# CLT for sums of Fourier transform of white noises r.v

Define

$$I_n(\lambda_j) = \frac{1}{2\pi n} |\sum_{t = 1}^n Z_t e^{it\lambda_j}|^2 = \frac{1}{2 \pi} \sum_{h = - \infty}^{\infty} \hat{\gamma}_n(h) e^{ih\lambda_j}$$

where $$Z_t$$ is a $$WN \sim (0, \sigma^2)$$ and $$\gamma(h) = cov(Z_{t}, Z_{t - h})$$ is the autocovariance of the process at lag $$h$$, $$\hat{\gamma}_n(h)$$ is sample counterpart. Denote, $$\boldsymbol{\gamma}_h = (..., \gamma(0), \gamma(1), ...)^T$$, we know that the vector $$\sqrt{n}(\hat{\boldsymbol{\gamma}_h} - \boldsymbol{\gamma}_h)$$ is asymptotically normally distributed with a covariance which is of order $$O(\frac{1}{n})$$.

I would like to prove a CLT for sums of the $$I_n(\lambda_j)$$, i.e.

\begin{aligned} \sum_{j=1}^m I_n\left(\lambda_j\right) & =\frac{1}{2 \pi n} \sum_{j=1}^m \sum_{h=-\infty}^{\infty} \hat{\gamma}(h) e^{i h \lambda_j} =\frac{1}{2 \pi n} \sum_{h=-\infty}^{\infty} \hat{\gamma}(h) \sum_{j=1}^m e^{i h \lambda_j} \end{aligned}.

Since the whole vector $$\boldsymbol{\hat{\gamma}_h}$$ is A.N. we can expect that the infinite sums is A.N as well, however I do not know how to prove it. Any ideas?

• This seems to rely basically on a theorem that would state that the asymptotic behaviour of a sum of variables is equal to the sum of the asymptotic limits of those variables. I don't recall a name for such theorem, but I would be surprised if something like that doesn't exist. Commented Mar 1, 2023 at 10:17
• I am not sure about it because convergence in distribution is quite weak. If $X \rightarrow^d Z$ and $Y \rightarrow^d W$ then it is not true that $X + Y \rightarrow^d Z + W$. Commented Mar 1, 2023 at 10:20
• "and $\gamma(h)$ is the autocovariance of the process at lag $h$, $\hat{\gamma}_n(h)$ is sample counterpart" the context here is not clear. Several people might be able to directly imagine the context, but could you explain this part to a wider public. What is 'the process'? Commented Mar 1, 2023 at 10:21
• This case when it is not true is that when $X_n$ and $Y_n$ are dependent? For independent variables I wouldn't see why it is untrue. Commented Mar 1, 2023 at 10:27
• I seem to be thinking about something like Slutsky's theorem. But that is indeed a more restricted case. However, I imagine that for sums of independent variables it can be extended. Commented Mar 1, 2023 at 10:30