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Let $\gamma_1,\gamma_2,\ldots$ be uncorrelated random variables with $\mathbb E[\gamma_k]=0, \mathbb E[\gamma_k^2]=c_k$ and $\sum_{k\geq 1} c_k < \infty$. Define

$$X(t) = \sum_{k=0}^\infty \gamma_k \exp(itk)$$

What is the spectral representation of $X(t)$?

$X(t)$ is weakly stationary since $\mathbb E[X(t)]\equiv 0$ and $CoV(X(t+s),X(t)) = \sum_{k\geq 1}c_k \exp(isk)=:b(s)$, which is a function of the time difference $s$ only. $b(t)$ is continuous and non-negative definite and thus according to the Bochner-Khintchine theorem has a representation of the form

$$b(t) = \int_{\mathbb R}e^{it\lambda}\mu(d\lambda),$$

where $\mu$ is a finite measure.

Now, according to some theorem in a script, $X(t)$ has a representation of the form

$$X(t) = \int_{\mathbb R} e^{it\lambda} dZ(\lambda),$$

where $Z(\lambda)$ is a stochastic process with $\mathbb E[Z] \equiv 0$, uncorrelated increments and $\mathbb E[\vert Z(\lambda_2) - Z(\lambda_1) \vert^2] = \mu(\lambda_1,\lambda_2]$.

I guess by asking for "the spectral representation of $X$" one wants to find $Z$, doesn't one? How to exactly determine this representation for this special case?

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I think, I figured it out myself.

I first repeat the construction of an isometry that seems to appear in the standard proof of the spectral representation:

Let $H_X$ be the closure in $L^2$ norm of the set of all linear combinations of the form $\sum_{j=1}^n c_j X_{t_j}$, where $H_X$ is a dense linear subspace of $L^2(\mu)$. Then define the isometry

$$I : L^2(\mu) \rightarrow H_X, \quad\sum_{j=1}^n c_j e^{i\lambda t_j} \mapsto \sum_{j=1}^n c_j X_{t_j}$$

To understand the derivation for this example, we consider two important cases for this isometry ($Z$ is the spectral process to $X$):

$$I(\mathbb I_{(-\infty,\lambda]}) = Z(\lambda)\qquad I(e^{it\lambda}) = X(t)$$

Therefore, in order to find $Z$, one wants to determine the Fourier representation of $\mathbb I_{(-\infty,\lambda]}$ and then apply the isometry $I$.

Applying Bochner's theorem to the covariance function for this specific example gives:

$$b(t) = \sum_{k=1}^{\infty} c_k e^{ikt} = \int_{-\infty}^{\infty} e^{it\lambda} \mu(d\lambda)$$

from which follows

$$\mu(d\lambda) = \sum_{k=1}^{\infty} c_k\delta_k(\lambda)d\lambda$$

We now want to write

$$\mathbb I_{(-\infty,\lambda]}(x) = \int_{\infty}^{\infty}d_t(\lambda) e^{itx} dt$$

where the coefficients are given by:

$$d_t(\lambda) = \frac{1}{C}\int_{-\infty}^{\infty} e^{-i t \lambda} \mathbb I_{(-\infty,\lambda]} \mu(d\lambda) = \frac{1}{C}\sum_{k=1}^{\lfloor \lambda \rfloor} c_k e^{-i k t} \qquad C:= \sum_{k=1}^{\infty} c_k < \infty$$

And thus one gets:

$$Z(\lambda) = I[\mathbb I_{(-\infty,\lambda]}] = \int_{-\infty}^{\infty}\sum_{k=1}^{\lfloor \lambda \rfloor} \frac{1}{C} e^{-i t k} c_k X(t) dt$$

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