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  1. MHB Apprentice

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    #1
    Assume two random variables X and Y are not independent,

    if P(X), P(Y) and P(Y|X) are all normal, then does P(X|Y) also can only be normal or not necessarily?

    thanks.

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    #2
    Quote Originally Posted by simon11 View Post
    Assume two random variables X and Y are not independent,

    if P(X), P(Y) and P(Y|X) are all normal, then does P(X|Y) also can only be normal or not necessarily?

    thanks.
    Because is $\displaystyle f_{Y} (y|X=x) = \frac{f_{X,Y} (x,y)}{f_{X} (x)}$ and $\displaystyle f_{X} (x|Y=y) = \frac{f_{X,Y} (x,y)}{f_{Y}(y)}$ if $f_{X}$, $f_{Y}$ and $f_{Y} (y|X=x)$ are all normal, for simmetry also $f_{X} (x|Y=y)$ is normal...


    Kind regards


    $\chi$ $\sigma$

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    Quote Originally Posted by chisigma View Post
    Because is $\displaystyle f_{Y} (y|X=x) = \frac{f_{X,Y} (x,y)}{f_{X} (x)}$ and $\displaystyle f_{X} (x|Y=y) = \frac{f_{X,Y} (x,y)}{f_{Y}(y)}$ if $f_{X}$, $f_{Y}$ and $f_{Y} (y|X=x)$ are all normal, for simmetry also $f_{X} (x|Y=y)$ is normal...


    Kind regards


    $\chi$ $\sigma$

    Thank you very much, but what do you by symmetry and why does it need to be symmetrical?

    Thanks

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    #4
    Quote Originally Posted by simon11 View Post
    Thank you very much, but what do you by symmetry and why does it need to be symmetrical?

    Thanks
    'Symmetry' means that if $f_{X}(x)$, $f_{Y}(y)$ and $f_{Y}(y|X=x)$ are gaussian, then $\displaystyle f_{X}(x|Y=y)= f_{Y}(y|X=x)\ \frac{f_{X}(x)}{f_{Y}(y)}$ is also gaussian...


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    #5 Thread Author
    Quote Originally Posted by chisigma View Post
    'Symmetry' means that if $f_{X}(x)$, $f_{Y}(y)$ and $f_{Y}(y|X=x)$ are gaussian, then $\displaystyle f_{X}(x|Y=y)= f_{Y}(y|X=x)\ \frac{f_{X}(x)}{f_{Y}(y)}$ is also gaussian...


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    $\chi$ $\sigma$

    oh i didn't know that thanks.

    I only knew $\displaystyle frac{f_{X}(x)}{f_{Y}(y)}$ is a cauchy distribution,
    so the product of a cauchy distribution and a normal distribution has to get you back to a normal distribution?


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    #6
    Quote Originally Posted by chisigma View Post
    Because is $\displaystyle f_{Y} (y|X=x) = \frac{f_{X,Y} (x,y)}{f_{X} (x)}$ and $\displaystyle f_{X} (x|Y=y) = \frac{f_{X,Y} (x,y)}{f_{Y}(y)}$ if $f_{X}$, $f_{Y}$ and $f_{Y} (y|X=x)$ are all normal, for simmetry also $f_{X} (x|Y=y)$ is normal...


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    $\chi$ $\sigma$
    You seem to be assuming rather more than I am happy with. You will need to prove, or give a reference to the proof, that conditional probability of Y given X, and that the marginals of X and Y are normal implies that the conditional probability of X given Y is normal.

    It may be true, I can't find a counter example.

    CB
    Last edited by CaptainBlack; October 20th, 2012 at 12:54.

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    #7
    Quote Originally Posted by chisigma View Post
    'Symmetry' means that if $f_{X}(x)$, $f_{Y}(y)$ and $f_{Y}(y|X=x)$ are gaussian, then $\displaystyle f_{X}(x|Y=y)= f_{Y}(y|X=x)\ \frac{f_{X}(x)}{f_{Y}(y)}$ is also gaussian...


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    $\chi$ $\sigma$
    May be that, as required from 'regulations', a formal proof of that has to be supplied. Very well!... if $f_{X}(x)$, $f_{Y}(y)$ and $f_{Y}(y|X=x)$ are gaussian that means that is...


