I'm trying to get the cdf from a pdf, which, according to this MIT open courseware video, is as simple as getting integral from -$\infty$ to $\infty$. Like this:

$$ F_z(z) = \int_{-\infty}^\infty f_z(z)dz\, = 1 $$

And that works for functions like this: $f_x(X)$

But what about a function like this $f(X;\beta)$?

Yeah, this is a homework problem, so I don't expect an exact answer, but hopefully some general principles. I have a hard time googling to learn how to do this because I don't know what to call this kind of function $f(X;\beta)$, and I don't know how to type a $\beta$ into google.

The pdf I have is:

$$ f(X;\beta)= \begin{cases} \frac {e^{-X/\beta}}{\beta}, & \ X\geq0 \\ 0, & \text{otherwise} \end{cases} $$

and the cdf it comes out to is: $$ f(X;\theta)=1-e^{-X/\beta} $$

Part of my question is also, why did the $f(X;\beta)$ become $f(X;\theta)$?

  • $\begingroup$ My guess is that $X$ is the variable while $\beta$ and $\theta$ are parameters of the probability distribution but without context it is impossible to tell. If $X$ is indeed the variable (and you understand the underlying theory) you will realize that you have to take the integral of the pdf w.r.t. $X$. $\endgroup$
    – Giskard
    Jan 15, 2016 at 7:13
  • 1
    $\begingroup$ maybe you can take more rigorous explanations on MathematicaSE for this question as there is not an economic context on your question. $\endgroup$ Jan 15, 2016 at 10:37
  • $\begingroup$ If we were to understand integrals as measuring areas under a curve, then cdfs are just special integrals, the area under which integrates to $1$ on $(-\infty,\infty)$. With this understanding, pdfs described by piece-wise functions are not that different from other "better behaved" functions. A simple sketch of the pdf will show you how such function may be integrated. $\endgroup$
    – Herr K.
    Jan 15, 2016 at 18:46
  • $\begingroup$ How do I move this to MathematicaSE? @optimalcontrol. That sounds like a good idea. $\endgroup$
    – user4207
    Feb 12, 2016 at 19:32
  • 1
    $\begingroup$ @ColeTrumbo Sorry, I should have written MathematicsSE and not MathematicaSE. I just flagged your question to be transferred to MathematicsSE. $\endgroup$ Feb 12, 2016 at 23:54

2 Answers 2


In addition to what John said (I would have commented on his answer, but don't have enough reputation). And this is not conditional, conditional would be $f(X|\beta)$

$$ F_X(x;\beta) = P(X \leq x; \beta=\theta) = \int_{-\infty}^{x} f_X(x;\beta) dx $$

We fix $x=u$ and $\beta=\theta$

$$ P(X \leq u; \theta)=\int_{-\infty}^u f_X(u;\theta) du $$

$$ = \int_{- \infty}^0 f(u;\theta) du + \int_0^x f(u; \theta) du $$

$$ = 0 + \int_0^x \frac{e^{-u/\theta}}{\theta} du $$

$$ = 1 - e^{-x/\theta}$$

Then over the set $X$ and and at particular values $\beta=\theta$

$$ F_X(X;\theta) = 1 - e^{-X/\beta}$$


CDF = integral of PDF from $-\infty$ to $x$, or:

$F\left(x\right) = \int_{-\infty}^{x} f\left(u\right) du$.

Functions like $f\left(x;\beta\right)$ take $\beta$ as the parameters, the $\beta$ index different probability density functions. You don't need to integrate over them to get the CDF.

$F\left(x;\beta\right) = \int_{-\infty}^{x} f\left(u;\beta\right)du$.


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