TransformedDistribution using $k$ iid random variables, but $k$ not fixed

Question

Can I create a TransformedDistribution that uses $k$ independent identically distributed (i.i.d.) random variables where $k$ is not fixed?

This question is closely related to TransformedDistribution using $k$ iid random variables.

The answer https://mathematica.stackexchange.com/a/65769/34820 states how the TransformedDistribution can be derived for an arbitrary, but fixed $k$:

iid[k_, dist_] := TransformedDistribution[
                    Sum[a[i], {i, k}],
                    Table[Distributed[a[i], dist], {i, k}]
                  ]

My question is whether I can also derive a general result for $k$ being any natural number.

To put a toy example: Let's take the sum of $k$ Bernoulli distributed variables with success probability $p$. This sum would be distributed according to a BinomialDistribution[k,p]. Using the code from above, I can get:

In[1]:= iid[5, BernoulliDistribution[p]]

Out[2]= BinomialDistribution[5, p]

But if I try a general $k$, I get an error (as expected):

In[2]:=  iid[k, BernoulliDistribution[p]]

During evaluation of In[2]:= Table::iterb: Iterator {i,k} does not have appropriate bounds.

I tried

Assuming[N \[Element] Integers && N > 0, iid[k, BernoulliDistribution[p]]]

but this gave the same error. Is there a way to get the general result BinomialDistribution[k,p] using Mathematica?

Answer 1

Still no automatic way seems possible for a variety of reasons.

Sometimes using the characteristic function (or the moment generating function) will result in a recognizable distribution. Well, recognizable to you and maybe not to Mathematica.

Here is an example. A Cauchy random variable has characteristic function

cf = CharacteristicFunction[CauchyDistribution[a, b], t]
(* E^(I a t - b t Sign[t]) *)

To find the characteristic function of the sum of $n$ iid (independent and identically distributed Cauchy random variables with parameters $a$ and $b$ one raises cf to the power $n$:

cfn = Simplify[cf^n, Assumptions -> n \[Element] PositiveIntegers] // ExpandAll
(* E^(I a n t - b n t Sign[t]) *)

That is recognized (by you) as the characteristic function of a Cauchy random variable with parameters $a n$ and $b n$.

Answer 2

How about the following?

Table[iid[k, BernoulliDistribution[p]], {k, 1, 10}]

{BernoulliDistribution[p], BinomialDistribution[2, p], BinomialDistribution[3, p], BinomialDistribution[4, p], BinomialDistribution[5, p], BinomialDistribution[6, p], BinomialDistribution[7, p], BinomialDistribution[8, p], BinomialDistribution[9, p], BinomialDistribution[10, p]}

Then

FindSequenceFunction[Table[iid[k, BernoulliDistribution[p]][[1]], {k, 2, 10}], n] /. 
n -> n - 1

n