Assume that a dT increase in temperature results in a dP increase in the transistor's power dissipation, and that a power increase of dP results in a temperature increase of dT'. Everything will be fine as long as dT' < dT.

We can calculate dT' by multiplying the power increase (dP) with the thermal resistance from junction to ambient, so:

dT' = R_{th,j-a} ∙ dP

So an amplifier will be thermally stable if:
R_{th,j-a}∙dP < dT, or:
R_{th,j-a}∙dP/dT < 1

Let's take a look at a very simple amplifier:

Without input signal the transistor (NPN or PNP) will dissipate:
P=V_{CE}∙I_{C}.

V_{CE}=V_{S} -
I_{C}∙RL => P=(V_{S} -
I_{C}∙RL)∙I_{C}=V_{S}∙I_{C}
- I_{C}^{2}∙RL.

dP/dI_{C}=V_{S} -
2∙I_{C}∙RL. => dP=(V_{S} -
2∙I_{C}∙RL)∙dI_{C}

The amplifier will therefore be stable if:

R_{th,j-a}∙(V_{S} -
2∙I_{C}∙RL)∙dI_{C}/dT <
1

This will always be true if: V_{S} <
2∙I_{C}∙RL => I_{C}∙RL >
V_{S}/2

I_{C}∙RL = V_{S} -
V_{CE} => V_{S} -
V_{CE} > V_{S}/2 =>
V_{CE} < V_{S}/2

So, the amplifier will be stable if V_{CE} is
smaller than V_{S}/2. For maximum output amplitude,
the collector voltage must be V_{S}/2. If the emitter
is grounded, V_{CE} will be equal to
V_{S}/2 and the amplifier may be thermally instable.
Therefore a small resistor is connected between emitter and ground. The
voltage across this resistor must be about V_{S}/5.
This is a compromise; less voltage may result into instability due to
component spread and aging. A higher voltage will decrease the maximum
output amplitude and thus the efficiency.