Noise power transient at output of RC filter
Consider a simple RC filter whose step response is $1-e^{-t/RC}$ for $t &gt; 0$,
and impulse response is
$$h(t) = \begin{cases} \frac{1}{RC}e^{-t/RC}, &amp; t \ge 0,\\ 0, &amp;t &lt; 0.\end{cases}$$
Then, when the input is white noise, the steady-state output variance is
$$\sigma^2 = \frac{N_0}{2} \int_{-\infty}^\infty h^2(t)\,\mathrm dt
= \frac{N_0}{2}\int_0^\infty \frac{1}{(RC)^2}e^{-2t/RC}\,\mathrm dt = \frac{N_0}{4RC}.$$
However, if the input noise is also $0$ for $t &lt; 0$, then at time $T$, the
noise variance at the output is
$$\begin{align}
\sigma_T^2 &amp;= E\left[\int_0^T N(t)\frac{1}{RC}e^{-(T-t)/RC}\,\mathrm dt
\int_0^T N(s)\frac{1}{RC}e^{-(T-s)/RC}\,\mathrm ds\right]\\
&amp;= \int_0^T\int_0^T E[N(t)N(s)]\frac{1}{(RC)^2}e^{-(2T-t-s)/RC}
\,\mathrm dt\,\mathrm ds\\
&amp;= \int_0^T\int_0^T \frac{N_0}{2}\delta(t-s)\frac{1}{(RC)^2}e^{-(2T-t-s)/RC}
\,\mathrm dt\,\mathrm ds\\
&amp;= \int_0^T \frac{N_0}{2}\frac{1}{(RC)^2}e^{-(2T-2s)/RC}
\,\mathrm ds\\
&amp;= \frac{N_0}{4RC}e^{-2T/RC}\int_0^T \frac{2}{RC}e^{2s/RC}\,\mathrm ds\\
&amp;= \frac{N_0}{4RC}e^{-2T/RC}[e^{2T/RC}-1]\\
&amp;= \left.\left.\frac{N_0}{4RC}\right[1 - e^{-2T/RC}\right]
\end{align}$$
showing that $\sigma_T^2$ increases to its asymptotic value 
$\displaystyle \sigma^2 = \frac{N_0}{4RC}$ more rapidly than the unit step
response increases to its asymptotic value $1$.

The time constant of the RC circuit is $RC$, not $1/RC$ as the OP
claims, and the time constant of the increase in the noise variance is
$RC/2$, half that of the RC circuit. The smaller the time constant,
the more rapid the decay to $0$  (or rise to the asymptotic value).