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Assume $X,Y$ to be random variables defined on $(\Omega,\Sigma,P)$, $Y = aX + b$ and that $X$ is Gaussian. I wanted to derive the conditional p.d.f. defined by:

$$f_{Y|X}(y|x) := \frac{f_{X,Y}(x,y)}{f_X(x)}$$

It is easy to show that $E[Y|X] = Y$ a.s. and also if $g(x) = \int_\mathbb{R} yf_{Y|X}(y|x)dy$ then $(g \circ X)(\omega) = g(X(\omega))$ is also a version of the conditional expectation (This is shown in the textbook titled "Probability With Martignales" by David Williams). Therefore,

$$g(X(\omega)) = aX(\omega) + b\text{ a.s.}$$

Is it possible to derive $f_{Y|X}(y|x)$ from the above relation? If not, what is the best way to derive this?

Any help is much appreciated. Thanks

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Conditioned on the value of $X$, $Y$ is a constant, and its conditional pdf is thus a discrete "distribution" assigning probability mass $1$ to the value $aX+b$. Note that $X$ and $Y$ are not jointly continuous random variables, and $f_{X,Y}(x,y)$ is a line density with all the probability mass on the line $y = ax+b$. –  Dilip Sarwate Mar 21 '12 at 15:11
    
Thanks! The situation is intuitively clear. I want to know if we can prove this rigorously. –  jpv Mar 21 '12 at 18:52
    
$Y = aX+b$ is either an identity: $Y(\omega) = aX(\omega) + b$ for all $\omega \in \Omega$ or holds almost surely ($Y(\omega) = aX(\omega) + b$ for almost all $\omega \in \Omega$) depending on what you mean when you write $Y = aX+b$. Thus, given that $X(\omega) = x$, either $Y(\omega) = ax+b$ in which case $P\{Y=ax+b|X=x\} = 1$ or $Y(\omega) = ax+b$ for almost all $\omega \in \Omega$ and so $P\{Y=ax+b|X=x\} = 1$ once again. Thus, given $X=x$, either $Y$ is a constant $ax+b$, or almost surely the constant $ax+b$ and in either case, $P\{Y=ax+b|X=x\} = 1$. –  Dilip Sarwate Mar 21 '12 at 19:17

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