Spectral density of fluctuations (white noise/delta-correlated process)

Let I be the current flowing across some junction as a result of N charge carriers of charge q. And let $langle I (t) rangle$ be its average.

Assume a particle number distribution such that its fluctuation is given by $langle (Delta N)^2 rangle=langle Nrangle$.

So $langle Irangle = q langle N rangle$ and by definition of the correlation function $K_I(tau)=langle (Delta I)^2 rangle$
(where $tau$ be the time difference $t’ – t$) we have

$$
K_f(tau)= q langle Irangle
$$

Simplifying this comes from the fact that the charge carriers flow randomly and independently. So we use the following:

Let the spectral density of fluctuations be defined as the Fourier transform of the correlation function $K_f(tau)$

$$
S_I(omega) = frac{1}{2pi} int_{- infty}^{infty} K_f(tau) e^{i omega tau} dtau
$$

The random and independent nature of the system means that this is a delta-correlated process, where we have $S_f(omega)= constant = S_f(0)$,

so that via an inverse fourier transform we have

begin{eqnarray}
K_I(tau) &=& 2pi S_I(0) delta(tau)
&=& 2 pi bigg(frac{1}{2pi} int_{- infty}^{infty} K_f(tau) e^{i 0 tau} dtau bigg) delta(tau)
&=& int_{- infty}^{infty} K_f(tau) dtau delta(tau)
&=& K_f(0) delta(tau)
end{eqnarray}

I am trying to understand if this next step I take is legitimate:

Could I not rearrange the first line of the above equation to say

begin{eqnarray}
K_I(tau) &=& 2pi S_I(0) delta(tau)
q langle Irangle &=& 2pi S_I(0) delta(tau)
frac{q langle Irangle}{2pi delta(tau)} &=& S_I(0)
end{eqnarray}

I hope this is more clear now.

My motivation behind the original question

if , for an average quantity $langle I rangle$, does

$$
frac{langle I rangle}{delta(tau)} = I
$$

Would $$S_I(omega)=S_I(0)= frac{q langle I rangle}{2 pi delta(tau)}$$ or $$S_I(omega)=S_I(0)= frac{q langle I rangle}{delta(tau)}$$

is that I know the answer to be

$$
S_I = q I
$$

with no average $langle I rangle$ or $2pi$.