# A Discrete Multivariate Probability Distribution

1. Apr 9, 2016

### squenshl

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. The attempt at a solution
Not sure exactly what we are being asked here. Is it $\{1,1,1\},\{2,2,2\},\{1,2,1\},\{1,2,2\},\{2,1,1\},\{2,2,1\},\{2,1,2\}$??

2. Apr 9, 2016

### 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).

3. Apr 10, 2016

### squenshl

Great thanks for that.

4. Apr 10, 2016

### Ray Vickson

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?

5. Apr 10, 2016

### squenshl

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.

6. Apr 10, 2016

### throneoo

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

7. Apr 10, 2016

### squenshl

Right of course so easy. Thanks.

8. Apr 10, 2016

### 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???

9. Apr 10, 2016

### throneoo

yes

10. Apr 10, 2016

### squenshl

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

11. Apr 10, 2016

### Ray Vickson

S
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:
$$\begin{array}{c|cc|c} & I_1 & I_2 & \\ \text{gender}& \ 100,000 & \ 200,000 & \text{Total} \\ \hline M & 120 & 130 & 250 \\ F & 75 & 125 & 200 \\ \hline \end{array}$$
Right click on the image to see the tex commands used to produce that.

12. Apr 10, 2016

### squenshl

Cool thanks.

13. Apr 10, 2016

### 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 $X_2=4$
$$\begin{split} \text{Pr}(X_2=4|X_1=2) &= \frac{\text{Pr}(X_2=4\cap X_1=2)}{\text{Pr}(X_1=2)} \\ &= \frac{f(2,4)}{\frac{1}{4}} \\ &= \frac{0}{\frac{1}{4}} \\ &= 0. \end{split}$$

14. Apr 11, 2016

### 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???

15. Apr 11, 2016

### squenshl

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 $E(X_1X_2)$???

16. Apr 12, 2016

### andrewkirk

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 $E(X_1X_2)$.

17. Apr 12, 2016

### squenshl

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

18. Apr 12, 2016

### andrewkirk

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

19. Apr 12, 2016

### squenshl

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

20. Apr 12, 2016

### 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???