I don't know where to start :(

please help me, thanks

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- Nov 8th 2011, 05:39 AMHelpmeplease1Random Variable.
I don't know where to start :(

please help me, thanks

http://i40.tinypic.com/143nj20.png - Nov 8th 2011, 08:13 AMemakarovRe: Random Variable VERY TRICKY
I assume that the two numbers are on the faces of the dice.

Let $\displaystyle E=\{1,\dots,6\}$ and $\displaystyle \mathcal{E}=2^E$; then $\displaystyle E$ is the domain of $\displaystyle X$. The definition says that $\displaystyle X:\Omega\to E, X(x,y)=\max(x,y)$ is an $\displaystyle (E,\mathcal{E})$-valued random variable if $\displaystyle X^{-1}(B)\in\mathcal{F}_2$ for all $\displaystyle B\in\mathcal{E}$. That is, for any subset $\displaystyle B$ of values that $\displaystyle X$ may have, the set of outcomes from $\displaystyle \Omega$ that produce an answer in $\displaystyle B$ is a legitimate event, i.e., belongs to the set of events $\displaystyle \mathcal{F}_2$. For this example, this fact is obvious because $\displaystyle \mathcal{F}_2$ includes all possible subsets of $\displaystyle \Omega$.

In contrast, $\displaystyle X$ is not a random variable with respect to $\displaystyle (\Omega,\mathcal{F}_1,\mathbb{P})$ and $\displaystyle (E,\mathcal{E})$. E.g., consider $\displaystyle X^{-1}(\{2\})$. - Nov 8th 2011, 08:45 AMHelpmeplease1Re: Random Variable VERY TRICKY
- Nov 8th 2011, 01:24 PMemakarovRe: Random Variable VERY TRICKY
No, here $\displaystyle X=\max$ and $\displaystyle \max\nolimits^{-1}(\{2\})=\{(1,2),(2,1),(2,2)\}$ because the maximum of all three pairs is 2. Now, $\displaystyle \{(1,2),(2,1),(2,2)\}\notin\mathcal{F}_1$.

- Nov 10th 2011, 11:31 AMHelpmeplease1Re: Random Variable VERY TRICKY
If I wanted to describe the distribution function Fx, I take it I do a step function?

- Nov 11th 2011, 05:59 AMHelpmeplease1Re: Random Variable VERY TRICKY
- Nov 11th 2011, 06:21 AMemakarovRe: Random Variable VERY TRICKY
Yes, so the cumulative distribution function $\displaystyle F_X$ is from {1, ..., 6} to [0, 1]; $\displaystyle F_X(n)=\mathop{\mbox{Pr}}(X\le n)$. So,

$\displaystyle F_X(1)=\mathop{\mbox{Pr}}(X=1)=1/36$,

$\displaystyle F_X(2)=\mathop{\mbox{Pr}}(X=1\mbox{ or }X=2)=1/36+3/36$,

$\displaystyle F_X(3)=\mathop{\mbox{Pr}}(X=1\mbox{ or }X=2\mbox{ or }X=3)=1/36+3/36+5/36$,

etc. - Nov 11th 2011, 06:28 AMHelpmeplease1Re: Random Variable VERY TRICKY
- Nov 11th 2011, 07:05 AMemakarovRe: Random Variable VERY TRICKYQuote:

Originally Posted by**emakarov**

$\displaystyle F_X(x)=0$ for $\displaystyle x<1$,

$\displaystyle F_X(x)=1/36$ for $\displaystyle 1\le x<2$,

$\displaystyle F_X(x)=1/36+3/36$ for $\displaystyle 2\le x<3$,

...

$\displaystyle F_X(x)=1$ for $\displaystyle x\ge 6$.

https://lh6.googleusercontent.com/-G...o/s800/cdf.png