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I'm intrigued by the following idea but I don't know how to do it.

If I have a r.v. $x$ with given distribution $f_X(x)$ and I have a second variable $y=2x$. The goal is to find $f_Y(y)$.

I know the traditional solution, but, can I say that from $y=2x$, I can conclude that $f(y\mid x)=2x \cdot \delta(y-2x)$ ?? If that is correct, then I can form the joint $f(x,y)$ then I can marginalize out $x$ to reach my goal.

I'm trying to do it but I'm missing something.

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    $\begingroup$ Maybe you mean $f_X(x)$ and $f_Y(y)$ instead of $f_x(x)$ and $f_y(y). \qquad$ $\endgroup$ – Michael Hardy Sep 10 at 0:23
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    $\begingroup$ @MichaelHardy Correct. Thanks. $\endgroup$ – John Deterious Sep 10 at 1:25
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    $\begingroup$ @JohnDeterious : If $x=3$ then $f(x) = f(3),$ and if $y = 3$ then $f(y) = f(3),$ so is $f(3)$ a value of the density function of (capital) $X$ or of (capital) $Y$? On the other hand $f_X(3)$ (with a capitlized subscript) and $f_Y(3)$ (likewise) are clear. So what you need is $f_X$ and $f_Y,$ not $f(x)$ and $f(y).$ This sort of thing can make a practical difference in writing some proofs that arise in exercises and elsewhere. $\qquad$ $\endgroup$ – Michael Hardy Sep 12 at 23:04
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    $\begingroup$ You must have meand $Y=2X$ rather than $y=2x. \qquad$ $\endgroup$ – Michael Hardy Sep 12 at 23:05
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    $\begingroup$ If one is to speak of this kind of joint density, it would need to be a generalized function $f$ having the property that for subsets $A$ of the plane $\mathbb R^2,$ you have $$ \iint\limits_A f(x,y)\,d(x,y) = \Pr\big( (X,Y)\in A\big). $$ $\endgroup$ – Michael Hardy Sep 13 at 4:02

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