1. ## Random Variable.

I don't know where to start

2. ## Re: Random Variable VERY TRICKY

Let X be the larger of the two numbers shown.
I assume that the two numbers are on the faces of the dice.

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

In contrast, $X$ is not a random variable with respect to $(\Omega,\mathcal{F}_1,\mathbb{P})$ and $(E,\mathcal{E})$. E.g., consider $X^{-1}(\{2\})$.

3. ## Re: Random Variable VERY TRICKY

Originally Posted by emakarov
I assume that the two numbers are on the faces of the dice.

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

In contrast, $X$ is not a random variable with respect to $(\Omega,\mathcal{F}_1,\mathbb{P})$ and $(E,\mathcal{E})$. E.g., consider $X^{-1}(\{2\})$.
Thanks so the inverse of {2} = (2,2) and you take the smallest one instead which is 2. And 2 isn't there, is this correct?

4. ## Re: Random Variable VERY TRICKY

No, here $X=\max$ and $\max\nolimits^{-1}(\{2\})=\{(1,2),(2,1),(2,2)\}$ because the maximum of all three pairs is 2. Now, $\{(1,2),(2,1),(2,2)\}\notin\mathcal{F}_1$.

5. ## Re: Random Variable VERY TRICKY

If I wanted to describe the distribution function Fx, I take it I do a step function?

6. ## Re: Random Variable VERY TRICKY

If I wanted to describe the distribution function Fx, I take it I do a step function?
Anyone, i am also struggling to do the step function

I know the chance of 1 = 1/36
2 = 3/36
3 = 5/26
4 = 7/36
5 = 9/36
6 = 11/36

7. ## Re: Random Variable VERY TRICKY

Yes, so the cumulative distribution function $F_X$ is from {1, ..., 6} to [0, 1]; $F_X(n)=\mathop{\mbox{Pr}}(X\le n)$. So,

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

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

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

etc.

8. ## Re: Random Variable VERY TRICKY

Originally Posted by emakarov
Yes, so the cumulative distribution function $F_X$ is from {1, ..., 6} to [0, 1]; $F_X(n)=\mathop{\mbox{Pr}}(X\le n)$. So,

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

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

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

etc.
So to plot a step function

would i draw a horizontal line between 0 and 1 on the x axis and 1 on the y -axis
then a horizontal line between 1 and 2 on the x axis and 4 on the y -axis?

or should i just put a dot instead of a line?

9. ## Re: Random Variable VERY TRICKY

Originally Posted by emakarov
Yes, so the cumulative distribution function $F_X$ is from {1, ..., 6} to [0, 1]
Correction: looking at the Wikipedia page, it seems that CDF is always defined on real numbers, even for discrete random variables. So,

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

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

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

...

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