    $\displaystyle f_{X}(x)= \frac{1}{\sigma_{x}\ \sqrt{2\ \pi}}\ e^{- \frac{(x-\mu_{x})^{2}}{2\ \sigma^{2}_{x}}}$ (1)

    $\displaystyle f_{Y}(y)= \frac{1}{\sigma_{y}\ \sqrt{2\ \pi}}\ e^{- \frac{(y-\mu_{y})^{2}}{2\ \sigma^{2}_{y}}}$ (2)

    $\displaystyle f_{X}(x|Y=y)= A\ e^{[a\ (x-\mu_{x})^{2} + 2\ b\ (x-\mu_{x})\ (y-\mu_{y}) + c\ (x-\mu_{x})^{2}]}$ (3)

    Now the p.d.f. of Y conditioned by X is...

    $\displaystyle f_{X}(x|Y=y)= f_{Y}(y|X=x)\ \frac{f_{X}(x)}{f_{Y}(y)}$ (4)

    ... and clearly it is similar to (3), i.e. is gaussian...


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    $\chi$ $\sigma$

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    #8
    Quote Originally Posted by chisigma View Post
    May be that, as required from 'regulations', a formal proof of that has to be supplied. Very well!... if $f_{X}(x)$, $f_{Y}(y)$ and $f_{Y}(y|X=x)$ are gaussian that means that is...


    $\displaystyle f_{X}(x)= \frac{1}{\sigma_{x}\ \sqrt{2\ \pi}}\ e^{- \frac{(x-\mu_{x})^{2}}{2\ \sigma^{2}_{x}}}$ (1)

    $\displaystyle f_{Y}(y)= \frac{1}{\sigma_{y}\ \sqrt{2\ \pi}}\ e^{- \frac{(y-\mu_{y})^{2}}{2\ \sigma^{2}_{y}}}$ (2)

    $\displaystyle f_{X}(x|Y=y)= A\ e^{[a\ (x-\mu_{x})^{2} + 2\ b\ (x-\mu_{x})\ (y-\mu_{y}) + c\ (x-\mu_{x})^{2}]}$ (3)

    Now the p.d.f. of Y conditioned by X is...

    $\displaystyle f_{X}(x|Y=y)= f_{Y}(y|X=x)\ \frac{f_{X}(x)}{f_{Y}(y)}$ (4)

    ... and clearly it is similar to (3), i.e. is gaussian...


    Kind regards


    $\chi$ $\sigma$
    (3) is wrong, it should contain the conditional mean and variance of x (conditioned on y) which may be functions of y.

    \[ f_{X|Y=y}(x) = \frac{1}{\sqrt{2 \pi}\sigma_{X|Y=y}} e^{-(x-\mu_{X|Y=y})^2/(2\sigma_{X|Y=y}^2)} \]

    CB
    Last edited by CaptainBlack; October 20th, 2012 at 23:12.

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    #9
    Quote Originally Posted by CaptainBlack View Post
    (3) is wrong, it should contain the conditional mean and variance of x (conditioned on y) which may be functions of y.

    \[ f_{X|Y=y}(x) = \frac{1}{\sqrt{2 \pi}\sigma_{X|Y=y}} e^{-(x-\mu_{X|Y=y})^2/(2\sigma_{X|Y=y}^2)} \]

    CB
    One hypothesis is that $\displaystyle f_{X}(x|Y=y)$ is normal so that it can be in any case written as...

    $\displaystyle f_{X} (x|Y=y) = A\ e^{[a\ (x-x_{0})^{2} + b\ (x-x_{0})\ (y-y_{0}) + c\ (y-y_{0})^{2}]}$ (1)

    ... where $A$, $a$, $b$, $c$, $x_{0}$ and $y_{0}$ are constants...


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    $\chi$ $\sigma$

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    #10
    Quote Originally Posted by chisigma View Post
    One hypothesis is that $\displaystyle f_{X}(x|Y=y)$ is normal so that it can be in any case written as...

    $\displaystyle f_{X} (x|Y=y) = A\ e^{[a\ (x-x_{0})^{2} + b\ (x-x_{0})\ (y-y_{0}) + c\ (y-y_{0})^{2}]}$ (1)

    ... where $A$, $a$, $b$, $c$, $x_{0}$ and $y_{0}$ are constants...


    Kind regards


    $\chi$ $\sigma$
    But that is not forced by the conditions in the problem, or rather it may be but must be demonstrated if it is (without evidence I don't believe it is forced). The normality of the conditional distribution of X does not entail such a result in itself.

    CB

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