# Donsker Theorem like result for sum of Sums

Recently I came up with the following question. Let's say that $$X_i$$ are i.i.d random variables with $$\mathbb{E}[X_1] = 0$$ and $$\mathbb{V}\text{ar}(X_1) = 1$$. We know by Donkser Theorem that: $$\frac{S_n}{\sqrt{n}} \to B(1) \sim N(0,1)$$ This case is Donsker with $$t=1$$ and using the special fact that $$\mathbb{E}[X_1] = 0$$ and $$\mathbb{V}\text{ar}(X_1) = 1$$. Now my question is how can I find the following: $$\lim_{n \to \infty} \frac{1}{n^{3/2}}\sum_{k=1}^n S_k$$ My first idea was passing $$n^{1/2}$$ inside and finding: $$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{S_k}{n^{1/2}} = \lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^n \frac{S_k}{k^{1/2}}\cdot \frac{k^{1/2}}{n^{1/2}}$$ But from here on out I have no idea how to procced. I may be wrong but it looks like I will have many Normal like distributions all scaled by a certain factor and then I take the mean of all of them. But I am lost in how I can find a more closed value for the limit. Any ideas?

As pointed out by @Mindlack, the convergence is possible only in distribution. Here are some alternative ways of computing the limit distribution:

1st Solution. For each $$n \geq 1$$ and $$1 \leq k \leq n$$, write

$$X_{n,k} = \frac{n+1-k}{n^{3/2}} X_k \qquad \text{and} \qquad T_n = \frac{1}{n^{3/2}} \sum_{k=1}^{n} S_k = \sum_{k=1}^{n} X_{n,k} .$$

Then $$\mathbf{E}[T_n] = 0$$ and $$\mathbf{Var}(T_n) = \frac{n(n+1)(2n+1)}{6n^3} \to \frac{1}{3}$$ as $$n\to\infty$$. Moreover, for each $$\varepsilon > 0$$,

\begin{align*} \sum_{k=1}^{n} \mathbf{E}\bigl[ X_{n,k}^2 \mathbf{1}_{\{|X_{n,k}| > \varepsilon\}} \bigr] &= \sum_{k=1}^{n} \mathbf{E}\biggl[ \frac{k^2 X_1^2}{n^3} \mathbf{1}_{\{k|X_1| > \varepsilon n^{3/2}\}} \biggr] \\ &\leq \sum_{k=1}^{n} \mathbf{E}\biggl[ \frac{n^2 X_1^2}{n^3} \mathbf{1}_{\{n|X_1| > \varepsilon n^{3/2}\}} \biggr] \\ &= \mathbf{E} \bigl[ X_1^2 \mathbf{1}_{\{|X_1| > \varepsilon n^{1/2}\}} \bigr] \\ &\to 0 \quad \text{as} \quad n \to \infty. \end{align*}

So by the Lindberg CLT, $$T_n$$ converges in distribution to a normal distribution with mean $$\lim_n \mathbf{E}[T_n] = 0$$ and variance $$\lim \mathbf{Var}(T_n) = \frac{1}{3}$$.

2nd Solution. Writing

$$T_n = \sum_{k=1}^{n} \frac{S_k}{\sqrt{n}} \cdot \frac{1}{n}, ​$$

an application of Donsker's invariance principle shows that this converges in distribution to

$$\int_{0}^{1} W_t \, \mathrm{d}t, ​$$

where $$(W_t)_{t\geq 0}$$ is a standard Brownian motion in 1D. Being a "linear combination" of jointly normal variables, this integral is again a normal variable. Then

$$\mathbf{E}\biggl[\int_{0}^{1} W_t \, \mathrm{d}t\biggr] = \int_{0}^{1} \mathbf{E}[W_t] \, \mathrm{d}t = 0$$

and

$$\mathbf{Var}\biggl(\int_{0}^{1} W_t \, \mathrm{d}t\biggr) = \int_{0}^{1} \int_{0}^{1} \mathbf{Cov}(W_s, W_t) \, \mathrm{d}s \mathrm{d}t = \int_{0}^{1} \int_{0}^{1} (s \wedge t) \, \mathrm{d}s \mathrm{d}t = \frac{1}{3},$$

hence we have $$\int_{0}^{1} W_t \, \mathrm{d}t \sim \mathcal{N}(0, \tfrac{1}{3})$$.

Unless I’m quite mistaken, the convergence of $$n^{-1/2} S_n$$ to $$B(1)$$ is only in distribution. So if we’re interested in the convergence of $$V_n=n^{-3/2}\sum_{k=1}^n{S_k}$$, it should probably work only in distribution.

Let $$f(t)$$ be the expected value of $$e^{itX_1}$$. Then the expected value $$e_n(t)$$ of $$e^{itV_n}$$ is $$\prod_{k=1}^n{f\left(\frac{kt}{n^{3/2}}\right)}$$.

So if $$t$$ is fixed and $$n$$ is large enough, $$\ln{e_n(t)}$$ is well defined and is the sum of the (well-defined) $$\ln{f(kt/n^{3/2})}$$ for $$1 \leq k \leq n$$. But as $$X_1$$ is $$L^2$$ with null expected value and variance one, $$\ln{f(kt/n^{3/2})}=-k^2t^2/n^3 + o(k^2/n^3)$$. It follows that for large enough $$n$$, $$\ln{e_n(t)}=o(1)-t^2/3$$, ie $$e_n(t) \rightarrow e^{-t^2/3}$$.

Thus $$V_n$$ seems to converge in distribution to a certain universal Gaussian.

• Oh my bad, I should've asserted convergence in distribution by Donsker. Just a question here. How could you conclude that $\ln f(kt/n^{3/2}) = -k^2t^2/n^3 + o(k^2/n^3)$ and the following result $\ln e_n(t) = o(1) - t^2/3$? Aug 29 at 22:11
• $X_1$ is $L^2$ with null expected value and variance $1$ so its characteristic function $c$ is $C^2$ and $c(u)=1-u^2/2+o(u^2)$ by Taylor as $u \rightarrow 0$. Then I set $u=kt/n^{3/2}$. That gives the expansion for $\ln{f(kt/n^{3/2})}$. To get to $\ln{e_n(t)}$, that’s just the sum of the $\ln{f(kt/n^{3/2})}$ for $1 \leq k \leq n$. Aug 30 at 9:20
• There is still a small thing that is strange to me. You still put the expansion $\ln{f(kt/n^{3/2})} = -k^2t^2/n^3 + o(k^2/n^3)$ without the $\ln$ on the other side. Is there a reson for this? Aug 30 at 14:13
• Yes: $\ln(1-x)=-x+o(x)$ when $x$ goes to zero. Aug 30 at 14:18