A Discrete Multivariate Probability Distribution

[squenshl]

Homework Statement

A fair coin has a $1$ painted upon one side and a $2$ painted upon the other side. The coin is tossed $3$ times.
Write down a sample space for this experiment.
Let $X_1$ be the sum of the numbers obtained on the first $2$ tosses and $X_2$ be the sum of the numbers obtained on the last $2$ tosses.

Homework Equations

The Attempt at a Solution

Not sure exactly what we are being asked here. Is it $\{1,1,1\},\{2,2,2\},\{1,2,1\},\{1,2,2\},\{2,1,1\},\{2,2,1\},\{2,1,2\}$??

 

[andrewkirk]

The sample space is the set of all possible outcomes. Your attempt is correct, except that you have only listed seven possibilities and there are eight ($2^3$).

$X_1$ and $X_2$ are random variables. A random variable is a map whose domain is the sample space and whose range is a specified set of numbers (in this case, the set of non-negative integers).

 

[squenshl]

The sample space is the set of all possible outcomes. Your attempt is correct, except that you have only listed seven possibilities and there are eight ($2^3$).

$X_1$ and $X_2$ are random variables. A random variable is a map whose domain is the sample space and whose range is a specified set of numbers (in this case, the set of non-negative integers).

Great thanks for that.

 

[Ray Vickson]

Homework Statement

A fair coin has a $1$ painted upon one side and a $2$ painted upon the other side. The coin is tossed $3$ times.
Write down a sample space for this experiment.
Let $X_1$ be the sum of the numbers obtained on the first $2$ tosses and $X_2$ be the sum of the numbers obtained on the last $2$ tosses.

Homework Equations

The Attempt at a Solution

Not sure exactly what we are being asked here. Is it $\{1,1,1\},\{2,2,2\},\{1,2,1\},\{1,2,2\},\{2,1,1\},\{2,2,1\},\{2,1,2\}$??

Is there also a question involving $X_1$ and $X_2$? For example, are you supposed to write down the bivariate distribution of $(X_1,X_2)$ or something like that?

 

[squenshl]

Is there also a question involving $X_1$ and $X_2$? For example, are you supposed to write down the bivariate distribution of $(X_1,X_2)$ or something like that?

Yup there is.
Let $X_1$ be the sum of the numbers obtained on the first $2$ tosses and $X_2$ be the sum of the numbers obtained on the last $2$ tosses.Verify that the entries in the table below specify the joint probability function, $f(x_1,x_2)$, of $X_1$ and $X_2$. I can't put the table in but the table has $x_1 = 2,3,4$ at the top and $x_2 = 2,3,4$ on the left hand side with $f(2,2) = \frac{1}{8}$, $f(3,2) = \frac{1}{8}$, $f(4,2) = 0$, $f(2,3) = \frac{1}{8}$, $f(3,3) = \frac{1}{4}$, $f(4,3) = \frac{1}{8}$, $f(2,4) = 0$, $f(3,4) = \frac{1}{8}$, $f(4,4) = \frac{1}{8}$. How do Get these?? I guess I only need to know how to get one and the rest are the same.

 

[throneoo]

I believe you just need to construct (X₁,X₂) for each possible outcome and count them to get the fractions

 

[squenshl]

I believe you just need to construct (X₁,X₂) for each possible outcome and count them to get the fractions

Right of course so easy. Thanks.

 

[squenshl]

To find the marginal probability function of $X_1$ is it just adding up the probabilities along $X_1$, i.e. for $X_1=2$, we get $\frac{1}{8}+\frac{1}{8}+0=\frac{1}{4}$ etc?

 

[throneoo]

yes

 

[squenshl]

yes

Thought so. It's just that word "function" threw me off a little.

