Recent content by d9dd9dd9d

Dear viraltux, In fact, f(x_2|x_1) and f(x_1|x_2) do not have the same expression, to be more specific, \begin{array}{l} f({x_2}|{x_1}) = \frac{1}{{\sqrt {2\pi } }}\exp \left( { - \frac{{{{({x_2} - {x_1})}^2}}}{2}} \right),\\ f({x_1}|{x_2}) = \frac{2}{{\sqrt {2\pi } }}\exp \left( { -...

Dear haruspex, I really appreciate your reply. Your tolling rice model is really helpful, now I know that f(x_1|x_2)=\frac{f(x_1)f(x_2|x_1)}{f(x_2)}. since we already know the "a priori":f(x_1). And this time f(x_2|x_1)f(x_1)==f(x_1|x_2)f(x_2) for sure. Thanks again.

Hi, all. I happened to think about a problem about conditional PDF: x_2=x_1+a, x_1 \approx \mathcal{N}(0,1), a \approx \mathcal{N}(0,1) so the conditional PDF of f(x_2|x_1), f(x_1|x_2) would both be f(x_2|x_1)=f(x_1|x_2)=\frac{1}{\sqrt{2\pi}}\exp{(-\frac{(x_1-x_2)^2}{2})} And it is clear...