# The transformation of a random discrete variable.

Let $X : (\Omega , \mathcal{A}) \to (\mathbb{R} , {\mathcal{B}}_{\mathbb{R}})$ be a random discrete variable (${\mathcal{B}}_{\mathbb{R}}$ denotes the Borel $\sigma$-algebra on $\mathbb{R}$) and let $g : \mathbb{R} \to \mathbb{R}$ be a measurable function. I want to prove that $Y = g(X)$ is a random discrete variable (I already know that $Y$ is a random variable, so I only want to prove that $Y$ is discrete). My attemps have no progress, though I have proved that $$P_Y(\{y\}) = \sum_{x \in D_X \cap g^{- 1}(\{y\})} P_X(\{x\}) \quad \mbox{ and } \quad F_Y(y) = \sum_{x \in D_X \cap g^{- 1}((- \infty , y])} P_X(\{x\})$$ for all $y \in \mathbb{R}$.

My attemps have been:

• to prove that $D_Y \neq \emptyset$ and $P_Y(D_Y) = 1$.
• to prove that $D_Y = g(D_X)$ and $S_Y = g(S_X)$, because we would have that $$D_Y = g(D_X) = g(S_X) = S_Y\mbox{,}$$ so $Y$ would be a random discrete variable.

Note: if $Z : (\Omega , \mathcal{A}) \to (\mathbb{R} , {\mathcal{B}}_{\mathbb{R}})$ is a random variable, then:

• $D_Z$ is the set of the points $z \in \mathbb{R}$ such that the distribution function associated to $Z$, $F_Z : \mathbb{R} \to \mathbb{R}$, is not continuous on $z$;
• $S_Z$ denotes the support of $Z$, $$S_Z = \{z \in \mathbb{R} : F_Z(z + \varepsilon) - F_Z(z - \varepsilon) > 0 \mbox{ for all } \varepsilon > 0\}\mbox{.}$$
• $Z$ is discrete iff $D_Z = S_Z$ iff $D_Z \neq \emptyset$ and $P_Z(D_Z) = 1$, being $P_Z : {\mathcal{B}}_{\mathbb{R}} \to [0 , 1]$ the probability induced by $Z$.
• Do you agree that $P(X\in D)=1$ implies $P(g(X)\in g(D))=1$? Further if $D$ is countable then $g(D)$ is countable? That's enough for discreteness of $g(X)$, isn't it? – drhab Jan 27 '18 at 10:42
• Yes, I agree. But why is that enough for discreteness of $Y$? – joseabp91 Jan 27 '18 at 10:57
• What is your definition for discreteness then? Mine is: $X$ is a discrete random variable if a countable set $D$ exists with $P(X\in D)=1$. – drhab Jan 27 '18 at 11:04
• My definition is $D_X \neq \emptyset$ and $P_X(D_X) = 1$. – joseabp91 Jan 27 '18 at 11:05
• $D_X$ is the set of the points $x \in \mathbb{R}$ such that $F_X$ is not continuous on $x$. Yes, $P_X(D_X) = 1$ implies that $D_X \neq \emptyset$. – joseabp91 Jan 27 '18 at 11:12

## 1 Answer

Let $D_X:=\{x\in\mathbb R\mid F_X\text{ is not continuous at }x\}$ or equivalently $D_X:=\{x\in\mathbb R\mid P_X(\{x\})>0\}$.

It is my aim to prove the following lemma which will make it more easy to prove that $g(X)$ is discrete if $X$ is discrete.

Lemma: $P_X(D_X)=1$ if and only if some countable set $D$ exists with $P_X(D)=1$.

First we claim that $D_X$ is a countable set.

To prove this let $D_X^{(n)}:=\{x\in\mathbb R\mid P_X(\{x\})>\frac1n\}$ and observe that this must be a finite set. Further we have $D_X=\bigcup_{n=1}^{\infty}D_X^{(n)}$ showing that $D_X$ - as a countable union of finite sets - is countable.

The necessity of the condition follows directly from the fact that $D_X$ is countable.

Conversely let $D$ be countable with $P(X\in D)=1$. It is immediate that $D_X\subseteq D$. Further $D-D_X$ is a subset of $D$ hence is countable so that: $$P_X(D-D_X)=\sum_{x\in D-D_X}P_X(\{x\})=0$$

Then from $1=P_X(D)=P_X(D_X)+P_X(D-D_X)$ it follows that $P_X(D_X)=1$ and we are ready.

Now an answer to your question:

If $X$ is discrete then (according to the lemma) $P_X(D)=1$ for some countable $D$. Then $g(D)$ is countable with $P_Y(g(D))=1$ so (according to the lemma) $Y=g(X)$ is discrete.

• Thank you very much, I understand. – joseabp91 Jan 27 '18 at 11:45
• You are very welcome. – drhab Jan 27 '18 at 11:48