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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0What is $F$ here? – 2011-10-19
2 Answers
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 $F(\cdot)$?
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
$0
Since $f(x)$ is a valid pdf,
$0<\frac{f(x)}{1-F(x_0)}<1$
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0Sorry, 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 :-) – 2011-09-19