Homework Help: 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???

21. Apr 12, 2016

andrewkirk

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

22. Apr 12, 2016

Ray Vickson

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.)

23. Apr 12, 2016

squenshl

Great. Cheers.