Exercise 2.18 (Random variable is dependent of itself)

In the following, let $(\mathrm{\Omega },F,P)$ denote the randomness source of $X$. The range $R$ of $X$ is $\sigma$-locally compact metric space, i.e.,

• ($\sigma$-compact) $R$ can be represented as a union ${\bigcup <!--\; FUNCTION\; APPLICATION\; -->}_{n=1}^{\infty }{C}_n$ of compact sets ${(C_n)}_{n\in N}$;
• (locally compact) every point $r$ of $R$ has a compact neighbourhood, i.e., there exists an open set $U$ and a compact set $K$, such that $r\in U\subseteq K$;
• (Hausdorff) every point $r$ of $R$ can be put into its own separate neighbourhood;

We will obviously use the Borel $\sigma$-algebra on $R$ which also contains all compact sets and all singleton sets in our case. We now demonstrate both sides of the equivalence.

• Suppose that $X$ is independent of itself, i.e, for all $S\in B$ we have

  $$P(X\in S)=P(X\in S)\cdot P(X\in S).$$

  Thus, from

  $$P(X\in S)=P{(X\in S)}^2,$$

  we conclude that either $P(X\in S)=1$ or that $P(X\in S)=0$ for all $S\in B$.
  Now suppose, for the sake of contradiction, that we have two values $a,b$ in the image of $X$ such that $P(X=a)=c$ and $P(X=b)=1-c$ for some $0<c<1$. By the assumption of local compactness, we can find compact sets $K,K^{\prime }\subseteq R$ such that $a\in K$ and $b\in K^{\prime }$. By monotonicity of the probability measure, this implies $P(X\in K)\ge c$ and $P(X\in K^{\prime })\ge 1-c$. But this leads to a contradiction with the previously derived fact claiming that any set has either a probability of zero or one because (1) we cannot have both $P(X\in K)=1\ge c$ and $P(X\in K^{\prime })=1\ge 1-c$; and (2) we cannot have one of theses probabilities to be zero as $c>0$ and $1-c>0$.

• Suppose that there exists $a\in R$ with $P(X=a)=1$ and $P(X\ne a)=0$. For any set $S\in B$ we have two cases:

  • $a\in S$, in which case $P(X\in S)\ge P(X=a)$ and so $1=P(X\in S)=P{(X\in S)}^2$.
  • $a\notin S$, in which case $P(X\in S)\le P(X\notin a)$ and so $0=P(X\in S)=P{(X\in S)}^2$.

  Hence, in any case we have $P(X\in S)=P(X\in S)\cdot P(X\in S).$