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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.

Taylor
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user3503589
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1 Answers1

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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.

Community
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Cm7F7Bb
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  • 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 – user3503589 Jun 03 '15 at 12:11
  • @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. – Cm7F7Bb Jun 03 '15 at 12:13
  • 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.... – user3503589 Jun 03 '15 at 12:18
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    @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. – Did Jun 03 '15 at 12:20