# Beginner conceptual question about the relationship between probability density and distribution functions

When introducing the relationship between a probability distribution function and a density function for a continuous RV, I see many books pointing this out:

$P(t<X<t+dt) = f(t)dt$

Questions

1. In very very simple terms, why is this important to know? How would you explain this to a beginner amateur who hasn't taken calculus in 10 years?

2. In practice would anyone ever use this for a calculation? For example let's say that $X$ is the amount of rain we'll get tomorrow in inches. I want to know the probability of getting 10 inches of rain tomorrow. Then I do $f(10)*dt$ which I'm guessing would equal zero and be useless (since a continuous RV has probability 0 for a specific value).

Note that the following relationship does make a lot more sense to me.

$\frac{d}{da}F(a)=f(a)$

The identity $$\frac{d}{dt} F(t) = f(t)$$ might be in practice more useful that the identity $$P(t<X<t+dt) = f(t)dt$$ since the first one gives you something you can compute. But these two are formally equivalent. The notation $\frac{d}{dt}$ stands for comparing the variation of a function (say $F$) with respect to small variations of time. So $$dF(t) =F(t + dt) - F(t)$$and thus formally $$f(t) = \frac{dF}{dt} = \frac{F(t + dt) - F(t)}{dt} = \frac{P(t<X<t+dt)}{dt}$$ multiplying everything with $dt$ gives us the first formula.

The positive thing about the second formula is that

1. you can see the mathematically important thing happening: the probability of falling in a box becomes smaller as we reduce the box's size. This is not always the case, say if $X = 0$ is deterministic, then for any box around zero the probability of falling in that box is constantly $1$.
2. you get a pictorial representation of your density function, since the formula tells you that you can approximate $f$ through a histogram.

For the mathematician the first reason is the most important one, and it suggests a change of paradigm. Whereas in analysis the derivative plays a central rôle, in modern probability theory it is integration that plays a central rôle. The idea is that a probability measure is a way of integrating. And the above formula tells you that this new way of integrating does not differ from the classical one. In some fancy notation one writes: $$P(s <X<t) = \int 1_{[s,t]}(X) dP = \int_s^t f(u) du.$$

• Thanks for your help! I read your response many times and will have to meditate on this... you offer many insights that for a slow one like me don't follow obviously from just mentioning the second formula (like it following that f can be approximated through a hist). Question... in a deterministic case wouldn't your "density function" only be a single point floating in space, like the coordinate (0,1) and then any box could only be the size of a point? Otherwise you end up with a continuous uniform? Commented Dec 7, 2017 at 16:26
• I don't believe that what I have said is easy, nor that I have expressed it in the best possible way. What is $F$ in the deterministic case? Can you compute it's derivative? The density should be "a single point", in the sense that it should be zero outside of zero. But what happens when you integrate such a function, following the last formula in my answer? What is the size of a point? The fact is that the classical integral does not see single points :) So all in all, does a deterministic random variable have a density? Commented Dec 7, 2017 at 17:00
• Yes you're right I suppose for that case it's not a random variable even and more like a constant. Thanks again, your response was very helpful. Commented Dec 7, 2017 at 17:14
• No, it's still a random variable (the simplest one). But it's a discrete random variable: it gives positive probability to single points. Whereas continuous random variables (those which admit a density in your sense) give zero mass to points (the probability of a point is proportional to it's size in the classical sense - namely zero). Similarly a random variable distributed like a binomial also does not admit a density (with respect to classical integration). But: classical integration is an arbitrary reference point. If you choose other reference points you end up in measure theory. Commented Dec 7, 2017 at 22:40
• Ah, okay well today I learned (a lot). I originally thought a Bernoulli RV was the "simplest" you could get. But it seems like $P(X=0)=1$ is the probability mass function of a discrete RV and $X$ in this case is not "just a constant"... Commented Dec 7, 2017 at 23:49