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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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    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