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.
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
$\chi$ $\sigma$
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.
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
$\chi$ $\sigma$
Last edited by CaptainBlack; October 20th, 2012 at 23:12.
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
$\chi$ $\sigma$