Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $-e^{-\lambda x} + C$

share|improve this question

2 Answers 2

up vote 5 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 $F(x)=0$.

share|improve this answer
    
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 $x>0$,

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.