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conditional normal distribution

simon11

New member
Oct 20, 2012
3
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.
 

chisigma

Well-known member
Feb 13, 2012
1,704
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


$\chi$ $\sigma$
 

simon11

New member
Oct 20, 2012
3
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
 

chisigma

Well-known member
Feb 13, 2012
1,704
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...


Kind regards


$\chi$ $\sigma$
 

simon11

New member
Oct 20, 2012
3
'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?

 

CaptainBlack

Well-known member
Jan 26, 2012
890
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


$\chi$ $\sigma$
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
 
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chisigma

Well-known member
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...


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...


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


$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 
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chisigma

Well-known member
Feb 13, 2012
1,704
(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...


Kind regards


$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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
 

chisigma

Well-known member
Feb 13, 2012
1,704
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
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)?...

Kind regards

$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
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)?...

Kind regards

$\chi$ $\sigma$
Already answered in post #8

CB
 

CaptainBlack

Well-known member
Jan 26, 2012
890
Last edited:

chisigma

Well-known member
Feb 13, 2012
1,704
Dante Alighieri.jpg

Durante degli Alighieri, better known as Dante, (1 june 1265 - 13 sepetember 1321) was an Italian Florentine poet. His greatest work, La divina commedia ( The Divine Comedy) is considered as one of the greatest literary statements produced in Europe in the medieval period and it is the basis of the modern Italian language. From Divine Comedy, Purgatorio, Canto III, line 78...


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


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

I'm sure that all friends of MHB now undestand why I decide to stop this boring and time wasting discussion...


Kind regards


$\chi$ $\sigma$
 

CaptainBlack

Well-known member
Jan 26, 2012
890
I'm sure that all friends of MHB now undestand why I decide to stop this boring and time wasting discussion...


Kind regards


$\chi$ $\sigma$
That would be because you can't justify your claim!

CB
 

chisigma

Well-known member
Feb 13, 2012
1,704
A definite and [I hope...] fully exhaustive answers to the problem proposed by simon11 can be found in...

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


$\chi$ $\sigma$

Perdertempo[4].jpg Perdertempo[3].JPG