# Question on Conditional PDF

1. Jun 21, 2012

### 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?

2. Jun 21, 2012

### 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.

3. Jun 22, 2012

### d9dd9dd9d

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.

4. Jun 22, 2012

### viraltux

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.

5. Jun 22, 2012

### 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,
$\begin{array}{l} f({x_2}|{x_1}) = \frac{1}{{\sqrt {2\pi } }}\exp \left( { - \frac{{{{({x_2} - {x_1})}^2}}}{2}} \right),\\ f({x_1}|{x_2}) = \frac{2}{{\sqrt {2\pi } }}\exp \left( { - \frac{{{{({x_2} - {x_1})}^2}}}{2} - \frac{{x_1^2}}{2} + \frac{{x_2^2}}{4}} \right). \end{array}$