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Suppose $X_{1}$ and $X_{2}$ and i.i.d random variables. Consider $K = X_{1}X_{2}$. Then does $f_{K}(k) = f_{X_{1}}(x_1) \cdot f_{X_{2}}(x_2)$?

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    No. To begin with, the equality does not even make sense (who are these $x_1$ and $x_2$ guys?).2012-03-12
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    @DidierPiau: I thought if $X$ and $Y$ are independent then $f_{XY}(x,y) = f_{X}(x) \cdot f_{Y}(y)$?2012-03-12
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    Yes--and this is unrelated to what your question says.2012-03-12
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    @DidierPiau: So to find this pdf, we would have to do some sort of transformation?2012-03-12
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    @alexm Distinguish between $f_{X,Y}$, the _joint_ density of random variables $X$ and $Y$, and $f_{XY}$, the density of the single random variable $XY$. It is true that $f_{X,Y}(x,y) = f_X(x)f_Y(y)$, but $f_{XY}(x,y)$ makes no sense (the density of a single random variable should have a single argument, not two arguments) and thus $f_{XY}(x,y) = f_{X}(x) \cdot f_{Y}(y)$ does not make sense since the left side is a function of one variable, not two.2012-03-12
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    @DilipSarwate: So to find the pdf of $XY$, how would you do that?2012-03-12
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    How about reading your probability textbook for a while and trying to understand what it says about solving such questions? Your several questions in the past few hours reveal wild misconceptions about standard basic notions, and seem to indicate that some serious self-study and trying to understand some basic ideas before asking another question on math.SE will probably serve you better.2012-03-12

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