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I have the following exponential distribution:

$$f(\lambda, x) = \begin{cases} \lambda e^{-\lambda x} &\text{if } x \geq 0 \\ 0 & \text{if } x<0. \end{cases}$$

I need to show that this expression integrates into

$$F(\lambda, x) = \begin{cases} 1-e^{-\lambda x} &\text{if } x \geq 0 \\ 0 & \text{if } x<0. \end{cases}$$

I know that the integral of a pdf is equal to one but I'm not sure how it plays out when computing for the cdf. I computed the indefinite integral of $\lambda e^{-\lambda x}$ and got $-e^{-\lambda x} + C$

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up vote 13 down vote accepted

For a fixed $\lambda$, let $X$ be the random variable in question. I will denote values of the density and distribution of $X$, simply, as $f(x)$ and $F(x)$, respectively.

By definition, $F(x)=P[X\le x]$. To compute this probability, you would integrate the density, $f$, from $-\infty$ to $x$.

For $x\ge0$:

$$ \eqalign{ F(x) =P[X\le x]&=\int_{-\infty}^x f(t)\,dt\cr &=\int_0^x \lambda e^{-\lambda t}\,dt\cr &= -e^{-\lambda t}\,\bigl|_0^x \cr &=-e^{-\lambda x}-(-e^0)\cr &=1-e^{-\lambda x}. }$$

Note that $f(x)=0$, for $x\le0$; hence the third equality above.

For $x\le0$, when computing $F(x)$, you would be integrating the zero function, and then you would conclude that $F(x)=0$.

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You guys are both awesome! Pardon my non-math language – Low Scores Apr 20 '12 at 0:40

By definition of the cumulative and probability density functions,

$$F(\lambda,x):= \int_{-\infty}^x f(\lambda,u)du$$

Thus, by elementary calculus, for $x\le 0$ we have $F=0$ and for $x>0$,

$$F=\int_{-\infty}^0 0\, du+\int_0^x \lambda e^{-\lambda u}du=[-e^{-\lambda u}]_{0}^{x}=1-e^{-\lambda x}. $$

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I like your answer a lot. I wish I could give both of you guys a check mark – Low Scores Apr 20 '12 at 0:45

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