Random Variable.

**Helpmeplease1** · November 8th 2011, 05:39 AM

I don't know where to start

please help me, thanks

**emakarov** · November 8th 2011, 08:13 AM

Originally Posted by Helpmeplease1

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 $X^{-1}(\{2\})$ .

**Helpmeplease1** · November 8th 2011, 08:45 AM

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?

**emakarov** · November 8th 2011, 01:24 PM

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, $\{(1,2),(2,1),(2,2)\}\notin\mathcal{F}_1$ .

**Helpmeplease1** · November 10th 2011, 11:31 AM

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

**Helpmeplease1** · November 11th 2011, 05:59 AM

Originally Posted by Helpmeplease1

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

**emakarov** · November 11th 2011, 06:21 AM

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.

**Helpmeplease1** · November 11th 2011, 06:28 AM

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?

**emakarov** · November 11th 2011, 07:05 AM

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 $x\ge 6$ .

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