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 7 & 6 & 1 & \text{} & \text{} & \text{} & \text{} \\ 1 & 15 & 25 & 10 & 1 & \text{} & \text{} & \text{} \\ 1 & 31 & 90 & 65 & 15 & 1 & \text{} & \text{} \\ 1 & 63 & 301 & 350 & 140 & 21 & 1 & \text{} \\ 1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1 \end{array}$$

After all the Stirling numbers of the second kind satisfy the recurrence: $$\left\{{n+1\atop k}\right\} = k \left\{{ n \atop k }\right\} + \left\{{n\atop k-1}\right\}$$

Where as the Pascal triangle satisfies the recurrence:

$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$

If such a distribution does exist I would like to find the x value for the maximum of the distribution for each row and see if it says anything about x/LambertW(x).

$\endgroup$

1 Answer 1

1
$\begingroup$

I think this is what you are looking for:

Stirling Behavior is Asymptotically Normal, L. H. Harper, The Annals of Mathematical Statistics Vol. 38, No. 2 (Apr., 1967), pp. 410-414

Link

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .