Why does Chinese Restaurant Table Distribution look like a Gaussian Distribution?

Question

The Chinese Restaurant Table Distribution describes the probability distribution for the number of non-empty tables in the Chinese Restaurant Process after $T$ customers have been seated. Specifically, if $K_T$ is the (random) number of non-empty tables after seating $T$ customers in a $CRP(\alpha)$, its PMF is defined as

$$P(K_T = k) = \frac{\Gamma(\alpha)}{\Gamma(\alpha + T)} |S(T, k)| \alpha^k$$

where $k \in \{1, ..., T\}$ and $|S(\cdot, \cdot)|$ are unsigned Stirling numbers of the first kind. I discovered that when I fix $\alpha$ and plot the distribution for increasing values of $T$, the distribution looks more and more like a Gaussian. For alpha=10.01, I plot the distribution from T=1 (red) to T=50 (blue):

My question is: why does the Chinese Restaurant Table distribution look like a Gaussian distribution for medium-to-large $T$?

Answer 1

We can apply the Lyapounov CLT to show that $K_T$ is normal for large $T$. Let $$ K_T = \sum_{i=1}^T I(\text{new table at $i$}) = \sum_{i=1}^T Z_i. $$ Then it is well-known that $Z_i \stackrel{\text{ind}}{\sim} \text{Bernoulli}(p_i)$ where $p_i = \alpha / (\alpha + i - 1)$. Now, $\sum_{i=1}^\infty p_i (1 - p_i) = \sum_{i=1}^\infty \frac{\alpha i}{(\alpha + i)^2} = \infty$ by comparison to the harmonic series. Hence we can apply the Lyapnouv CLT to conclude that $$ \frac{K_T - E(K_T)}{\sqrt{\text{Var}(K_T)}} \to N(0,1) $$ in distribution, using the argument outlined here to verify the Lyapounov condition for Bernoulli random variables.