Convergence of $\operatorname{Binomial}(1/\sqrt n, n)$ Let $Z_n \sim \operatorname{Binomial}(1/\sqrt n, n)$. Is the following expression asymptotically normally distributed for some normalization rate $r_n$?
$$r_n\left(Z_n - \mathbb E Z_n \right )$$
It's easy to show that $\mathbb E Z_n = \sqrt n$. The typical rate is $\sqrt n$, but a rate other than that might be required here.
What I've tried
Computing the moment generating function of the desired expression:
$$\mathbb E \exp \left ( \lambda r_n\left(Z_n - \mathbb E Z_n \right )\right ) = e^{-\lambda r_n \sqrt n}\left( 1 + \frac {r_n} {n} \left( e^{\lambda r_n /n} - 1 \right )\right)^n$$
where the terms in parens on the RHS grows exponentially. I dont think there is any rate $r_n$ such that the expression under consideration is asymptotically normal.
 A: The log-moment generating function under consideration is given by:
$$-\lambda r_n n \theta_n + n \log(1-\theta_n + \theta_n e^{\lambda r_n})$$
Taking a taylor expansion of the exponential we have the following inside the argument of the logarithm:
$$1 - \theta_n + \theta_n ( 1 + \lambda r_n + (\lambda r_n)^2/2 + o(r_n^2)) = 1 +  \theta_n ( \lambda r_n + (\lambda r_n)^2/2 + o(r_n^2))$$
Now returning to our log-mgf and taking a taylor expansion of the $\log(1 + y)$:
$$-\lambda r_n n \theta_n + n\left(\theta_n\left[\lambda r_n + (\lambda r_n)^2/2 + o(r_n^2)\right] - y^2/2 + y^3/3 + o(y^3)\right) = 
n\left(\theta_n \left[(\lambda r_n)^2/2 + o(r_n^2))\right] - y^2/2 + y^3/3 + o(y^3)\right)$$
Notice that $y^2 = o(\theta_n^2 r_n^2)$ and that $n \, \theta_n = \sqrt n$ so we'll choose the norming rate $r_n = n^{-1/4}$ so that the expression converges to a non-zero value. Now, the expression reduces to:
$$\lambda^2 /2 + o(1)$$
Which is the log-mgf of a normal with mean 0 and variance 1. Computer simulations verify this:
n = 10000
outs = [np.power(n, 1/4) * ((np.random.rand(n) < 1/np.sqrt(n)).sum()/np.sqrt(n) - 1) for i in range(100000)]
plt.hist(outs, bins=30, alpha= 0.5, density=True)
plt.hist(np.random.normal(size=10000), alpha= 0.5, density=True)
plt.show()


And as expected it converges faster than a binomial (whose norming rate $r_n=1$)
plt.hist((np.random.binomial(n, 0.5, size=len(outs)) - n/2)/np.sqrt(n/4), bins=30, density=True, alpha=0.5, label='bin')
plt.legend()


