# Map of random variable

• MHB
• mathmari
In summary, the mapping for a random variable is unique if and only if the probability function and the outcome space are given.

#### mathmari

Gold Member
MHB
Hey! :giggle:

What does it mean to give the mapping for a random variable? Do we have to give the outcome space and the probability function? Does it hold that $X: ( \Omega, P)\mapsto \mathbb{R}$ ? :unsure:

mathmari said:
What does it mean to give the mapping for a random variable? Do we have to give the outcome space and the probability function? Does it hold that $X: ( \Omega, P)\mapsto \mathbb{R}$ ?
Hey mathmari!

From the wiki defintion:
A random variable $X$ is a measurable function $X \colon \Omega \to E$ from a set of possible outcomes $\Omega$ to a measurable space $E$.
[...]
The probability that $X$ takes on a value in a measurable set $S\subseteq E$ is written as
$$\operatorname{P}(X \in S) = \operatorname{P}(\{ \omega \in \Omega \mid X(\omega) \in S \})$$

To give a mapping means that we need to characterize that mapping uniquely.

Klaas van Aarsen said:
Hey mathmari!

From the wiki defintion:
A random variable $X$ is a measurable function $X \colon \Omega \to E$ from a set of possible outcomes $\Omega$ to a measurable space $E$.
[...]
The probability that $X$ takes on a value in a measurable set $S\subseteq E$ is written as
$$\operatorname{P}(X \in S) = \operatorname{P}(\{ \omega \in \Omega \mid X(\omega) \in S \})$$

To give a mapping means that we need to characterize that mapping uniquely.

The exercise statement is :

An urn 1 contains 2 red and 8 white balls. An urn 2 contains 4 red and 6 white balls. A ball is drawn from each urn.

(a) Give a suitable probability space.

(b) Tim receives 1 Euro if the ball from urn 1 is red. Lena receives 1 euro if the Ball from urn 2 is white. Give the mapping rule for a random variable X that describes the profit of Tim, and a random variable Y, which describes Lena's profit. Find the joint distribution of X and Y. Are X and Y independent?

At (a) I have found the outcome space $\Omega =\{ (R,R), (R,W), (W,R),(W,W)\}$ and the probabilities \begin{align*}&p((R,R))=\frac{2}{10}\cdot \frac{4}{10}=\frac{2}{25} \\ &p((R,W))=\frac{2}{10}\cdot \frac{6}{10}=\frac{3}{25} \\ &p((W,R))=\frac{8}{10}\cdot \frac{4}{10}=\frac{8}{25} \\ &p((W,W))=\frac{8}{10}\cdot \frac{6}{10}=\frac{12}{25}\end{align*}

At (b) we have \begin{align*}&X(R,R)=1 \\ &X(R,W)= 1\\ &X(W,R)=0 \\ &X(W,W)=0\end{align*} and so \begin{align*}&P(X=1)=P(R,R)+P(R,W)=\frac 2{25}+\frac 3{25}=\frac 15 \\ &P(X=0)=P(W,R)+P(W,W)=\frac{8}{25}+\frac{12}{25}=\frac{4}{5}\end{align*} So is the map that we are looking for the $X$, the $P$ or both of them or something completely else? :unsure:

The map of the random variable $X: \Omega \to \text{Euros}$ is given by what you've already found:
\begin{align*}&X(R,R)=€ 1 \\ &X(R,W)= € 1\\ &X(W,R)=€ 0 \\ &X(W,W)=€ 0\end{align*}
This fully identifies the mapping of $X$. (Nod)

The probability map is a different map that needs to be identified separately.

Klaas van Aarsen said:
The map of the random variable $X: \Omega \to \text{Euros}$ is given by what you've already found:
\begin{align*}&X(R,R)=€ 1 \\ &X(R,W)= € 1\\ &X(W,R)=€ 0 \\ &X(W,W)=€ 0\end{align*}
This fully identifies the mapping of $X$. (Nod)

The probability map is a different map that needs to be identified separately.

Ahh ok! Thank you for the clarification!