4.10 Properties of the Cumulative Distribution Function of Discrete Random Variables in Sheldon Ross 8th

Question

I don't understand those explanations that $F(a)$ is a subset of $F(b)$, so $F(a)$ is $\leq F(b)$. And A event is contained in the event B, so $F(a)$ is $\leq F(b)$

function

Thanks for the help!

Andy

Answer 1

What this is saying is that for the distribution function $F$ of $X$, $F(b)$ denotes the probability that the random variable takes on a value less than or equal to $b$. Formally, this is written $$ F(b) = \mathbb P (\omega\in\Omega:X(\omega)\leqslant b\}\tag1 $$ where it is understood that we aer working over a probability space $(\Omega,\mathcal F,\mathbb P)$ where $\Omega$ is a (possibly infinite) set called the sample space, $\mathcal F$ is a $\sigma$-algebra on $\Omega$ (which satisfies the properties that $\varnothing\in \mathcal F$ and $\mathcal F$ is closed under complementation and countable union), and $\mathbb P$ is a probability measure. Here the random variable $X$ is a measurable function from $\Omega$ to $\mathbb R$, meaning that for every Borel set $B$ in the collection of Borel sets $\mathcal B(\mathbb R)$, the preimage $X^{-1}(B)\in\mathcal F$. Now, we normally write $$ F(b)= \mathbb P(X\leqslant b) $$ as shorthand for the notation in (1), as the sample space is not usually explicitly defined, nor is $X$ explicitly defined as a function. Instead we consider the distribution of $X$, namely $F$, which allows us to recover all information about $X$. This is because the Borel $\sigma$-algebra over $\mathbb R$ is generated by the sets $\{(-\infty,x]:x\in\mathbb R\}$. So knowing the value of $F(b)$ for every $b\in \mathbb R$ is equivalent to knowing the distribution of $X$.

As for the property that if $a<b$ then $F(a)\leqslant F(b)$, then is readily apparently from $(-\infty,a] \subset (-\infty,b]$ and so $X^{-1}(-\infty,a]\subset X^{-1}(-\infty,b]$ from which $$ F(a) = \mathbb P(\omega\in\Omega:X(\omega)\leqslant a) = \mathbb P\circ X^{-1}(-\infty,a] \leqslant \mathbb P\circ X^{-1}(-\infty,b] = F(b), $$ by monotony of probability measures (if $A\subset B$ then $\mathbb P(A)\leqslant \mathbb P(B)$).

For a direct proof that $A\subset B \implies \mathbb P(A)\leqslant \mathbb P(B)$, consider that $B = (B\setminus A) \cup A$, and these events are disjoint, so by countably additivity of probability measures, $$ \mathbb P(B) = \mathbb P ((B\setminus A)\cup A) = \mathbb P(B\setminus A) + \mathbb P(A) \geqslant \mathbb P(A), $$ since $\mathbb P(B\setminus A)\geqslant0$.

Answer 2

Recall that, for a real-valued random variable $X$, and a real number $a$, we define the event $$(X\leq a) := \{\omega\in\Omega :\, X(\omega)\leq a\}.$$ Suppose that $a<b$ and we claim that $(X\leq a) \subseteq (X\leq b)$. To see this, let $\omega\in (X\leq a)$. Then $X(\omega)\leq a$, and since $a<b$, also $X(\omega)\leq b$, meaning that $\omega\in(X\leq b)$.

So, $(X\leq a)$ is contained in $(X\leq b)$, and by the monotony$^1$ of the probability, $$\textsf P(X\leq a) \leq \textsf P(X\leq b),$$ that is $F(a)\leq F(b)$, since $F(x)$ is defined as $F(x) := \textsf P(X\leq x)$.

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$^1$ You can see the proof of this fact :

If $E\subset F$, then $\textsf P(E)\leq \textsf P(F)$.

in the section Axioms of Probability of your book, in the part of Some Simple Propositions.