Calculating probability mass functions with constraints from cumulative distribution

Question

This is a self-study question. The name of the book is called: Applied Statistics and Probability for Engineers by Montgomery and Runger. This problem is on page 73. It's exercise 3-41.

The entire problem is listed as the following:

Given the following cumulative distribution function: $$ F(x)=\left\{\begin{matrix} 0 & & x<-10 \\ 0.25 & & -10\leq x< 30 \\ 0.75 & &30\leq x< 50 \\ 1 & & 50 \leq x \end{matrix}\right. $$ Determine each of the probabilities: a) $P(X<50)$ b) $P(0 \leq X < 10)$ c) $P(-10 < X <10) $... etc etc. The question I have is this:

Why does the following probability mass function evaluate to 0? $$ P(0 \leq X < 10) = 0 $$

Isn't this set of outcomes a subset of $-10\leq x< 30$ and therefore should be evaluated to 0.25?

Answer 1

The cumulative probability distribution function $F_X(x)$ tells us how much probability mass there is to the left of $x$ or at $x$ for each $x$ on the real line. (The choice of notation, though almost universally used is truly dreadful for use in a classroom setting! How on earth does one read out aloud $F_X(x)$ or $P\{X\leq x\}$? F-sub-big X of little x? probability that random variable $X$ is no larger than lower-case x?) Formally, the value of $F_X(x)$ is just $P\{X \leq x\}$. As Glen_b's comment says, you really should start by sketching the function $F_X(x)$ at the very least.

When $X$ is a discrete random variable taking on values $x_1, x_2, \ldots$ with probabilities $p_1, p_2, \ldots $ respectively, a little thought (instead of rote memorization of the definition) reveals that $F_X(x)$ must be what can be described as a staircase function, increasing from $0$ to $1$ as $x$ increases, with steps of heights $p_1, p_2, \ldots$ at points $x_1, x_2, \ldots$ etc. The function is discontinuous at each $x_i$, and is constant in each interval $[x_i, x_{i+1})$ (please be sure to note the $[$ and $)$ in the description of the intervals). Note that $F_X(x_i)$ includes $p_i$ so that the value of $F_X(x)$ at the point $x=x_i$ (where the function is discontinuous) is the value on the right. Since you are studying from a text intended for engineers, you might find this written as $F_X(x) = F_X(x^+)$. Thus, $$F_X(x) = P\{X \leq x\} = F_X(x^+) ~ \text{and} ~ P\{X < x\} = F_X(x^-).$$

In fact, for any random variable (not necessarily a discrete random variable or an integer-valued random variable as in Rusan's answer) and for any real numbers $a$ and $b$ such that $a \leq b$, $$\begin{align} P\{a < X \leq b\} &= F_X(b^+) - F_X(a^+) = F_X(b)-F_X(a),\tag{1}\\ P\{a \leq X \leq b\} &= F_X(b^+) - F_X(a^-) = F_X(b) - F_X(a^-),\tag{2}\\ P\{a \leq X < b\} &= F_X(b^-) - F_X(a^-) = F_X(b^-) - F_X(a^-),\tag{3}\\ P\{a < X < b\} &= F_X(b^-) - F_X(a^+) = F_X(b^-)-F_X(a).\tag{4} \end{align}$$

For the special case when $b = a$, $(2)$ above becomes $$P\{X=a\} = F_X(a^+)-F_X(a^-),$$ that is, $P\{X=a\}$ is the jump (if any) in the value of $F_X(x)$ at $x=a$. If $F_X(x)$ is continuous at $x=a$, then $P\{X=a\}=0$.

With this as prologue, note that your given $F_X(x)$ is a staircase function with jumps of $\frac 14, \frac 12, \frac 14$ at $x=-10, 30, 50$ respectively; that is, $X$ takes on values $-10, 30, 50$ with probabilities $\frac 14, \frac 12, \frac 14$ respectively, and once you have that, the answers to the questions asked are easy to compute directly, or, if you prefer to read the $F_X(0^-)$ etc off the graph that you have drawn as you apply $(1)$-$(4)$, that is fine too.

Answer 2

Since this question relates specifically to a discrete random variable (as your book says...), I will answer it as such. Dilip Sarwate's answer provides the general result for a discrete random variable.

With those health warnings, the cumulative distribution function (CDF), $F$, between $x=-10$ and $x=30$ does not vary. This implies the probability mass strictly between these points is $0$. In particular, the interval $0\leq x <10$ lies strictly within the interval $-10\leq x <30$ so $\mathbb{P}(0\leq X<10)=0$.

To see why this is, assume that the CFD does change at some $x$. Because $F$ is a non-decreasing function, and $X$ is a discrete random variable, we can find a sufficiently small $\epsilon>0$ so that $$F(x-\epsilon)<F(x)$$ holds, and so can be rearranged as $$0<F(x)-F(x-\epsilon)=\mathbb{P}(X\leq x)-\mathbb{P}(X\leq x-\epsilon).$$ Because we are free to choose $\epsilon$ as small as we like, subject to $0<\epsilon$, we rewrite the inequality as $$0<\mathbb{P}(X\leq x)-\mathbb{P}(X<x)=f(x)=\mathbb{P}(X=x).$$ So the points for which the CDF changes (steps) are those that have positive probability mass. By reversing the argument, you can see the converse is also true.