The content of this question is about rigorously proving something which is otherwise considered easily correct intuitively.

Let's assume we have a multivariate distribution $g(x_1,x_2,...,x_n)$ over the variables $x_{1:n}$. Let's assume that we know how to sample from that distribution, too. We draw samples $x^{1}_{1:n},x^{2}_{1:n}, ... ,x^{N}_{1:n}$ from this distribution.

Then we assume that we have the distributions $f_{1}(x_1), f_{2}(x_2|x_1),f_{3}(x_3|x_2,x_1),...,f_n(x_n|x_{1:n-1})$ which is equal to $g(x_1,x_2,...,x_n) = f_{1}(x_1)f_{2}(x_2|x_1)f_{3}(x_3|x_2,x_1),...,f_n(x_n|x_{1:n-1})$ by the chain rule of probabilities. Now, for each sample $x^{i}_{1:n}$ we first sample $x^{i}_{1}$ from $f_{1}(x_1)$, then $x^{i}_{2}$ from $f_{2}(x_2|x_1)$ up to $x^{i}_{N}$ from $f_n(x_n|x_{1:n-1})$. We again obtain $N$ samples.

Intuitively we know that the first $N$ samples which come from $g(x_1,x_2,...,x_n)$ and second $N$ samples each of which sequentially come from $f_{1}(x_1),f_{2}(x_2|x_1),f_{3}(x_3|x_2,x_1),...,f_n(x_n|x_{1:n-1})$ are identically distributed. But how can we show this fact in a mathematically rigorous way? I could not think of any procedure and became stuck.

Thanks in advance

  • 1
    $\begingroup$ No sorry, it was a typo, now I corrected it. $\endgroup$ – Ufuk Can Bicici May 13 '14 at 5:27

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