In a non-traditional type of derivation of Maxwell's speed distribution for gases,I happen to face the following problem:
They say since P(v_x),P(v_y),P(v_z) are independent,so the combined probability wil be P=P(v_x)P(v_y)P(v_z).
This much is OK.Then they say the only function having the property f(a+b+c)=f(a)f(b)f(c)
is an exponential function.So, consider the P(v_x) as to have exponential dependence P(v_x)=K exp[-L*(v_x)^2].This makes me uncomfortable.Did we have P=P(v_x+P_y+P_z)?I am a bit new to statistical ideas,so really cannot be sure when we said the joint probability is P,it means P=P(v_x+P_y+P_z).
Please help.

This makes me uncomfortable.Did we have P=P(v_x+P_y+P_z)?


No, you have P=P\left(v_x^2+v_y^2+v_z^2\right).

OK,so it is possible to write:
P[((v_x)^2)+((v_y)^2)+((v_z)^2)]=P(v_x)P(v_y)P(v_z) ?

First of all, you are trying to find the speed distribution as a function of energy. That is, you are looking for P(E) So when there is no external field and the gas molecules are non-interacting, the energy is entirely kinetic. Hence, P(E)=P(K)=P(\vec{v}\cdot\vec{v})=P(v_x^2+v_y^2+v_z ^2).

Now on the other hand, the distribution of the components of velocity should be independent of one another, as momentum is conserved separately along all 3 coordinate axes. So the probabilities for finding a particle whose i^{th} component of velocity is in the interval [v_i,v_i+dv_i] is independent for all i\in\{x,y,z\}. So we can write P(E)=Q(v_x^2)Q(v_y^2)Q(v_z^2). (I use a different letter Q here because the individual factor functions are slightly different from the product P. They have a different normalization constant.)