1
$\begingroup$

Good morning Math Stack Exchange

In my journey to learn Measure Theory, I wish to understand with clarity the link between a random variable's distribution function (df) and it's distribution. I believe some would call the distribution a density but in Resnick's "A Probability Path", the book I'm using, he calls it distribution.

I've been reading the chapter on Integration and Expectation and oddly enough, there isn't a theorem on how we can get the distribution from its df. Undergraduate classes would be quick to say "differentiate the df to get the distribution". But I'm sure there are numerous df which are not continuous hence not differentiable, i.e., $F(x)=\sum\frac{1}{2^n}1_{x\le n}$. For such df's, is it possible, and under what conditions, for us to get the df?

To me, I feel that we can easily get the df from the measure on defined on $(\Omega,\mathcal{F})$ namely $F(x)=P[X. The problem comes in getting the distribution. Most of the time, I need it so I can integrate in $\mathbb{R}$ to prove that, say, something is in $L_1$. Can we just differentiate $F(x)$?

Thank you!

  • 0
    What do you mean when you say "getting the distribution"? The density? Are you aware of the fact that many distributions do not have a density function?2011-10-23
  • 0
    Hello Rasmus. I do mean the density and am also aware the fact that many distributions do not have a density function. My question then is that 1. If the distribution function is differentiable, is it the density and 2. If a random variable does not have a density, does that mean I can't integrate in R to get expectation?2011-10-23

2 Answers 2