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Question on Conditional PDF

  1. Jun 21, 2012 #1
    Hi, all. I happened to think about a problem about conditional PDF:
    [itex]x_2=x_1+a, x_1 \approx \mathcal{N}(0,1), a \approx \mathcal{N}(0,1)[/itex]
    so the conditional PDF of [itex]f(x_2|x_1), f(x_1|x_2)[/itex] would both be
    And it is clear that [itex]f(x_1)[/itex] and [itex]f(x_2)[/itex] are not identical, so
    [itex]f(x_1)f(x_2|x_1) \neq f(x_2)f(x_1|x_2)[/itex]

    How does this occur?

    Thanks in advance.
  2. jcsd
  3. Jun 21, 2012 #2


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    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.
  4. Jun 22, 2012 #3
    Dear haruspex, I really appreciate your reply.
    Your tolling rice model is really helpful, now I know that [itex]f(x_1|x_2)=\frac{f(x_1)f(x_2|x_1)}{f(x_2)}[/itex]. since we already know the "a priori":[itex]f(x_1)[/itex]. And this time [itex]f(x_2|x_1)f(x_1)==f(x_1|x_2)f(x_2)[/itex] for sure.

    Thanks again.
  5. Jun 22, 2012 #4
    Hi d9,

    Actually [itex]f(x_2|x_1)≠f(x_1|x_2)[/itex], they just happen to have the same expression, that's why your confusion.
  6. Jun 22, 2012 #5
    Dear viraltux,
    In fact, [itex]f(x_2|x_1) and f(x_1|x_2)[/itex] do not have the same expression, to be more specific,
    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).
    Anyway, thanks for your reply.
  7. Jun 22, 2012 #6
    More even so then :smile:
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