# 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 $\int_{x_0}^\infty{f(x)/(1-F(x_0))} = 1$.

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HINT $(1-F(x_0))$ is a constant and comes out of the integral. – Srivatsan Sep 19 '11 at 19:46
What is $F$ here? – user5137 Oct 19 '11 at 23:11

You should verify two things:

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

2. $g$ 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 $F(\cdot)$?

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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,

$0<\frac{f(x)}{1-F(x_0)}<1$

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Sorry, but a PDF need not be less than $1$. It should only integrate to $1$. (Think about a $\delta(\cdot)$ distribution; it has "infinite" height.) And I don't think the algebra in the last step is correct :-) – Srivatsan Sep 19 '11 at 22:37