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Based on the formula in:

https://en.wikipedia.org/wiki/Product_distribution

It seems that Z = XY = YX. However, if X is a bounded uniform distribution, one could have either

$f_Y(z/x)$ or $f_Y(z)$, producing two different values for the Z integral

I am missing something?

Alex
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    The two integrals $\int_{-\infty}^\infty f_X(x) f_Y(z/x) \frac{1}{|x|} , dx$ and $\int_{-\infty}^\infty f_Y(y) f_X(z/y) \frac{1}{|y|} , dy$ will be the same. – angryavian Jul 22 '21 at 16:21
  • What does $f_Y(z \mid x)$ mean and same question for $f_Y(z)$? – William M. Jul 22 '21 at 16:35
  • Also... use \mid for the "given" condition in probability, it looks better. – William M. Jul 22 '21 at 16:35
  • @WillM, they’re not intended to be conditionals—actual division. – A rural reader Jul 22 '21 at 17:31
  • +1 @angryavian, correct. – A rural reader Jul 22 '21 at 17:41
  • Oh, okay. In that case you can use \frac{}{} which is a tiny fraction, perhaps clearer. – William M. Jul 22 '21 at 18:12
  • @angryavian, I see, I was able to get the second integral by a change of variables y=z/x, except for a -ve sign (https://math.stackexchange.com/questions/856654/why-absolute-values-of-jacobians-in-change-of-variables-for-multiple-integrals-b) – Alex Jul 22 '21 at 19:14
  • That is why we take the absolute determinant for the Jacobian transition matrix. – Graham Kemp Jul 23 '21 at 04:55
  • @Graham Kemp I don't think you do in 1D: https://math.stackexchange.com/questions/856654/why-absolute-values-of-jacobians-in-change-of-variables-for-multiple-integrals-b – Alex Jul 23 '21 at 17:35

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