 

[Ray Vickson]

S

Yup there is.
Let $X_1$ be the sum of the numbers obtained on the first $2$ tosses and $X_2$ be the sum of the numbers obtained on the last $2$ tosses.Verify that the entries in the table below specify the joint probability function, $f(x_1,x_2)$, of $X_1$ and $X_2$. I can't put the table in but the table has $x_1 = 2,3,4$ at the top and $x_2 = 2,3,4$ on the left hand side with $f(2,2) = \frac{1}{8}$, $f(3,2) = \frac{1}{8}$, $f(4,2) = 0$, $f(2,3) = \frac{1}{8}$, $f(3,3) = \frac{1}{4}$, $f(4,3) = \frac{1}{8}$, $f(2,4) = 0$, $f(3,4) = \frac{1}{8}$, $f(4,4) = \frac{1}{8}$. How do Get these?? I guess I only need to know how to get one and the rest are the same.

Since you are using LaTeX, you can enter tables using "array" in this Forum. (This Forum's version of tex does not seem to support the TeX/LaTeX "tabular" command or the "multicolumn" command, so is a bit limited.) For example:
\begin{array}{c|cc|c}<br /> &amp; I_1 &amp; I_2 &amp; \\<br /> \text{gender}&amp; \$ 100,000 &amp; \$ 200,000 &amp; \text{Total} \\ \hline<br /> M &amp; 120 &amp; 130 &amp; 250 \\<br /> F &amp; 75 &amp; 125 &amp; 200 \\ \hline<br /> \end{array}<br />
Right click on the image to see the tex commands used to produce that.

 

[squenshl]

SSince you are using LaTeX, you can enter tables using "array" in this Forum. (This Forum's version of tex does not seem to support the TeX/LaTeX "tabular" command or the "multicolumn" command, so is a bit limited.) For example:
\begin{array}{c|cc|c}<br /> &amp; I_1 &amp; I_2 &amp; \\<br /> \text{gender}&amp; \$ 100,000 &amp; \$ 200,000 &amp; \text{Total} \\ \hline<br /> M &amp; 120 &amp; 130 &amp; 250 \\<br /> F &amp; 75 &amp; 125 &amp; 200 \\ \hline<br /> \end{array}<br />
Right click on the image to see the tex commands used to produce that.

Cool thanks.

 

[squenshl]

If I wanted to calculate the conditional probability function of $X_2$ given $X_1 = 2$, do i just do for $X_2=2$
$$\begin{split}
\text{Pr}(X_2=2|X_1=2) &= \frac{\text{Pr}(X_2=2\cap X_1=2)}{\text{Pr}(X_1=2)} \\
&= \frac{f(2,2)}{\frac{1}{4}} \\
&= \frac{\frac{1}{8}}{\frac{1}{4}} \\
&= \frac{1}{2}.
\end{split}$$
For $X_2=3$,
$$\begin{split}
\text{Pr}(X_2=3|X_1=2) &= \frac{\text{Pr}(X_2=3\cap X_1=2)}{\text{Pr}(X_1=2)} \\
&= \frac{f(2,3)}{\frac{1}{4}} \\
&= \frac{\frac{1}{8}}{\frac{1}{4}} \\
&= \frac{1}{2}.
\end{split}$$
Finally, for $X_2=4$
$$\begin{split}
\text{Pr}(X_2=4|X_1=2) &= \frac{\text{Pr}(X_2=4\cap X_1=2)}{\text{Pr}(X_1=2)} \\
&= \frac{f(2,4)}{\frac{1}{4}} \\
&= \frac{0}{\frac{1}{4}} \\
&= 0.
\end{split}$$

 

[squenshl]

How would I calculate the probability function of $X_1X_2$ as well as $E(X_1X_2)$ as I know that $X_1$ and $X_2$ are not independent?

 

[squenshl]

How would I calculate the probability function of $X_1X_2$ as well as $E(X_1X_2)$ as I know that $X_1$ and $X_2$ are not independent?

