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In a textbook I'm reading:

A random variable $X$ is continuous if there exists a function $f_X$ such that $f_X(x) \ge 0$ for all $x$, $\int_\infty^\infty f_X(x) dx = 1$ and for every $a \le b$,

$$ \mathbb{P}(a < X < b) = \int_a^b f_X(x)dx. $$

This is different than the definition provided for on Wikipedia, which seems to define a continuous random variable $X$ as any random variable with an uncountably infinite range.

  1. Are these two definitions equivalent?
  2. Does it matter if the left-side of the equition above uses $\le$ instead of $<$? That is do we have the following:

$$ \mathbb{P}(a < X < b) = \int_a^b f_X(x)dx = \mathbb{P}(a \le X \le b)? $$

Edit: Wikipedia link on random variables. Here is the quote:

If the image is uncountably infinite then $X$ is called a continuous random variable.

leonbloy
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user1770201
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  • @carmichael561 Thanks for the correction, let me make a more technically correct comment. – Ian May 31 '16 at 18:31

2 Answers2

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First, some references would prefer the term "absolutely continuous" for what you are calling continuous. They would instead use the word "continuous" to refer to r.v.s whose CDF is continuous; the Cantor distribution is "continuous" in this sense but not in your sense.

At any rate, there are r.v.s with uncountably infinite range that are not even continuous in this weaker sense. For instance you can have $X=B U + 1-B$, where $B$ is Bernoulli(1/2), $U$ is uniform on $(0,1)$, and the two are independent. Then $X$ can take on any value between $0$ and $1$ but its CDF is not continuous at $1$.

As for your second question, $<$ vs. $\leq$ does not matter because in this situation single points have probability zero.

Ian
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  1. These two definitions are equivalent.
  2. It doesn't matter whether $<$ or $\leq$ is used in a continuous distribution, because continuous distributions have no mass on any one point. (e.g. $Pr(x=k)=0$ in a continuous case)
Daniel
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  • On #2: sure you can, it's just zero. On #1 you are just wrong, even though in most practical situations the two are equivalent, they are not actually equivalent, you can have all sorts of weird mixed distributions. – Ian May 31 '16 at 18:42
  • Indeed, #2 is right but #1 is awfuly ill-informed. – Did May 31 '16 at 18:49