# Gamma mixture distributions

I suppose that I have a mixture of two Gamma distributions: $$f(x) = p*f_1(x) + (1-p)*f_2(x)$$ where $$f_i(x) = \frac{b_i^{a_i}}{\Gamma(a_i)} x^{a_i-1} e^{-b_i x},\; x>0$$ I take specific values for $a_i,\; b_i$ and I change the value of $p$. I can see that all graphs pass from a specific point. Does anybody knows which point is that? I have made a code at Mathematica. See the result. Thank you.

a1 = 2; b1 = 1; a2 = 6; b2 = 2;
f1[x_, p_] = p*b1^a1/Gamma[a1]*x^(a1 - 1)*E^(-b1*x) + (1-p)*b2^a2/Gamma[a2]*
x^(a2 - 1)*E^(-b2*x);
Plot[{f1[x, 0.1], f1[x, 0.3], f1[x, 0.5], f1[x, 0.7], f1[x, 0.9]}, {x,0,10}]


• Since it doesn't matter what $p$ is, take $p=0$ and $p=1$ -- it's then obviously the place where the two component densities cross. Dec 8, 2016 at 5:15

You've found a point $x$ where $p_1f_1(x) + (1-p_1)f_2(x) = p_2f_1(x) + (1-p_2)f_2(x)$ for all $p_1, p_2$. Let's do some algebra:

$p_1(f_1(x)-f_2(x)) + f_2(x) = p_2(f_1(x) - f_2(x)) + f_2(x)$

and subtract the right hand side from the left:

$(p_1-p_2)(f_1(x)-f_2(x)) = 0$

which will be true for any $p_1$ and $p_2$ at $x$ such that $f_1(x) = f_2(x)$. So your intersection occurs at $x$ such that $f_1(x) = f_2(x)$.

Note that there will always be such an $x$, as both $f_1$ and $f_2$ integrate to 1 and are continuous positive functions on $(0, \infty)$. If $f_1(x) < f_2(x)$ for all $x$, the integral would be too, and they couldn't both integrate to 1. Therefore they must be equal everywhere or there must be a region where $f_1 < f_2$ and another region where $f_1 > f_2$. Continuity ensures that, consequently, there is a point where $f_1 = f_2$.

Try setting just two of the mixture distributions equal:

$$p_1f_1(x) + (1-p_1)f_2(x) - p_2f_1(x) - (1-p_2)f_2(x) = 0$$

Set $p_1 = 0.5$ and $p_2 = 0.4$ if you like, and use Newton's Method to find the root.

• I think that you have not understand my question. I do not care of $a_i, b_i$. If I change the value of $p$ I take two versions of the distribution and I what to find whether their intersection is an interesting point. Dec 8, 2016 at 1:09
• I edited my previous answer to reflect that the two probabilities should be distinct.
– JTH
Dec 9, 2016 at 6:31