# For a continuous random variable, why does $P(a < Z < b) = P(a \leq Z < b) = P(a < Z \leq b) = P(a \leq Z \leq b)$

My textbook puts this in a sidebox with the heading "Note" and doesn't explain why. Could you tell me why this statement holds?

$P(a < Z < b) = P(a \leq Z < b) = P(a < Z \leq b) = P(a \leq Z \leq b)$

• One could almost take these assertions as definitions of "continuous." What, then, is your definition of a continuous random variable? – whuber Jan 11 '14 at 18:14
• For a continuous random variable, what is $P(Z=a)$ and $P(Z=b)$? – Glen_b Jan 11 '14 at 23:38
• The Wikipedia article on probability distributions does a pretty good job of explaining this. Ultimately it invokes the fact that the CDF is continuous, whence the probability of any single point must be zero. – whuber Jan 14 '14 at 15:44

Nothing formal to add to this, but an analogy that really helped me to understand this came from a calculus text. Imagine you have an iron pipe of a certain length and weight. And you wish to cut it into two pieces. If the pipe is say 1 m long you might want to cut it in half at the 0.5 mark. Now think of the pipe's weight as some constant times the length of the pipe, (we assume that all cross-sections of equal length has the same weight).

Cutting the pipe in half at the 0.5 m mark - how much weight do you lose? Remember that the only cross-section you are removing is the 0.5 m mark itself. So what is the length of this cross-section? Consider that 0.49999999... is not apart of it, and neither is 0.5000000000...1, or any other point that is close to, but not equal to 0.5 - so the length of this cross-section is technically zero. Which means you're not really removing any weight at all.

This would explain why $\leq$ and $<$ are basically the same for continuous variables - including or excluding the endpoint really doesn't change anything - for any point you pick close to the endpoint, there is still an infinite amount of points between them.

Does this make any sense?

First I will give the definition of an (absolutely) continuous random variable $Z$.

(Advanced probability is needed, you many skip it!)

Let $(\Omega,F, P)$ be a probability space and let $Z:\Omega\rightarrow R^n$ be a random vector. The probability $P_X$ on $B(R^n)$ defined by $P_Z(A)=P\{Z\in A\}$, $A \in B(R^n)$ is called the distribution of $Z$. Now if $P_Z\ll \mu,$ where $\mu$ is Lebesgue measure on $R^n$, (i.e. $P$ is absolutely continuous with respect to $\mu$) then we say that $Z$ is an (absolutely) continuous random vector. Now by using Radon–Nikodym theorem, there exists a function $f: R^n\to[0,+\infty]$ such that $P_Z(A)=\int_{A}fd_{\mu}$ for all $A \in B(R^n)$. We call $f$ the density function of $Z$.

Now define the cumulative distribution function (CDF) of an absolutely continuous random variable $Z$ as: $$F_Z(z)=P(Z\leq z).$$

Before I give a formal proof, let's have an example of a continuous random varible that is uniformly distributed i.e. with a probability density function of $f(z)=1$ for $0\leq z \leq 1$ and 0 otherwise. Now let's try to find $P(z=0.5)$. We have $$P(z=0.5)\sim P(0.4< z\leq 0.6)=\int_{0.4}^{0.6}f(z)d_z=0.2.$$ We can shrink that interval to get a better approximation as follows: $$P(z=0.5)\sim P(0.49< z\leq 0.51)=\int_{0.49}^{0.51}f(z)d_z=0.02,$$ $$P(z=0.5)\sim P(0.499< z\leq 0.501)=\int_{0.499}^{0.501}f(z)d_z=0.002.$$ As you can see, these probabilities are converging to zero as we shrink the length of the interval. Now let's prove it formally. I am goig to show that for any continuous random variable $Z$, we have: $$P(Z=a)=0,$$ by using the CDF. $$P(Z=a)=\lim_{\epsilon\to 0}P(a-\epsilon< Z\leq a+\epsilon)=\lim_{\epsilon\to 0}F_Z(a+\epsilon)-\lim_{\epsilon\to 0}F_Z(a-\epsilon)\\=F_Z(a)-F_Z(a)=0,$$ since the CDF function, $F$, is a "continuous" function for the continuous random variable $Z$. Similarly $P(Z=b)=0$.
Finally note that $$P(A\bigcup B)=P(A)+P(B)-P(A\bigcap B).$$ So $$P(a\leq Z<b)=P\Big(\{Z=a\}\bigcap \{a<Z<b\}\Big)=P(Z=a)+P(a<Z<b)\\=0+P(a<Z<b)=P(a<Z<b).$$ You can use the same argument for other equalities.

• It appears this argument would apply equally well to any discrete distribution with a countably infinite set of possible values, but then the conclusions are obviously wrong, so something is the matter. – whuber Jan 11 '14 at 18:27
• It would be great if you can show us how "the conclusions are obviously wrong" ... – Stat Jan 11 '14 at 18:46
• In a discrete distribution--even one (like a Poisson or Negative Binomial) with countably infinite support--every value has nonzero probability, whereas your argument implies they all have zero probability. – whuber Jan 11 '14 at 19:44
• I changed my answer. – Stat Jan 11 '14 at 21:24
• I disagree with your assertion that $F_Z(z)$ being a right-continuous function gives us that $$\lim_{\epsilon \to 0} F_Z(a-\epsilon) = F_Z(a).$$ The desired result requires left-continuity of $F_Z(z)$, which of course holds since $F_Z(z)$ is both right-continuous and left-continuous for continuous random variables $Z$. Also, you are using the fact that probability is a set-continuous function when you compute $\lim_{\epsilon\to 0}P(a-\epsilon < Z \leq a+\epsilon)$ and assert that it is the same as $P\{\lim_{\epsilon\to 0} (a-\epsilon<Z<a+\epsilon)\}$. – Dilip Sarwate Jan 12 '14 at 4:29

Perhaps a more intuitive explanation is that for a continuous variable the contribution of the edges (e.g., $a$ or $b$) to the cumulative probability in the surrounding intervals (or semi-intervals) is negligibly small.