A useful figure of merit must be independent of the measurement conditions, and that is exactly what disqualifies statement 1 — so the answer is 2, 3 and 4, option 3.

Why input noise voltage fails as a comparison. A raw noise voltage means nothing until the bandwidth is stated, since thermal noise grows as

\(V_{n}=\sqrt{4kTBR}\)

Measure the same receiver over 1 kHz and over 1 MHz and the voltage differs by a factor of about 32, though the receiver has not changed. Two receivers quoted at different bandwidths, source impedances or gains cannot be ranked from their noise voltages at all. It is a measurement, not a figure of merit.

The other three are all normalised, each in its own way.

Noise figure is the most widely quoted:

\(F=\dfrac{\left(S/N\right)_{in}}{\left(S/N\right)_{out}}\)

A perfect amplifier would have F = 1, or 0 dB; a real one adds noise of its own and so degrades the ratio. Because it is a ratio of ratios, gain cancels out and any two receivers can be compared directly.

Noise temperature expresses the same information on a scale better suited to very quiet systems:

\(T_{e}=\left(F-1\right)T_{0},\qquad T_{0}=290\ \text{K}\)

The advantage is resolution at the low end: noise figures of 0.5 dB and 0.6 dB look almost identical, whereas the corresponding temperatures of 35 K and 43 K differ plainly. Satellite and radio-astronomy receivers are therefore always specified in kelvin.

Equivalent noise resistance restates the noise as the value of a fictitious resistor at the input that would generate it, which is convenient when matching a source impedance for minimum noise.

Why the distinction matters in a cascade. Friis's formula shows the first stage dominates:

\(F=F_{1}+\dfrac{F_{2}-1}{G_{1}}+\dfrac{F_{3}-1}{G_{1}G_{2}}+\cdots\)

so a low-noise amplifier of high gain placed first fixes the noise performance of the whole receiver — a conclusion that can only be drawn from normalised measures.

Hence, the useful measures are 2, 3 and 4.