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I know that if $X_1$ and $X_2$ follow a Gamma distribution with parameters $(p_1,a)$ and $(p_2,a)$ respectively, then $\frac{X_1}{X_2}$ will follow a Beta prime distribution of parameters $(p_1,p_2)$.

Is the reverse also true? I mean if I know the ratio is distributed as a Beta prime, can I infer that the numerator and denominator variables are distributed as Gammas? And how could I prove it?

More in general, are the relationship between distributions valid in both directions?

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  • $\begingroup$ Please contemplate the similar situation where the $X_i$ each have a $1/2$ chance of being either $1$ or $2.$ Their ratio can equal $1/2,$ $1,$ or $2.$ When you multiply these values by $X_2$ you will therefore obtain one of the values $1/2,$ $1,$ $2,$ or $4,$ all with nonzero probability. $\endgroup$ – whuber♦ Nov 14 '18 at 23:31
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What you're neglecting there is the dependence between variables.

Since the question specifies nothing about dependence, if we knew we had a beta prime and that it was a ratio of two random variables, we don't have any good reason to anticipate that they were gammas in the numerator and denominator. It might be the case (you can certainly construct a case where it's true) but might not be the case for the variables at hand.

If we also knew for certain that the numerator and denominator were independent it might be possible to construct a proof that they were gamma; it might, for example, be possible to do something with characteristic functions of their logs - but it's not immediately obvious this would necessarily be the case; (I wouldn't be surprised if it were the case but I don't have a demonstration of it).

Not quite what was asked, but related to a common kind of question I get asked: If you take a beta prime and multiply it by an independent gamma, you shouldn't expect to get a gamma back out. That would correspond to saying that $X_1 X_3/X_2$ (for all three being independent gammas) is itself gamma, which isn't the case -- for example typically it is considerably more heavy-tailed than a gamma.

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  • $\begingroup$ Then your answer would be no, right? I mean once I know Y's distribution, I cannot infer those of $X_1$ and $X_2$ because of dependence $\endgroup$ – PhDing Nov 14 '18 at 23:25
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    $\begingroup$ Re "shouldn't expect:" one way to demonstrate this is to consider the case $p_2\le 1.$ The ratio has infinite mean, whence the ratio times any Gamma variable must also have infinite mean, proving it cannot have a Gamma distribution. $\endgroup$ – whuber♦ Nov 14 '18 at 23:36
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    $\begingroup$ I have the impression that this answer is only tangential to what the OP asks. For another example, If I am given a Cauchy, I can certainly say that I can model it as the ratio of two independent Normals. I think this is what the OP asks. $\endgroup$ – Alecos Papadopoulos Nov 14 '18 at 23:40
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    $\begingroup$ @Alecos yes, it was, but I realized after posting that the OP asked something else and edited a more direct answer in at the top. I will leave the tangential answer at the end, though because I expect that people who turn this up in search will often be looking for that. $\endgroup$ – Glen_b Nov 14 '18 at 23:48
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If we are told that the ratio of two random variables follows a Beta prime, then we can certainly say that the numerator and the denominator can be modeled as independent Gammas with the parameter mapping shown in the OP's post.

Whether this "backwards decomposition" is unique it is another matter.

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  • $\begingroup$ Exactly! What I am asking (tell me if I need to be more precise) is if given $Y$ Beta prime and knowing $Y=\frac{X_1}{X_2}$ is enough to say that $X_i$ are distributed as Gammas $\endgroup$ – PhDing Nov 14 '18 at 23:43
  • $\begingroup$ As presently stated the answer is no, that's not enough. It's not even enough to say "$X_1$ and $X_2$ are Gamma" in order to conclude that the ratio is beta prime. $\endgroup$ – Glen_b Nov 14 '18 at 23:45
  • $\begingroup$ @Glen_b Would it be enough if I assume independence between $X_1$ and $X_2$, right? I mean that the ratio is Beta prime $\endgroup$ – PhDing Nov 14 '18 at 23:49
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    $\begingroup$ yes, then it follows. $\endgroup$ – Glen_b Nov 14 '18 at 23:54

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