I am a bit confused.

The probability law regarding a random variable is defined as mapping $\mathcal{P} : \mathbb{B} \to [0,1]$, where $\mathbb{B}$ is the regular Borel set; that is, a probability law is an image measure.

On the other hand, the probability distribution is $\mathbb{D} : \mathbb{R} \to [0,1]$, where $\mathbb{D}$ is defined as $\mathbb{D}(x) = P(\omega | X (\omega) < x)$.

Although they are very similar, they are not equivalent, since the latter has a very specific structure, but, in the former, one can find the measure of any Borel set. Why, then, do a lot of texts informally claim that they are the, in fact, equivalent?

It is clear that, for any measurable function, for example, a random variable, one can write its distribution using the probability law mapping, but, conversely, for example, assigning a probability measure to an arbitrary Borel set using the distribution function is not very clear to me.

I have never taken a formal course on probability theory, so forgive me if this question seems too stupid, or does not make much sense.

  • 1
    $\begingroup$ You should start by saying what $\mathbb B$ and $X$ is before using them. For example, if $X$ is a set, then $X(\omega)$ doesn't make sense, while if $X$ is a function then $\mathbb D:X\to[0,1]$ doesn't make sense. $\endgroup$ – Thomas Andrews Jun 3 '15 at 11:58
  • $\begingroup$ @ThomasAndrews My bad. Does the question make sense now? $\endgroup$ – user3503589 Jun 3 '15 at 12:08
  • 3
    $\begingroup$ Probability law of X = Law of X = Distribution of X. $\endgroup$ – Did Jun 3 '15 at 12:09
  • $\begingroup$ @Did Thank you everybody. $\endgroup$ – user3503589 Jun 3 '15 at 12:13

The distribution or the law of a random variable $X$ is a probability measure $\mathcal L$ on $(\mathbb R,\mathcal R)$, where $\mathcal R$ is the Borel $\sigma$-algebra on $\mathbb R$, such that $\mathcal L:\mathcal R\to[0,1]$. The cumulative distribution function (CDF) of a random variable $X$ is the function $F_X:\mathbb R\to[0,1]$ such that $F_X(x)=\Pr\{X\le x\}$ for $x\in\mathbb R$. If we know the distribution of the random variable $X$, then we also know the CDF of the random variable $X$. It also true that the CDF uniquely determines the distribution of the random variable $X$ (see this question).

I hope this helps.

  • $\begingroup$ so I think I was getting confused between the distribution and the cumulative distribution function of a random variable X. So the conclusion is that CDF gives me all the probability from $(-\infty , x)$ where as the distribution function is computing the probability measure of any arbitrary borel set in the sigma algebra $\endgroup$ – user3503589 Jun 3 '15 at 12:11
  • $\begingroup$ @user3503589 Yes, you are precisely right. But if you know the CDF of the random variable $X$, you can also determine the measure of any borel set. $\endgroup$ – Cm7F7Bb Jun 3 '15 at 12:13
  • $\begingroup$ But then it could be very nasty to use just the cdf to actually compute the measure of a pathological borel set. By pathological i mean say countable union of measurable sets.... $\endgroup$ – user3503589 Jun 3 '15 at 12:18
  • 1
    $\begingroup$ @user3503589 Indeed. Your very right (and too rare on this forum) intuition is that the basic object is the distribution, not its translation in terms of CDF. $\endgroup$ – Did Jun 3 '15 at 12:20

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.