If $X_1X_2$ is the product of the sum of the first $2$ tosses and last $2$ tosses does that make, for example, $\{1,1,1\}$ become $\{1,1\}$ and if so how do we make that into a probability function and in turn calculate $E(X_1X_2)$?

 

[andrewkirk]

How would I calculate the probability function of $X_1X_2$ as well as $E(X_1X_2)$ as I know that $X_1$ and $X_2$ are not independent?

In this example it's very easy, because the sample space is so small. For every one of the eight elements of the sample space you work out what $X_1X_2$ is. Then $Pr(X_1X_2=n)$ is just the number of cases in the sample space where $X_1X_2=n$, divided by the number of elements in the sample space.
Then you use that probability function to calculate $E(X_1X_2)$.

 

[squenshl]

Okay so is post 15 the way to go?
Also what is $n$? is it $n=2,3,4$?
Thanks.

 

[andrewkirk]

what is $n$? is it $n=2,3,4$?

$n$ is a positive integer (not excluding 1).

 

[squenshl]

$n$ is a positive integer (not excluding 1).

Oh right. So say, for example, if we have $\{1,1,1\}$, then $X_1=X_2=2$. Thus, $X_1X_2=4$.

 

[squenshl]

I get
$\{1,1,1\}$, $X_1X_2=4$,
$\{2,2,2\}$, $X_1X_2=16$,
$\{1,1,2\}$, $X_1X_2=6$,
$\{1,2,1\}$, $X_1X_2=9$,
$\{1,2,2\}$, $X_1X_2=12$,
$\{2,1,1\}$, $X_1X_2=6$,
$\{2,1,2\}$, $X_1X_2=9$,
$\{2,2,1\}$, $X_1X_2=12$.
so $\text{Pr}(X_1X_2=4)=\text{Pr}(X_1X_2=16)=\frac{1}{8}$ and $\text{Pr}(X_1X_2=6)=\text{Pr}(X_1X_2=9)=\text{Pr}(X_1X_2=12)=\frac{1}{4}.$
Therefore $E(X_1X_2) = 4\times \frac{1}{8} + 16\times \frac{1}{8} + 6\times \frac{1}{4} + 9\times \frac{1}{4} + 12\times \frac{1}{4} = \frac{37}{4}.$
Am I even on the right track?

 

[andrewkirk]

Am I even on the right track?

I didn't check the arithmetic, but the method all looks correct to me.

 

[Ray Vickson]

I get
$\{1,1,1\}$, $X_1X_2=4$,
$\{2,2,2\}$, $X_1X_2=16$,
$\{1,1,2\}$, $X_1X_2=6$,
$\{1,2,1\}$, $X_1X_2=9$,
$\{1,2,2\}$, $X_1X_2=12$,
$\{2,1,1\}$, $X_1X_2=6$,
$\{2,1,2\}$, $X_1X_2=9$,
$\{2,2,1\}$, $X_1X_2=12$.
so $\text{Pr}(X_1X_2=4)=\text{Pr}(X_1X_2=16)=\frac{1}{8}$ and $\text{Pr}(X_1X_2=6)=\text{Pr}(X_1X_2=9)=\text{Pr}(X_1X_2=12)=\frac{1}{4}.$
Therefore $E(X_1X_2) = 4\times \frac{1}{8} + 16\times \frac{1}{8} + 6\times \frac{1}{4} + 9\times \frac{1}{4} + 12\times \frac{1}{4} = \frac{37}{4}.$
Am I even on the right track?

Yes, that is exactly correct. However, why go to the bother of repeating what you have already done? In post #5 you had already written out the probabilities of all the $(X_1,X_2)$ pairs, and all you need to do now is to multiply the components together to get the $X_1 X_2$ values. (My main concern is that you do not seem to realize that you can build up the solution, and so use previously-developed results to further the analysis. Somehow, you are missing something, or maybe you just lack confidence in your own abilities.)

 

[squenshl]

I didn't check the arithmetic, but the method all looks correct to me.

Great. Cheers.