Let F and G be two distribution functions, does the product FG still a distribution function?
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$\begingroup$If, as the caps hint at, you mean that $F$ and $G$ are cumulative distribution functions, the answer is yes. We need to verify that the product has the required properties. So we want to show that $F(x)G(x)$ is continuous from the right, that $\lim_{x\to\infty}F(x)G(x)=1$, that $F(x)G(x)$ is non-decreasing, that $\lim_{x\to -\infty} F(x)G(x)=0$. The verifications are straightforward.
$\endgroup$ 1 $\begingroup$If $F$ and $G$ are two distribution functions, the product $FG$ is still a distribution function, as demonstrated in this answer by André Nicolas. I would like to expand on his comment.
By definition, $F(x) = P(X<x)$, $G(x) = P(Y<x)$ , where $X$ and $Y$ are random variables. Let's prove that, if $X$ and $Y$ are independent, $P(\max(X,Y)<x) = F(x)G(x)$.
$\max(X,Y) < x \implies X < x \land Y < x$
$P(\max(X,Y) < x) = P(X < x \land Y < x) = P(X < x) P( Y<x | X <x)$.
If $X$ and $Y$ are independent, $P( Y<x | X <x) = P( Y<x)$, and the line above becomes $P(\max(X,Y) < x) = P(X < x) P( Y<x) = F(x)G(x)$
If you call $Z=\max(X,Y)$ and the distribution function of Z is $H(x) = P(Z<x) = F(x)G(x)$.
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