# relation between probability and probability density function

To translate the probability density $ρ(x)$ into a probability, imagine that $I_x$ is some small interval around the point $x$. Then, assuming $ρ$ is continuous, the probability that $X$ is in that interval will depend both on the density $ρ(x)$ and the length of the interval:

$Pr(X∈I_x)≈ρ(x) ×$ Length of $I_x$

We don't have a true equality here, because the density $ρ$ may vary over the interval $I_x$. But the approximation becomes better and better as the interval $I_x$ shrinks around the point $x$, as $ρ$ will be come closer and closer to a constant inside that small interval.

My questions are:

1. Why is $Pr(X∈I_x)≈ρ(x) ×$ Length of $I_x$? Shouldn't it simply be $Pr(x∈A)=\int_{A}ρ(x)dx$?

2. Why as $I_x$ shrinks around the point $x$, will $ρ$ become closer and closer to a constant inside that small interval? Is it because as we reduce the interval, we zero in on $x$?

3. How and why the density $ρ$ may vary over the interval $I_x$?

• Does stats.stackexchange.com/questions/4220/… or stats.stackexchange.com/questions/133369/… answer your questions? – Tim Jul 17 '15 at 8:53
• All of your questions belong to real analysis, not probability. A background in real analysis is necessary for studying statistics. – Aksakal Jul 17 '15 at 11:30
• (1) and (2) are justified by the Mean Value Theorem for Integration. What do you mean by (3)? If the density does not vary over any interval, it must be constant, but then it could not be a density (because the total probability would be either $0$ or infinite rather than $1$). – whuber Jul 17 '15 at 16:14

1. Yes; the description is however useful for mathematically naive audience (perhaps an audience more naive than you); it's simply motivating an understanding like the one you already have, by using a concept akin to that of one term in a middle Riemann sum.

$$\qquad\qquad$$

If the audience has had exposure to the mean value theorem, or would at least be able to understand it, that would be a better motivation.

2. Imagine we're dealing with a sufficiently "nice" pdf, say one that's Lipschitz continuous. Then in a direct sense, as the interval in $$x$$ gets smaller, the interval in $$f$$ must also get smaller.

$$\qquad\qquad$$

3. It's not quite clear what you're after here. Clearly for "2." to hold we must have some restrictions on the way $$f$$ behaves, but density functions are not necessarily required to be Lipschitz continuous.

However, we needn't overly interpret the meaning of the phrases in question; it's intended as motivation in building a naive understanding of the relationship between density and probability, not proof of anything.

Another observation is that this is almost a restatement of the definition of the density function. Recall that $$f(x) \equiv \lim_{\delta \downarrow 0} \frac{P(x \leq X \leq x + \delta)}{\delta}$$ which means $\delta f(x) \cong P(x \leq X \leq x + \delta)$ for sufficiently small $\delta$.