Showing that $g(x)$ is a valid PDF 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 $\int_{x_0}^\infty{f(x)/(1-F(x_0))} = 1$.
 A: You should verify two things:


*

*$g$ is nonnegative. I assume you can show this. 

*$g$ integrates to $1$. 
$$
\int_{\mathbb R} g(x) dx = \int_{-\infty}^{x_0} g(x) dx + \int_{x_0}^{\infty} g(x) dx.
$$
The first integral is obviously $0$. The second integral can be written as
$$
\int_{x_0}^{\infty} \frac{f(x)}{1-F(x_0)} dx.
$$
Recognize that $1-F(x_0)$ is constant with respect to $x$ and can be taken out. You will then get the integral
$$
\int_{x_0}^{\infty} f(x) dx,
$$
with some factor outside the integral. Can you write this integral in terms of $F(\cdot)$?
A: One thing to see is that 
$F(x_0)=\int_{-\infty}^{x_0}f(x)dx$ 
and 
$\int_{x_0}^{\infty}f(x)dx=1-F(x_0)=1-\int_{-\infty}^{x_0}f(x)dx=1 \cdot \Bigg(1-F(x_0) \Bigg)$
since $f(x)$ is a valid probability density. Then do a very simple algebra to prove that $\int_{x_0}^{\infty}g(x)dx$ integrates to 1.
But you also need to verify that $0<g(x) <1$. This is easy, since $F(x_0)$ is a cumulative distribution function, so it is bounded:
$0<F(x_0)<1, \ 0<1-F(x_0)<1$ and $\frac{1}{1-F(x_0)}>1$. 
Since $f(x)$ is a valid pdf, 
$0<\frac{f(x)}{1-F(x_0)}<1$
