The noise analysis of the inverting 'virtual earth' amplifier is perhaps less well known than that of the non-inverting version. For my MJR7 mosfet power amplifier I usually just mention the 2 uV noise of the 10k input resistor and ignore other sources, so here I will attempt to show why this is accurate enough for most purposes. Initially, to keep it simple I will use a jfet input stage assumed to have insignificant current noise, and just a 1 uV noise voltage. The noise voltages are of course dependent on bandwidth and temperature, but to keep the figures simple I will assume these have been chosen such that the 10k resistor has exactly 2uV rms noise voltage, which approximately corresponds to a 25kHz bandwidth and 20 deg.C. The noise voltage of the input jfet is taken to be 1 uV under the same conditions, corresponding to an equivalent noise generating series resistance of 2k5 as shown next:

It is useful to include Vs as the noise voltage at the gate end of the equivalent noise generating resistor, but this is almost entirely fictitious, there is no separate resistor accounting for all the jfet noise so no actual location with voltage Vs, but it is still useful to include this in our approximate equivalent circuit.
The important point is that for a high open-loop amplifier gain, shown here as -10,000, Vs is very small. If we cheat a little by looking further down the page we find the value of output voltage Vo is 46 uV, so Vs must be that value divided by the voltage gain, so it is only 4.6 nV. This is insignificant compared to the 1uV noise voltage, so Vx must be almost exactly 1 uV. The 'almost exactly' is important, if it was exactly 1uV then the noise voltage of the 10k input resistor would have no effect on Vx and therefore no effect on the output Vo. The effect is however found as a contribution to the 'insignificant' Vs.

To work out the total output noise voltage Vo we need to calculate the contribution from the three noise generators individually, in each case setting the other two to zero.
The 1uV source requires Vx to be 1 uV, and the 10k and 200k form a potential divider making Vx = Vo x 10/210, and so Vo = 21 Vx = 21 uV.
The 2 uV source effect is calculated assuming the 9 uV and 1 uV sources are zero, and therefore that Vx is zero (again not exactly, Vs still there). To make Vx zero the 2 uV must be inverted by the amplifier and fed back through the 200k. This requires Vo = 40 uV.
The 9 uV source must also leave Vx zero, requiring zero feedback current through the 200k, so the 9uV must appear at the output, and so Vo = 9uV.
To add the three contributions to Vo we must add the noise powers, not the voltages. The voltages are random and not additive. The total voltage is calculated as the square root of the sum of the squares of the three individual voltages.
Vo = SQRT( 21 2 + 40 2 + 9 2)
= SQRT (441 + 1600 + 81) = 46.065 uV

If we just used the 2uV input resistor noise and multiplied by the closed loop gain 20 then we get Vo = 40uV, so not very accurate, but only out by a little over 1dB. Most of the error comes from the assumed 1uV noise voltage of the input jfet. Had we used the MJR7's 2SC2240BL input transistor we would need to include both voltage and current noise, or we could just look up the 'noise figure' for a 10k source impedance at the 0.5mA collector current, and find a figure of something like 0.5dB, which saves a lot of time by telling us how much noise is added to the 10k resistor noise, actually less than we calculated above for the jfet input. There are lower noise jfets available, so using something like a BF862 would also keep the noise low, with noise voltage around 0.15 uV instead of 1 uV.

Suppose we applied an input signal to the amplifier, if we applied 1V (AC) then the output would be 20V and the signal to noise, assuming 40 uV output noise would be 20V / 40uV = 114dB. What then is the signal to noise ratio at the input of the amplifier input stage, where the noise is Vx?
The noise is still 1 uV, but the signal is the 20v output divided by the 100,000 open loop gain, so only 0.2 mV, giving a signal to noise ratio of just 46dB. If we could find the fictitious location of Vs and measure the signal to noise there then we would find about the same 114dB as at the output.

It is perhaps counter-intuitive that the signal to noise ratio can be far higher at the output of the amplifier than at the input of the input device. Also, we often assume the way to reduce noise is to reduce the series resistances, but if we increase the 10k input resistor the noise voltage at the output will fall, and it will fall even further if we leave the input open-circuit. In the above example with open-circuit input the 1 uV and 9 uV are in series and appear unamplified at the output, giving:
Vo = SQRT( 1 2 + 9 2 ) = 9.05 uV, so reduced by a factor of five.
Unfortunately the closed-loop gain falls with increased series input resistance, and for a given input signal the output signal falls faster than the output noise as the 10k is increased, so the signal to noise ratio gets worse. Increasing the 200k feedback resistor to maintain the closed-loop gain then increases both signal and noise by about the same factor, so higher resistor values are not helpful. Increasing both 10k resistor and the input signal level does however give a better output signal to noise ratio.