Let us consider an orthonormal basis, e.g. $\{\psi_n(t)\}_{n\in\mathbb Z}$ of the space of real-valued functions that are square integrable $L^2(\mathbb R)$. Does the linear combination of the elements in $\{\psi_n(t)\}_{n\in\mathbb Z}$ with standard normal random coefficients always give a white noise? In other words, consider coefficients $\epsilon_{n}$ that are i.i.d. $N(0,1)$, is it true that $$\sum_{n=-\infty}^{\infty} \epsilon_{n} \psi_n (t)$$ defines a white noise process?
If the answer is "yes", is this still true if we replace $\{\psi_n(t)\}_{n\in\mathbb Z}$ with a redundant set, i.e. of elements that are not linearly independent but still able to generate the entire space $L^2(\mathbb R)$?
Proposed answer for the second question. Let us assume that the answer to the forst question is yes. The system $\{\psi_n(t)\}_{n\in\mathbb Z}\cup \{\phi\}$ where $\phi\in L^2(\mathbb R)$ is not a basis, anyway if we consider: $$\sum_{n=-\infty}^{\infty} \epsilon_{n} \psi_n (t)+\epsilon\phi$$ such that $\{\epsilon_{n}\}\cup\{\epsilon\}$ are i.i.d. $N(0,1)$, then $\sum_{n=-\infty}^{\infty} \epsilon_{n} \psi_n (t)+\epsilon\phi$ still defines a white noise.
Edit.
If $X(\cdot)$ is any real-valued stochastic process with $E\left(X^2(t)\right)<\infty$ for all $t \geq 0$, we define $$ r(t, s):=E(X(t) X(s)) \quad(t, s \geq 0) $$ the autocorrelation function of $X(\cdot)$. If $r(t, s)=c(t-s)$ for some function $c: \mathbb{R} \rightarrow \mathbb{R}$ and if $E(X(t))=E(X(s))$ for all $t, s \geq 0, X(\cdot)$ is called stationary in the wide sense. A white noise process $\xi(\cdot)$ is by definition Gaussian, wide sense stationary, with $c(\cdot)=\delta_0$.
In general we define $$ f(\lambda):=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i \lambda t} c(t) d t \quad(\lambda \in \mathbb{R}) $$ to be the spectral density of a process $X(\cdot)$. For white noise, we have $$ f(\lambda)=\frac{1}{2 \pi} \int_{-\infty}^{\infty} e^{-i \lambda t} \delta_0 d t=\frac{1}{2 \pi} \quad \text { for all } \lambda $$
