Determining the CDF of the Laplace PDF

**statmajor** · November 21st 2009, 06:22 AM

$\frac{1}{2} \int^x_{- \infty} e^{-|t|} dt= \frac{1}{2} \int^x_{ \infty} e^{t}dt + \frac{1}{2} \int^x_{- \infty} e^{-t} dt$

I'm pretty sure I bounded the second integral incorrectly. What would be the correct way?

**Moo** · November 21st 2009, 06:42 AM

Hello,

First you have to determine if x is positive or negative...
Let's suppose it's negative.
Then $t\in(-\infty,x)\implies |t|=-t$

So $\int_{-\infty}^x e^{-|t|} ~dt=\int_{-\infty}^x e^t ~dt=\int_{-x}^\infty e^{-t} ~dt$ (if you substitute t=-t)

If x is positive, then $t\in(-\infty,0) \implies |t|=-t$ and $t\in(0,x)\implies |t|=t$

So $\int_{-\infty}^x e^{-|t|} ~dt=\int_{-\infty}^0 e^{t} ~dt+\int_0^x e^{-t} ~dt=1+\int_0^x e^{-t} ~dt$

**statmajor** · November 21st 2009, 06:52 AM

Thank you once again.

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