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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...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...
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$\chi$ $\sigma$
'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...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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$\chi$ $\sigma$
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.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
$\chi$ $\sigma$
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...'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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$\chi$ $\sigma$
(3) is wrong, it should contain the conditional mean and variance of x (conditioned on y) which may be functions of y.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
$\chi$ $\sigma$
One hypothesis is that $\displaystyle f_{X}(x|Y=y)$ is normal so that it can be in any case written as...(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
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.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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$\chi$ $\sigma$
In the original post one hypothesis extablishes that $\displaystyle \varphi(x,y)=f_{X} (x|Y=y)$ is normal in (x,y)... well!... what does it mean that a $\displaystyle \varphi(x,y)$ is normal in (x,y)?...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
Already answered in post #8In the original post one hypothesis extablishes that $\displaystyle \varphi(x,y)=f_{X} (x|Y=y)$ is normal in (x,y)... well!... what does it mean that a $\displaystyle \varphi(x,y)$ is normal in (x,y)?...
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$\chi$ $\sigma$
It seems that 'Monster Wolfram' doesn't agree with You...Already answered in post #8
CB
We don't necessarily have a bi-variate normal distribution. That is the mistake you have been making all along, assuming that we do, it would have to be proven.It seems that 'Monster Wolfram' doesn't agree with You...
Bivariate Normal Distribution -- from Wolfram MathWorld
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$\chi$ $\sigma$
That would be because you can't justify your claim!I'm sure that all friends of MHB now undestand why I decide to stop this boring and time wasting discussion...
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$\chi$ $\sigma$