I don't understand why in the case of continuous R.V. (hence, continuous sample spaces) we don't even care about defining a Probability on the sample space.

By analogy with the discrete case (where we define first a probability on the sample space, then an induced one on the event space of the R.V.) we should have defined $P(X\in A)$ as $P(X^{-1}(A))$ (and not $P(X\in A)=\int _A f(t)dt$)!

So, first:

  1. why don't we care about defining a probability on the sample space?


  1. Why the probability distribution is defined in the event space of the R.V. and not in the sample space ?

PS: If the probability distribution had been defined on the sample space that would have given the sample space a Probability and then everything would have been perfectly analogous to the discrete case

EDIT: My question is not about understanding what is a probability distribution, nor about defining the $\sigma$-algebra of the events:

it is about understanding why the distribution is defined in the arrival space (aka event space) and not in the departure space (aka sample space).

In other words: why don't we define $f$ as $$P(B)=\int _B f(t)dt$$ for $ B \in \mathscr B(\Omega)$

instead of:

$$P(X\in A)=\int _A f(t)dt$$

  • $\begingroup$ In formal treatments of probability in my experience, even for discrete distributions the probability function is defined on the event space, not the sample space. The theory for continuous distributions is analogous to that for discrete distributions. Where did you get the idea that it is not? $\endgroup$
    – David K
    Commented Nov 30, 2022 at 3:34
  • $\begingroup$ I think you need to clarify what you mean by event space and sample space in this instance. $\endgroup$
    – Wintermute
    Commented Nov 30, 2022 at 5:20
  • $\begingroup$ What do you expect integrating over an element of the sample space to mean? Suppose we're talking about a standard normal distribution: what do you expect the value of, say, $P(0.5)$ to be? $\endgroup$
    – N. Virgo
    Commented Nov 30, 2022 at 6:17
  • 1
    $\begingroup$ In $P(B)=\int_Bf(t)\,dt,$ the LHS has nothing to do with the random variable and the RHS does. This doesn't make sense to me—why do you think this is true or useful? (I interpret the RHS as implying that $f:\Omega\to\mathbb R,$ i.e. a random variable, and $\int_Bf(t)\,dt$ has the interpretation $E(f\mid B)$). Further, the proposed definition for the probability distribution of random variables, $P(X\in A)=P(X^{-1}(A))$ is the definition you'll find on e.g. Wikipedia. So what really is the question? $\endgroup$
    – HTNW
    Commented Nov 30, 2022 at 6:59

3 Answers 3


Short answer (leaving out all the technical stuff). For continuous probability the probability of a single point is $0$ and you can't get the probability of an event (say an interval) by summing the uncountably many $0$ probabilities of the the individual points. You really do need to integrate a density to get the probability of a measurable set.

  • $\begingroup$ I'm sorry but my question addresses an other problem. $\endgroup$
    – niobium
    Commented Nov 29, 2022 at 19:10
  • 2
    $\begingroup$ @niobium ... Then you should edit the question to clarify just what the your problem is. $\endgroup$ Commented Nov 29, 2022 at 19:47

The probability space is the source of all randomness. If there are multiple random variables defined on the same sample space, then any probability distribution on the sample space determines the joint distribution of all random variables at once. If you only want to focus on a particular random variable $X$, then it is much simpler to just look at the distribution supported on the output values of $X$.

For example, suppose we pick a random point in the interior of the unit circle. Our sample space is $\mathbb R^2$, and the distribution function is $1/\pi$ when $(x,y)$ is inside the circle, and zero otherwise. If $X$ and $Y$ are the $x$ and $y$-coordinates of this random point in the circle, then we have given a distribution on the same space, but it is still unclear what the distribution of $X$ looks like. Only when we find the pdf of $X$ to be $$ f_X(x)=\frac2{\pi}\sqrt{1-x^2},\qquad -1<x<1, $$ do we really grasp what the distribution of $X$ is. With this formulation, we can compute the variance and higher moments of $X$ easily, where that was not as obvious from the joint distribution. Note this is a distribution on $\mathbb R$, the outputs of $X$, not on the sample space $\mathbb R^2$.

  • $\begingroup$ I just did it on a sheat of paper, that works great! I have one last question: here we have deduced the distribution of $X$ from the distribution $\frac{1}{\pi}$. Can we conversely deduce the distribution of the sample space ($\frac{1}{\pi}$) from the distribution of $X$ ? My guess would be it is impossible, but I ask just in case my intuition is wrong ^^ $\endgroup$
    – niobium
    Commented Nov 30, 2022 at 12:48
  • 1
    $\begingroup$ @niobium Absolutely not! When you go from the circle distribution for $(X,Y)$ to the $2\pi^{-1}(1-x^2)^{-1/2}$ distribution for $X$, you lose all of the information about $Y$, so you cannot go back the other way. Even if you had the marginal distributions for both $X$ and $Y$, it would be insufficient to recover the joint distribution of $(X,Y)$. $\endgroup$ Commented Nov 30, 2022 at 18:44

For any continuous R.V., the value can be a decimals. For example, there are infinite amount of deciamls between 0 and 1, e.g. it can be 0.9 or 0.99999999 or 0.99999999999999. So the $P[X=x]$ for any continuous R.V. X and any value x would close to 0. We don't bother listing and defining the sample space of continuous R.V. cause there are infinite amount of samples.


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