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Im considering the double exponential with parameter $\lambda$, $$g(x)=\frac{\lambda}{2}e^{\lambda x}, x<0; \frac{\lambda}{2}e^{-\lambda x}, \geq 0$$

Just simple one looking for the c.d.f, I can't for some reason get the expression for $\geq$ 0 as I get $1/2 -\frac{1}{2}e^{-\lambda x}$.

Many thanks in advance.

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Sorry but what is $c.d.f.$? –  1015 Feb 2 '13 at 14:36
    
cdf = cumulative distribution function. Contrast this with pdf, probability distribution function –  bryansis2010 Feb 2 '13 at 14:52
    
The c.d.f. $G(x)$ is the integral of $g(t)$ over the interval $(-\infty,x)$. This will be a piecewise function changing its rule at $x=0$; remember for $x>0$ to include the result of integrating from $-\infty$ to $0$ plus the integral from $0$ to $x$. –  coffeemath Feb 2 '13 at 14:56
    
@coffeemath Got you forget these simple things, many thanks! –  user24930 Feb 2 '13 at 15:33

1 Answer 1

up vote 0 down vote accepted

For the sake of having an answer: $G(x)=\frac12\mathrm e^{\lambda x}$ if $x\leqslant0$ and $G(x)=1-\frac12\mathrm e^{-\lambda x}$ if $x\geqslant0$.

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