Partition function for Gaussian white noise Problem
I'm trying to understand, motivate, or derive from first principles, the partition function for gaussian white noise, namely
$$
Z = \int \mathcal{D}\eta(t)\exp\left[-\frac{1}{2D}\int dt \eta(t)^2\right].
$$
I thought I could work out the steps, but I'm unable to pass to the continuum limit from the discrete case.
My attempt
I assumed that, for gaussian white noise on a lattice (discrete points in time), the probability of a noise of any amplitude $\eta(t_i) = \eta_i$ is given by
$$
P(\eta_i) = \frac{1}{\sqrt{2\pi D}}\exp\left(-\frac{\eta_i^2}{2D}\right).
$$
I am stuck trying to pass to the continuum limit from here. If you try to calculate the partition function from this, you get
$$
Z \stackrel{?}{=} \left(\frac{1}{2\pi D}\right)^{N/2}\prod_i^N \int d\eta_i\exp\left(-\frac{\sum_i^N \eta_i^2}{2D}\right).
$$
But ideally you'd have an epsilon in the exponent so that you could do
$$
\epsilon \sum_i \eta_i^2 \rightarrow \int dt \eta(t)^2.
$$
Could someone help steer me in the right direction? Thanks for any help.
 A: White noise is characterised by the autocorrelation function 
$$\langle \eta(t) \eta(t') \rangle = D \delta(t-t'),$$ 
(here I'm assuming that $\eta$ is dimensionless so $[D] = [t]$). This means that (1) the noise should be uncorrelated with itself at different times, but also that (2) the variance of $\eta(t)$ must be infinite. Condition (2) is the essential missing ingredient here. 
Before showing how to incorporate (2) into a discretised model of the noise, let us first motivate why we need $\eta(t)$ to have infinite variance. This is necessary in order that the integrated signal, defined by
$$ W(\tau) = \int_0^\tau{\rm d}t\;\eta(t), $$
be non-zero. Indeed, we have that
$$ \langle W(\tau)^2\rangle = \int_0^\tau{\rm d}t\int_0^\tau{\rm d}t'\;\langle \eta(t) \eta(t') \rangle = D\tau,$$
i.e. the variance of the integrated signal grows linearly with time, with "diffusion constant" $D$, just like a continuous random walk where the walker's position as a function of time is given by $W(\tau)$. This is no coincidence: the random variable $W(\tau)$ describes the Wiener process, which is fundamental to the definition of white noise. In particular, if ${\rm d}W(t) = W(t+{\rm d}t)-W(t)$, then the variable $\eta(t)$ is actually defined as
$$ {\rm d}W(t) = \eta(t){\rm d}t.$$
Note that if the variance of $\eta(t)$ were finite, $W(\tau)$ would vanish identically. Suppose that instead $\langle \eta(t) \eta(t') \rangle = f(t-t')$, where $f(0)$ is finite and $f(t) = 0$ for $t\neq 0$. Hence, it follows that $\langle W(\tau)^2\rangle = \langle W(\tau)\rangle = 0$. This implies that the integrated signal $W(\tau)=0$ with probability one. In other words, the finite instantaneous fluctuations of $\eta(t)$ combined with its zero correlation time leads the integrated effect of the noise signal to cancel to zero over any finite time.
In order to define the noise signal as the continuum limit of a discrete process, we start from a discrete (unbiased) random walk. In each time increment $\delta t = t_{i+1} - t_i$ the walker is displaced by the amount $\delta W_i = \eta_i \delta t$ (c.f. $ {\rm d}W(t) = \eta(t){\rm d}t$), where $\langle \delta W_i\rangle = 0$. Making each step size proportional to $\delta t$ ensures that the step size tends to zero in the continuum limit $\delta t\to 0$, so that $W(t)$ describes a continuous (nowhere differentiable) function. Moreover, each of these steps must be statistically independent in order to recover zero correlation time (white noise) in the continuum limit. We also assume that the set of steps $\delta W_i$ are identically distributed. Now, the variance of the walker position after $N$ steps, corresponding to a time $\tau = N \delta t$, is given by
\begin{align}
\langle W(\tau)^2\rangle & = \left \langle \left ( \sum_{i=1}^N \delta W_i \right)^2 \right\rangle\\
& = (\delta t)^2 \left(\sum_{i=1}^N \langle \eta_i^2 \rangle + \sum_{i\neq j} \langle \eta_i\eta_j\rangle\right) \\ 
& = (\delta t)^2 \sum_{i=1}^N \langle \eta_i^2 \rangle,
\end{align}
where the third equality follows from statistical independence of the steps: $\langle \eta_i\eta_j\rangle = 0$ for $i\neq j$.
Now, in order to recover the correct scaling $\langle W(\tau)^2\rangle = D \tau$, we must have $$\langle \eta_i^2 \rangle = D/\delta t.$$ 
The variance of the discretised white-noise variable $\eta_i$ therefore tends to infinity in the continuum limit, as required. Since we also require the noise statistics to be zero-mean Gaussian (i.e. the only non-zero cumulant is the variance), it follows that the unique probability distribution function is
$$ P(\eta_i) \propto \exp \left ( -\frac{\eta_i^2 \delta t }{2D} \right), $$
up to an unspecified normalisation constant. The correct result for the partition function follows after writing down the path integral and taking the continuum limit $\delta t\sum_i \to \int\mathrm{d} t$.
