How can the probability of each point be zero in continuous random variable? [duplicate]

Question

I know this is duplicated but I think the question is a bit different and needs different answer.

How can CDF be continuous and have derivative at each point that is not equal to zero but the probability at each point is zero?

Why not say for example if you want to choose a real number between 0 and 1 that the probablity is 1/N? Where this is just infinitesimal. (Just an example to describe the situation.)

It can neither be finite nor zero

The difference is: Why not assign it an infinitesimal instead of zero and to say that an event with P = 0 can happen?

Answer 1

A guess at your point of confusion:

Zero probability does not mean an event cannot occur! It means the probability measure gives the event (a set of outcomes) a measure zero.

As @Aksakai's answer points out, the union of an infinite number of zero width points can form a positive width line segment and similarly, the union of an infinite number of zero probability events can form a positive probability event.

More explanation:

• Our intuition from discrete probability is that if an outcome has zero probability, then the outcome is impossible. If the probability of drawing the ace of spades from a deck is equal to zero, it means that ace of spades is not in the deck!

• With continuous random variables (or more generally, an infinite number of possible outcomes) that intuition is flawed.

  • Probability measure zero events can happen. Measure one events need not happen. If an event has probability measure 1, you say that it occurs almost surely. Notice the critical word almost! It doesn't happen surely.
  • If you want to say an event is impossible, you may say it is "outside the support." What's inside and outside the support is a big distinction.
  • Loosely, an infinite sum of measure zero events can add up to something positive. You need an infinite sum though. Each point on a line segment has zero width, but collectively, they have positive width.

Answer 2

It's really not a statistics question. It's a real analysis question. For instance, it's almost the same as asking "what's the width of a point in line?" (the answer is zero, by the way)

This is an interesting situation though. In mathematics the line is defined as a set of points. There are certain geometric constraints on the points, so that they form a line and not a circle, for instance. However, that's not what's important.

What's important is this. If the width of each point is zero, and the line is a set of points, how come the sum of widths of all its points is NOT zero? You add two zeros and it gives you a zero. If I keep adding this way shouldn't the length of a line be zero? Apparently, not!

This is the same question you're asking. How is it that each point's probability is zero, yet the total probability is one? The reason why is this question the same is because probabilities are intimately linked to the concept of the length of a line between two points. The central concept of the modern probability theory is the concept of a measure. Unsurprisingly, it has its roots in the simplest of all measures: the length in geometry.

If you want a shortcut in understanding this bind boggling stuff then look up the concept of countable and uncountable sets. Note the difference between infinite countable sets and uncountable sets. Both have infinite number of points in them, yet the latter has more points in it (totally crazy!). So discrete and continuous random variables (and their distributions) are related to these two kinds of sets.

UPDATE
Example: In English there are countable and uncountable nouns such as apple vs. milk. I could ask you how much does an apple weigh? And you could say that it's half a pound in this batch. However, if I asked how much does milk weigh, it wouldn't make a sense without specifying an amount such as a pint or a quart.

In this regard the discrete random variables and their probabilities are like apples and their weights. You could say that the probability of Poisson variable 1 is 10%, for instance.

The continuous random variables are like milk. It's pointless asking what's the probability of a given value, you need to specify the bucket. Say, for a standard normal (Gaussian) variables you could ask what's the probability that their values are between 0 and 1, and the answer would be something like 34%. However, the probability of 1 is pretty much meaningless in practical sense. You can calculate the density at $x=1$ but what are you going to do with it? It's not the probability. In the same way if you're interested in the weight of milk the density of milk is not an answer, you need to specify the container size then we can tell you the weight using its density. That's why probability density function is actually called density, it originates from densities of bodies.

Answer 3

I think it is helpful to imagine the area under the point. The probability for a continuous distribution is the integral of the PDF from (a,b). If you pick a single point (a,a) is there any area? Imagine a simple PDF like the uniform distribution do the math.

PS No, but if you ask enough mathematicians 1/20 will say yes. However, I'll accept the null with an $\alpha$ of 5%.