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Given a joint probability density distribution $P(\boldsymbol{\alpha})$ with $\boldsymbol{\alpha}=(\alpha_1,\alpha_2,\alpha_3)$ a variable in $\mathbb{R}^{3}$, such that all marginals $P_i$ are Dirac-delta functions i.e. : \begin{align} P_{1}(\alpha_{1})&=\int d\alpha_{2}\int d\alpha_{3}P(\boldsymbol{\alpha})=\delta(\alpha_{1}-\gamma_{1}),\\ P_{2}(\alpha_{2})&=\int d\alpha_{1}\int d\alpha_{3}P(\boldsymbol{\alpha})=\delta(\alpha_{2}-\gamma_{2}),\\ P_{3}(\alpha_{3})&=\int d\alpha_{1}\int d\alpha_{2}P(\boldsymbol{\alpha})=\delta(\alpha_{3}-\gamma_{3}), \end{align} is it possible to prove that the only joint distribution is \begin{equation} P(\boldsymbol{\alpha})=P_{1}(\alpha_{1})P_{2}(\alpha_{2})P_{3}(\alpha_{3})\,\,? \end{equation} In general, I do expect that knowing just the marginals is not enough to uniquely determine the joint probability distribution. However, given that Dirac-deltas are a very special case, it feels more reasonable that the only possible joint distribution is the product of the three.

If working with continuous distribution creates problems, it would be very helpful also an analogue proof using discrete variables. Also references are very welcome.

Thanks in advance to everyone for the help!

Sandro
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    You seem to be assuming the only value each $\alpha_i$ can take on (with nonzero probability) is $\gamma_i.$ If so, that's the same thing as saying the only value $(\alpha_1,\alpha_2,\alpha_3)$ can have is $(\gamma_1,\gamma_2,\gamma_3).$ By the probability axioms, its chance is $1.$ I hope that's what you mean by your formula. Otherwise, please explain your notation. – whuber Mar 15 '22 at 17:44
  • Yes, I meant exactly what you wrote. And I agree with your conclusion, I was just wondering if there is a better way to formalize it – Sandro Mar 16 '22 at 16:58
  • You could make it more formal by arguing that the $\alpha_i$ must be independent when their marginal distributions are singular. Then, for independent distributed variables the joint distribution is the product of the marginal distributions. – Sextus Empiricus Sep 22 '22 at 16:33

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