Normal Conditional Distribution

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If two random variables X and Y are marginally normal and the conditional distribution of Y given X is normal, then the conditional distribution of X given Y is also normal, even if X and Y are not independent. This conclusion is supported by the relationship between their probability density functions (PDFs), where the product of two normal PDFs results in another normal PDF. The proof involves demonstrating that the posterior distribution remains normal under these conditions. However, the covariance matrix will reflect any dependencies between X and Y, affecting the integration limits for probability calculations. Ultimately, normality is preserved in the conditional distributions as long as the initial conditions are met.
learner928
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Anyone know answer to below.

If two random variables X and Y are both marginally normal, and conditional distribution of Y given any value of X is also normal.

Does this automatically mean the conditional distribution of X given any value of Y also has to be normal? or not necessarily.
 
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Hey learner928 and welcome to the forums.

GIven P(Y|X) is Normal, P(X) is Normal, P(Y) is normal so P(X|Y) is proportional to P(Y|X)*P(X) (using a posterior) and if P(Y|X) has normal PDF with P(X) having a normal PDF, then the posterior will also have a normal PDF.

You can prove this by starting off with two PDF expressions P(A) = Normal_PDF_1, P(B) = Normal_PDF_2 and then multiply the two together and show that the product is also a Normal PDF and you're done.
 
chiro said:
Hey learner928 and welcome to the forums.

GIven P(Y|X) is Normal, P(X) is Normal, P(Y) is normal so P(X|Y) is proportional to P(Y|X)*P(X) (using a posterior) and if P(Y|X) has normal PDF with P(X) having a normal PDF, then the posterior will also have a normal PDF.

You can prove this by starting off with two PDF expressions P(A) = Normal_PDF_1, P(B) = Normal_PDF_2 and then multiply the two together and show that the product is also a Normal PDF and you're done.


Thanks Chiro, just to confirm, so your argument still applies even if P(X) and P(Y) are not independent right,

so in conclusion,
Even if P(X) and P(Y) are not independent, if P(X), P(Y) and P(Y|X) are all normal, then P(X|Y) also has to be normal.
 
If they are not independent normal then you will have a covariance matrix with off-diagonal positions, or you will have in general, a relationship where A = f(B) [A and f(A) both normal] so what you should do in this case is use limits that reflect the dependencies (which is what happens in dependent distributions).

You should however be able to prove that the product is normal (even if they are dependent or related) by assuming both have normal PDF and thus the product has a normal PDF.

The big difference will be the region of integration, and how these limits allow you to calculate an actual probability for your final variable.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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