I have a Poisson process where new elements arrive to a set with Poisson intensity $\lambda$. Initially, there are $N\_0$ elements in the set. The probability that there are $N\_0 + M$ elements in the set at time $t$ is $Pr[N(t) = N\_0+M] = \frac{(\lambda t)^M}{M!} e^{-\lambda t}$.


I'm interested in the expected number of subsets of size $k$. That is, I want to compute


$$f(k,t) = \sum\_{M=0}^{\infty} \frac{(\lambda t)^M}{M!} e^{-\lambda t} \binom{N\_0+M}{k}$$


Is there some formula or approximation for this expectation when $N\_0$ is large?



Expected size of binomial coefficient with Poisson arrivals?
Suppose that a real $c:=\lambda t>0$ and a natural $k$ are fixed, whereas $n:=N\_0\to\infty$.
Take any real $m>0$. Then
$$f(k,t)=g\_m(k,t)+h\_m(k,t),$$
where
$$g\_m(k,t):=e^{-c}\sum\_{0\le j<m}\frac{c^j}{j!}\,\binom{n+j}k,$$
$$h\_m(k,t):=e^{-c}\sum\_{j\ge m}\frac{c^j}{j!}\,\binom{n+j}k.$$
For each $j$, $\binom{n+j}k\sim n^k/k!$, whence
$$g\_m(k,t)\sim P\_m\frac{n^k}{k!},$$
where
$$P\_m:=e^{-c}\sum\_{0\le j<m}\frac{c^j}{j!}.$$
Also, $\binom{n+j}k\le(n+j)^k/k!=O(n^k+j^k)$ for all $j$, whence
$$h\_m(k,t)=O\Big((1-P\_m)n^k+\sum\_{j\ge0}\frac{c^j}{j!}\,j^k\Big)
=O\Big((1-P\_m)n^k+1\Big)
=O\big((1-P\_m)n^k\big).$$
Letting now $m\to\infty$ and noting that then $P\_m\to1$, we conclude that
$$f(k,t)\sim\frac{n^k}{k!}.$$


