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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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$\mathbb{P}(\omega _{1})X(\omega_{1}) + \mathbb{P}(\omega _{2})X(\omega_{2}) + \mathbb{P}(\omega _{3})X(\omega_{3})$ 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
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