Talk:Compound probability distribution

Statistics Mid‑importance

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Mathematics Mid‑priority

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Exponential-inverse gamma

The article contains the text "Compounding an exponential distribution with parameter distributed according to an inverse gamma distribution yields a pareto distribution." I am not sure this works and https://math.stackexchange.com/questions/646852/compound-of-gamma-and-exponential-distribution suggests that it should be an exponential-gamma mixture. I don't have a reference for this though. 217.44.78.244 (talk) 07:52, 27 September 2016 (UTC)[reply]

The proof is muddled

In the subsection Proof, there are unnecessary and confusing simplifications and misleading or mistaken notation.

1) There is no good reason to write $F$ and $G$ as if they are parametrized by mean and standard deviation. In fact, it is misleading, because we are given only that $F$ is parametrized by $\theta$ , and $G$ is fixed. The other three quantities are outcomes, not parameters.

Furthermore, there is no reason to assume that the relation between $\theta$ and $E[F|\theta ]$ is bijective, so reparametrization to make them equal might not be possible.

2) Writing the variance $\operatorname {Var} _{F}(X\mid \theta )$ as $\tau ^{2}$ is incorrect, since $\operatorname {Var} _{F}(X\mid \theta )$ is a random variable and $\tau ^{2}$ is a constant. (If it is intended to be a random variable, this has not been declared, and moreover, it can't be pulled out of an integral as the quantity $\tau ^{2}$ is.)

It would be better if

a) The expression $E_{F}[X|\theta ]$ were substituted for $\theta$ wherever the latter is being used to denote the mean of $(X|\theta )$ , and

b) The variables $\mu$ , $\sigma$ , and $\tau$ were deleted, and replaced by the expressions

E_{G}[E_{F}[X|\theta ]],\quad

\operatorname {Var} _{G}(E_{F}[X|\theta ]),\quad

\operatorname {Var} _{F}(X\mid \theta )

.

In fact, these quantities are all calculated at some point in the proof as it is written, so the logic and sequencing of the proof don't have to change, just the symbols in the calculations.

2A02:1210:2642:4A00:C9E4:9F6:5E99:526E (talk) 20:07, 23 June 2023 (UTC)[reply]