1
$\begingroup$

I've just started with various types of continuous distributions and the very first one is the uniform continuous distribution. One can justify that the function given as its pdf is indeed A pdf, but how does one justify that pdf is indeed THE pdf of the distribution. One should then first define its cdf to be the improper integral of that function from negative infinity till t, for each t in R. But here I run into another problem. Is such a definition possible, without any inconsistencies with the axioms of probability? I've learned that the pdf is unique as it is also defined to be continuous on its support. Thus using the Fundamental Theorem of Calculus, one can see its uniqueness.

$\endgroup$
9
  • $\begingroup$ "the pdf is unique as it is also defined to be continuous on its support" No, PDFs are not unique. And I sincerely hope you were not told either that PDFs are continuous on their support. $\endgroup$ – Did Mar 12 '15 at 17:37
  • $\begingroup$ @Did I was told that this is the assumption one makes. As then the PDF is a unique function, as the CDF is differentiable on the support of the PDF, and by taking the derivative one gets the pdf on the support, and elsewhere it is zero. $\endgroup$ – Student Mar 15 '15 at 11:45
  • $\begingroup$ Argh... No, the CDF is not always differentiable on the support of the PDF. Example: $F(x)=0$ if $x\geqslant0$, $F(x)=x/2$ if $0<x\leqslant1/2$, $F(x)=(3x-1)/2$ if $1/2<x\leqslant1$, $F(x)=1$ if $x>1$. Not a legit CDF? (What is your source on this?) $\endgroup$ – Did Mar 15 '15 at 11:51
  • $\begingroup$ The CDF in my previous comment is not continuous, really? Sure about that? $\endgroup$ – Did Mar 15 '15 at 13:06
  • $\begingroup$ Sorry, elementary calculation error. But the way I've been taught to define the pdf is as a function such that the value of the cdf at t equals the integral of that function from negative infinity to t. Also, it must be continuous on its support. So yeah. And then, when such a function exists then the distribution is called a continuous distribution. $\endgroup$ – Student Mar 15 '15 at 13:55
0
$\begingroup$

If $F$ is the distribution of a random variable $X$, i.e. $$F(x) = \mathbb P(X\leqslant x)$$ for $x\in\mathbb R$, and $F$ is continuous (i.e. $X$ is a continuous random variable), then a probability density function of $X$ is any function $f$ satisfying $$\mathbb P(X\in B) = \int_B f(x)\mathsf dx $$ for all Borel sets $B$. This function need not be unique; for example, for the uniform $[0,1]$ distribution, we have $$ F(x) = \begin{cases} 0,& x\leqslant 0\\ x,& 0<x<1\\ 1,& x\geqslant 1.\end{cases}$$ One density would be $$ f(x) = 1_{(0,1)}(x).$$ But we could also use the density $$g(x) = 1_{[0,1]}(x), $$ as $f=g$ a.e. and so $$\int_B f(x)\mathsf dx = \int_B g(x)\mathsf dx$$ for any Borel set $B$.

$\endgroup$
0
$\begingroup$

A pdf is not unique. In measure theoretic probability you find that the pdf is determined only up to a set of measure zero (on the line, which intuitively means it has no length). If you alter the pdf on a set of measure zero, you obtain a new pdf, which has the same corresponding cdf.

This means that for example $1_{(0,1)}$ and $1_{[0,1]}$ are both valid pdfs of the variable which is uniform on $[0,1]$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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