# Multiple random variables with multiple events

I'm getting rather confused with the random variable concept and its distribution in probability, especially when it gets abstract with no actual example to base my understanding on.

Take for example a sample space where $\Omega = \left \{ \omega _{1} ,\omega _{2}, \omega _{3} \right \}$ and $\mathbb{P}(\omega _{1})=\frac{1}{2}$, $\mathbb{P}(\omega _{2})=\mathbb{P}(\omega _{3})=\frac{1}{4}$. $X, Y, Z$ are defined as such:

$X(\omega_{1}) = 1$ , $X(\omega_{2}) = 2$ , $X(\omega_{3}) = 2$

$Y(\omega_{1}) = 2$ , $Y(\omega_{2}) = 1$ , $Y(\omega_{3}) = 1$

$Z(\omega_{1}) = 1$ , $Z(\omega_{2}) = 2$ , $Z(\omega_{3}) = 1$

To show: That $X$ and $Y$ have the same distribution.

The question is then, how do I intuitively interpret the whole idea? And how would the probability of $X, Y$ or $Z$, or even say, where arithmetic operations are used, $X+Y, XY$, be understood?

(My idea is that the distribution of $X$ is simply defined by the $\mathbb{P}(\omega _{1})X(\omega_{1}) + \mathbb{P}(\omega _{2})X(\omega_{2}) + \mathbb{P}(\omega _{3})X(\omega_{3})$, but as to whether that's true or why it is true, if true, I have no clue. )

Thanks!

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## 1 Answer

$\mathbb{P}(\omega _{1})X(\omega_{1}) + \mathbb{P}(\omega _{2})X(\omega_{2}) + \mathbb{P}(\omega _{3})X(\omega_{3})$ is the expected value of $X$. Or rather, the expected value is $\Bbb P(\{\omega_1\})X(\omega_1)+\Bbb P(\{\omega_2\})X(\omega_2)+\Bbb P(\{\omega_3\})X(\omega_3)$: one should talk about the probabilities of subsets of $\Omega$, not of elements.

The distribution of $X$ is the function $f(x)=\Bbb P(X=x)$, which is given by $$f(x)=\begin{cases}\frac12,&\text{if }x=1\\\\\frac12,&\text{if }x=2\\\\0,&\text{otherwise}\;:\end{cases}$$ $\Bbb P(X=1)=\Bbb P(\{\omega_1\})=\frac12$, and $\Bbb P(X=2)=\Bbb P(\{\omega_2,\omega_3\})=\frac14+\frac14=\frac12$. $Y$ has exactly the same distribution, because $\Bbb P(X=1)=\Bbb P(\{\omega_2,\omega_3\})=\frac12$ and $\Bbb P(X=2)=\Bbb P(\{\omega_1\})=\frac12$. Intuitively, the distribution simply tells you how likely each possible value of the random variable is.

To find the distribution of $XY$, you’d observe that $X(\omega)Y(\omega)=2$ for each $\omega\in\Omega$ and conclude that $\Bbb P(XY=2)=1$; thus, the probability distribution is just $$f(x)=\begin{cases}1,&\text{if }x=2\\0,&\text{otherwise}\;.\end{cases}$$ I’ll leave the probability distribution for $X+Y$ to you; it will be much like the one for $XY$.

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@Didier: Yes, that was sloppy. –  Brian M. Scott May 15 '12 at 2:42
@BrianM.Scott XY and X+Y do seem to be special cases where $X(\omega)Y(\omega) = k$ or $X(\omega)+Y(\omega) = k$, for each $\omega \epsilon \Omega$. I infer then that in the case of the probability distribution of XZ, there are now 3 different probabilities and hence the distribution $f(x)=\left\{ \begin{matrix} \frac{1}{3}, {if x = 1}\\ \frac{1}{3}, {if x = 2}\\\frac{1}{3}, {if x = 4}\\0, {otherwise} \end{matrix}\right.$ as depicted in the above probability distribution? Also, must the sum in the function always be 1? –  mercurial May 15 '12 at 8:11
@mercurial: you have the right conditions, but the probabilities should be $1/2$, $1/4$, $1/4$, and $0$, corresponding to the probabilities of the outcomes $\omega_1,\omega_2$, and $\omega_3$. (By the way, the code is \text{if }.) –  Brian M. Scott May 15 '12 at 8:14
@BrianM.Scott thanks! I kinda accidentally hit enter instead of shift-enter as i was trying to add the distribution. –  mercurial May 15 '12 at 8:17