Having a difficult proving that $g(x)= f(x)/(1-F(x_0))$, $x \geq x_0$ and 0 otherwise is a valid PDF. I have shown the first to criteria for it to be a PDF, in which that all values $x \leq x_0$ are 0, and that for all $x \geq x_0$, since $F(x_0) < 1$, then $1 - F(x_0) > 0$ and $f(x)$ is another valid PDF. The trouble is showing that $\int_{x_0}^\infty{f(x)/(1-F(x_0))} = 1$.
Showing that $g(x)$ is a valid PDF
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probability
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3HINT $(1-F(x_0))$ is a constant and comes out of the integral. – 2011-09-19
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0What is $F$ here? – 2011-10-19