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If $X \in \mathbb{R}$ is a continuous random variable then its density $f$ (if it exists) is defined as

$$ f(x) = \lim_{\epsilon \rightarrow 0} \frac{F(x + \epsilon) - F(x)}{\epsilon}, $$

where $F(x) = \Pr(X \leq x)$.

It is sometimes more convenient to think of it as $f(x) = \Pr(X = x)$. However, this cannot be written!

Is it more rigourous (i.e., can it be written) to write $f(x) = \Pr(X \in I_x)$, where $I_x$ is a neightbourhood of $x$?

user7064
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    Actually, it's probably *not* a good idea to think of $f(x)$ as $\Pr(X=x)$ because, when a density exists, $\Pr(X=x)=0$ for all $x$. This way of thinking makes unnecessary mysteries of some common phenomena, such as the [existence of values](http://stats.stackexchange.com/q/4220) for which $f(x) \ge 1$, an impossibility for any actual probability. For a better alternative, please see http://stats.stackexchange.com/q/14483. – whuber Mar 23 '12 at 21:48
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    Assuming that $f(x)$ is continuous at $x$, it is more rigourous to write $\Pr(X \in I_x) \approx f(x)|I_x|$ where $I_x$ is a neighbourhood of $x$ of _length_ $|I_x|$ (in the one-dimensional case) where the approximation increases in accuracy as $|I_x| \to 0$. – Dilip Sarwate Mar 24 '12 at 02:05

2 Answers2

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Do you have a particular probability space in mind where you want to do this calculation? For a random variable $X$ that admits a probability density function $f(x)$, you can think of $(X\in I_{x})$ as the event of being in the neighborhood $I_{x}$ of some specific point $x$. The then probability of that event is just

$$ P(X\in I_{x}) = \int_{I_{x}}f(y)dy.$$

Do you want something more than this? For the one-dimensional case, I suppose you can think of the density as being the differential probability of being in an infinitesimal neighborhood that has the point $x$ as its left endpoint:

$$ P(X\in [x,x+h)) = F(x+h) - F(x) = \int_{x}^{x+h}f(y)dy \approx f(x)\cdot{}h$$ for small $h$. Note that you can get the same thing by computing a Taylor series expansion of $F(x)$ at the point $x$, and then evaluating the resulting expansion at $x+h$, and considering only the linear approximation.

ely
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I think the closest you can get to the concept you want to express without risking confusion is

$f(x)=\frac{\Pr(x<X \le x+dx)}{dx}$.

This is a bit of an abuse of notation unless $\Pr(x<X \le x+dx)$ is defined to be equivalent to the differential $dF(x)$.

Cyan
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