Suppose that each random variable $B_i, i \in \mathbb{N}$ in an iid sequence of random variables $B_n$ has a Poisson($i$) distribution. How could I find the limiting distribution of $G_n = \sqrt{B_n}$?
My first idea was to see if $B_n$ converges in distribution to some limiting distribution $B$, and then use that fact to show that $G_n$ converges to some limiting distribution $G = \sqrt{B}$. However, $B_n$ does not converge to anything, so I can't use that line of reasoning.
How could I approach this problem?
$$ \frac{B_i - i}{\sqrt i} \overset{\mathcal L}\longrightarrow N(0,1) \text{ as } i\to\infty. $$ $$ f(B_i) \approx f(i) + f'(i)(B_i-i) $$ $$ \frac{f(B_i) - f(i)}{ f'(i) \sqrt i} \approx N(0,1) $$ $$ f(x) = \sqrt x $$ $$ f'(x) = \frac 1 {2\sqrt x} $$ $$ f'(i)\sqrt i = \frac 1 2 $$ $$ 2(\sqrt{B_i} - \sqrt i\,) \approx N(0,1) $$ Notice that this implies there is no limiting distribution of $\sqrt{B_i}.$ There is, however, an "asymptotic distribution" of $\sqrt{B_i}.$ It is interesting that the variance of $\sqrt{B_i}$ remains bounded as $i$ grows.
The argument above is not fully rigorous; you need to cite the appropriate theorems. They can be found in Serfling's Limit Theorems of Mathematical Statistics.