What is the relationship between conditional PDFs of x1 and x2?

[d9dd9dd9d]

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 that f(x_1) and f(x_2) are not identical, so
f(x_1)f(x_2|x_1) \neq f(x_2)f(x_1|x_2)

How does this occur?

Thanks in advance.

 

[haruspex]

I think P(x1|x2) is more subtle than that. Consider a simpler model. X1 and A are dice rolls. The distribution of x2 = x1+a given x1=3 is uniform 4 to 9. The distribution of X1 given X2 = 9 is uniform 3 to 6, not uniform 3 to 8. I.e. the distribution of X1 given X2 is a consequence of both the distribution of A and the a priori distribution of X1.

 

[d9dd9dd9d]

I think P(x1|x2) is more subtle than that. Consider a simpler model. X1 and A are dice rolls. The distribution of x2 = x1+a given x1=3 is uniform 4 to 9. The distribution of X1 given X2 = 9 is uniform 3 to 6, not uniform 3 to 8. I.e. the distribution of X1 given X2 is a consequence of both the distribution of A and the a priori distribution of X1.

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.

 

[viraltux]

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 that f(x_1) and f(x_2) are not identical, so
f(x_1)f(x_2|x_1) \neq f(x_2)f(x_1|x_2)

How does this occur?

Thanks in advance.

Hi d9,

Actually f(x_2|x_1)≠f(x_1|x_2), they just happen to have the same expression, that's why your confusion.

 

[d9dd9dd9d]

Hi d9,

Actually f(x_2|x_1)≠f(x_1|x_2), they just happen to have the same expression, that's why your confusion.

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}<br /> f({x_2}|{x_1}) = \frac{1}{{\sqrt {2\pi } }}\exp \left( { - \frac{{{{({x_2} - {x_1})}^2}}}{2}} \right),\\<br /> f({x_1}|{x_2}) = \frac{2}{{\sqrt {2\pi } }}\exp \left( { - \frac{{{{({x_2} - {x_1})}^2}}}{2} - \frac{{x_1^2}}{2} + \frac{{x_2^2}}{4}} \right).<br /> \end{array}
Anyway, thanks for your reply.

 

[viraltux]

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,

More even so then