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I was wondering how to calculate the following joint distribution by assuming that $x_i$'s are continuous observations from normal distirbution $N(\mu, \sigma)$ with mean $\mu$ and variance $\sigma$ and $p(\mu, \sigma)$ is the joint distribution of $\mu$ and $\sigma$: $$ p(x_1,x_2,..., x_n, x_{n+1}) = \iint {\displaystyle \prod_{j=1}^{n+1} N(\mu,\sigma)|_{x_j}} . p(\mu, \sigma) \,d\mu \, d\sigma $$ I know that this can be accomplished by solving the intergration numerically (e.g. with rectangular integration), however, the thing that puzzles me is that the products of $N(\mu,\sigma)|_{x_j}$ would be always zero since the probability of the point $x_i$ in a continuous distribution $N(\mu,\sigma)$ is zero (or I'm wrong?).

In addition, what is the best method to calculate (or estimate) this distribution in python?

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Ok, this looks like deep learning related stuff, maybe an exercise.

$N(\mu,\sigma)|_{x_j}$ is not the probability of a point $x_j$, this is the density of point $x_j$ given the parameters $\mu, \sigma$. The total integral is not the probability either, it is the likelihood of sorts or a density function. The whole thing is in fact never a zero, because the density $N$ of normal distribution is never zero, it can be very small of course.

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  • $\begingroup$ Thanks. Do you have any suggestion for the implementation in python? I've used dblquad for the integration but it takes a lot of time for evaluating a series of let's say 20 observations. I think the slowness of the integration is related to the fact that the p(μ,σ) is calculated using the product of two log-normal distribution but I simply use linear (not exponential) interval for the integrands. $\endgroup$ – farzad Feb 18 '18 at 16:40
  • $\begingroup$ I'm not sure someone intended to integrate this numerically. They may have written the equation to get across the concept. Multidimensional integration is very difficult beyond 2-3 dimensions, it's heavy computationally. You may want to integrate analytically or re-consider whether you need to integrate at all. If need help on integration though, it's better to post a separate question. There's a chance the question would be considered off-topic and moved to other SE or SO sites $\endgroup$ – Aksakal Feb 18 '18 at 17:37
  • $\begingroup$ But this formula comes from a paper in which the authors used numerical integration via a simple quadrature rule. Since the integrands are represented by log-normal distributions, they have used a rectangle rule in which the intervals between interpolation points vary exponentially. But I don't know how to use the rectangular rule of integration with exponential intervals for the interpolations. $\endgroup$ – farzad Feb 18 '18 at 17:54
  • $\begingroup$ What was dimensionality? $\endgroup$ – Aksakal Feb 18 '18 at 17:55
  • $\begingroup$ The observations have only one dimension. $\endgroup$ – farzad Feb 18 '18 at 18:12

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