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Given the joint probability distribution for three dependent continuous random variables, I want to find a formula to compute the probability distribution for the product of these 3 random variables. I am currently using a formula but I am not sure it is correct.

To find my formula I used this relevant thread but it is for 2 dependant random variables (Probability distribution of the product of two dependent random variables). What it is in case of 3 dependent random variables ?

In mathematical terms:

Let's E be a random variables such that e=abc, were A, B and C are dependent random variables. Is it correct to use the following formula to compute the pdf of E ?

$f_E(e)=∫^∞_{−∞} ∫^∞_{−∞} f_{A,B,C}(a,b,e/ab) \space 1/|ab| \space da \space db$

or in another notation:

$P(abc=e)=∫^∞_{−∞} ∫^∞_{−∞} P(A=a,B=b,C=e/ab) \space 1/|ab| \space da \space db$

where:

e/ab means the value of e divided by ab

1/|ab| means one divided by the absolute value of the product ab.

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  • $\begingroup$ Looks good to me. $\endgroup$ May 16, 2023 at 13:07
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    $\begingroup$ And now for an opposing viewpoint to that of Doctor Milt.... No, I don't think that you can express $f_E$ as that integral. There are five other similar integrals (corresponding to the other five permutations of $\{a,b,c\}$). Can you provide some justification as to why all six integrals should evaluate to the same $f_E$? $\endgroup$ May 16, 2023 at 13:44
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    $\begingroup$ You apply the same principles and techniques given in the referenced thread. If you have trouble at any step, please edit your post to indicate where that occurs. // Another approach is to proceed in two steps: first compute the distribution of $D=BC$ using the referenced thread and then compute $E=AD$ in the same way. $\endgroup$
    – whuber
    May 16, 2023 at 14:07
  • $\begingroup$ I just would like to get the confirmation that what I wrote (i.e. the modification of the formula given in the referenced thread for three variables), is correct. $\endgroup$
    – MadMax2048
    May 19, 2023 at 11:11
  • $\begingroup$ Performing numerical simulations seems to validate the formula. I am going to use it but I would have prefer to get a mathematical demonstration of its validity. $\endgroup$
    – MadMax2048
    May 26, 2023 at 10:05

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