That's well within error bound of the scope.
Errors aside, RMS will always be higher for a raw signal than for filtered. We can decompose a signal into AC and DC components,
\$v(t) = V + f(t)\$
where in general f(t) varies with time, and has an average value of zero over the period \$\Delta t\$ being considered. V is a constant.
RMS is defined as,
\$ \sqrt{ \frac{1}{\Delta t} \int_{t_0}^{t + \Delta t} v(t)^2 dt} \$
so,
\$ \sqrt{ \frac{1}{\Delta t} \int_{t_0}^{t + \Delta t} \left( V + f(t) \right)^2 dt} \$
\$ \sqrt{ \frac{1}{\Delta t} \int_{t_0}^{t + \Delta t} \left( V^2 + 2 V f(t) + f(t)^2 \right) dt} \$
\$ \sqrt{ \frac{1}{\Delta t} \left[ \int_{t_0}^{t + \Delta t} V^2 dt + \int_{t_0}^{t + \Delta t} 2 V f(t) dt + \int_{t_0}^{t + \Delta t} f(t)^2 dt \right] } \$
\$ \sqrt{ \frac{1}{\Delta t} \left[ V^2 \int_{t_0}^{t + \Delta t} dt + 2 V \int_{t_0}^{t + \Delta t} f(t) dt + \int_{t_0}^{t + \Delta t} f(t)^2 dt \right] } \$
\$ \sqrt{ \frac{1}{\Delta t} \left[ V^2 (\Delta t) + 2 V (0) + \int_{t_0}^{t + \Delta t} f(t)^2 dt \right] } \$
\$ \sqrt{ \frac{V^2 \Delta t}{\Delta t} + \frac{1}{\Delta t} \int_{t_0}^{t + \Delta t} f(t)^2 dt } \$
\$ \sqrt{ V^2 + {f(t)_\textrm{rms}}^2 } \$
Which is just taking the RMS of two orthogonal components,
\$ V_\textrm{rms} = \sqrt{v_1(t)^2 + v_2(t)^2 } \$
the DC and AC parts are orthogonal for \$\Delta t \rightarrow \infty\$ at the least.
Further, if we decompose f(t) into a sum of sinusoidal waves (Fourier transform), we have Parseval's theorem: simply sum up the squares of the amplitudes of all components, take the average, and square-root.
When orthogonality doesn't hold, we might get a residual. Besides rounding error, this may be what the scope is reading. For periodic signals, the orthogonality also holds for any multiple of that period. For aperiodic signals, it only holds at infinity, or at special cases. The scope can't buffer to infinity, its window is always finite of course. Basically, there's likely some residual blip that's unbalanced over that window, and this causes the RMS to measure differently. (And your signal looks fairly nonperiodic.)
Finally, basic statistics! You're asking about a very small difference -- the AC peak-to-peak is a small fraction of the DC value, so the DC dominates. What's the difference in length between the hypotenuse and long side, of, say, a 1° right triangle? (Recall what else uses a sum-of-squares formula!) It's fuck all, and if you measured that with anything less than a micrometer, chances are, you're as likely as not to find the hypotenuse shorter instead -- what ought to be a Euclidian contradiction.
If you're not familiar with signals and calculus, this probably isn't all that helpful; there are some excellent, simple explanations out there of what integration means, which may prove illuminating. If nothing else, since the scope is doing it with DSP (digital signal processing), what it really means is, not calculus at all, but the stepwise equivalent (Riemann summation), literally taking the square of individual points in time, and averaging over that.
Tim
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