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$X$ has an exponential distribution, $Pr(X>0)=1$, p.d.f is $f$, c.d.f is $F$. $h(x)=\frac{f(x)}{1-F(x)}$ for $x>0$. Prove that $h(x)$ is constant for $x>0$.

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    Welcome to math.SE: since you are a new user, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say what your thoughts on the problem are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. Also, many find the use of imperative ("prove") to be rude when asking for help; please consider rewriting your post.2012-06-08
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    Also, please see [here](http://meta.math.stackexchange.com/a/464/264) and [here](http://meta.stackexchange.com/a/70559/161783) for how to format your mathematics expressions with LaTeX. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own [TeX.SE](http://tex.stackexchange.com/) site. If you see a piece of LaTeX you want to know the code for on the site, you can right click on it, go to "Show Math As", then choose "TeX Commands" - this is a good way of picking up how to do things.2012-06-08

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We have $f(x)=\lambda e^{-\lambda x}$ (for $x\ge 0$). For the cumulative density function $F(x)$, we have $$F(x)=\int_0^x \lambda e^{-\lambda t}\,dt=1-e^{-\lambda x}$$ (if $x\ge 0$).

So $1-F(x)=e^{-\lambda x}$, and therefore $$h(x)=\frac{f(x)}{1-F(x)}=\frac{\lambda e^{-\lambda x}}{e^{-\lambda x}}=\lambda,$$ a constant, (if $x\ge 0$).

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    oh,I misinterpret the exponential distribution.thank you!2012-06-08