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- Thread starter simon11
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- Feb 13, 2012

- 1,704

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.

Kind regards

$\chi$ $\sigma$

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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...

Kind regards

$\chi$ $\sigma$

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

Thanks

- Feb 13, 2012

- 1,704

'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

Kind regards

$\chi$ $\sigma$

- Thread starter
- #5

'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...

Kind regards

$\chi$ $\sigma$

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?

- Jan 26, 2012

- 890

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

$\chi$ $\sigma$

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

CB

Last edited:

- Feb 13, 2012

- 1,704

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...

Kind regards

$\chi$ $\sigma$

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

$\chi$ $\sigma$

- Jan 26, 2012

- 890

(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

$\chi$ $\sigma$

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

CB

Last edited:

- Feb 13, 2012

- 1,704

One hypothesis is that $\displaystyle f_{X}(x|Y=y)$ is(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

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

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

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

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

$\chi$ $\sigma$

CB

- Feb 13, 2012

- 1,704

In the original post one hypothesis extablishes that $\displaystyle \varphi(x,y)=f_{X} (x|Y=y)$ isBut 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

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

Already answered in post #8In the original post one hypothesis extablishes that $\displaystyle \varphi(x,y)=f_{X} (x|Y=y)$ isnormalin (x,y)... well!... what does it mean that a $\displaystyle \varphi(x,y)$ is normal in (x,y)?...

Kind regards

$\chi$ $\sigma$

CB

- Feb 13, 2012

- 1,704

It seems that 'Monster Wolfram' doesn't agree with You...Already answered in post #8

CB

Bivariate Normal Distribution -- from Wolfram MathWorld

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

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

Kind regards

$\chi$ $\sigma$

CB

Last edited:

- Feb 13, 2012

- 1,704

'... because the more you know, the less you like wasting time...'

Kind regards

$\chi$ $\sigma$

- Jan 26, 2012

- 890

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...

Kind regards

$\chi$ $\sigma$

CB

- Feb 13, 2012

- 1,704

http://www.mathhelpboards.com/f19/some-considerations-about-conditional-normal-distribution-2189/

... a thread I opened at this scope on request of the Administration. It is fully evident that the level of discussion exceeds the knowledge required in the 'Basic Probability and Statistic forum' but of course that is a minor problem respect to give correct and appropriate answers in any case...

Now I suggest to close this discussion with a smile spending some word about the 'infamous and insulting' verse...

'... che' perder tempo a chi piu' sa piu' spiace...'

This sentence is incised on a tablet on the wall of porter's lodge of 'Collegio Ghislieri' of Pavia...

Ghislieri College - Wikipedia, the free encyclopedia

... of which I have been 'schoolboy' in the distant years of university. The pitcures below show the tablet and myself under the tablet in a meeting of some years ago. The tablet has a long story because in the centuries has be lost and regained several time in competition with other Colleges of Pavia. In some sense the sentence is in DNA of the 'Ghislerians'...

Kind regards

$\chi$ $\sigma$