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Bayes formula and picking variance from distribution

  1. Sep 29, 2011 #1
    This should be rather simple bayesian problem, but I can't figure it out for myself.

    If i pick numbers from normal distributionS, where the variance of the distribution at each pick v1, is in turn picked out of a normal distribution with variance v2.

    What is then the distribution of the random number? i tried this on my calculator a bit, and it looks as if it is normal itself, but what is the variance of x?

    This is not homework, but something I would like to understand how to calculate.

    //Cal
     
  2. jcsd
  3. Sep 29, 2011 #2

    mathman

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    You need to clarify this.
     
  4. Sep 29, 2011 #3

    Mute

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    If you have two continuous random variables (for example), X and Y, with joint pdf [itex]\rho_{X,Y}(x,y)[/itex], this can be written in terms of the conditional distribution [itex]\rho_{X|Y}(x|y)[/itex] or [itex]\rho_{Y|X}(y|x)[/itex] as:

    [tex]\rho_{X,Y}(x,y) = \rho_{X|Y}(x|y)\rho_Y(y) = \rho_{Y|X}(y|x)\rho_X(x),[/tex]

    where the individual distributions are

    [tex]\rho_X(x) = \int_{-\infty}^\infty dy~\rho_{X,Y}(x,y),[/tex]
    and similarly for y. So, if you knew the distribution for y and you knew the condition distribution for x given y, you can calculate the distribution of x as

    [tex]\rho_X(x) = \int_{-\infty}^\infty dy~\rho_{X|Y}(x|y)\rho_Y(y).[/tex]

    Amusingly enough, I was dealing with this concept myself this week, although in the context of a much more intractable problem.

    This said, I think you need to tweak your suggested problem. The variance of a distribution is positive, so it can't be normally distributed. If you use the standard deviation, I don't think you'll get a result, because you'll have a factor of [itex]1/\sigma[/itex] in your integration, why of course blows up at the origin, but the exponential will also be even in sigma, so the principal value might be zero.
     
    Last edited: Sep 29, 2011
  5. Sep 30, 2011 #4
    Thank you mute, all you said seems very right and true to me. with large enough mu compared to sigma it will become pretty normal. I might have to use some other distribution to pick the variance if i use this for something. But this was more of a though experiment.
     
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