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A Discrete Multivariate Probability Distribution

  1. Apr 9, 2016 #1
    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. jcsd
  3. Apr 9, 2016 #2

    andrewkirk

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    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).
     
  4. Apr 10, 2016 #3
    Great thanks for that.
     
  5. Apr 10, 2016 #4

    Ray Vickson

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    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?
     
  6. Apr 10, 2016 #5
    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.
     
  7. Apr 10, 2016 #6
    I believe you just need to construct (X1,X2) for each possible outcome and count them to get the fractions
     
  8. Apr 10, 2016 #7
    Right of course so easy. Thanks.
     
  9. Apr 10, 2016 #8
    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???
     
  10. Apr 10, 2016 #9
  11. Apr 10, 2016 #10
    Thought so. It's just that word "function" threw me off a little.
     
  12. Apr 10, 2016 #11

    Ray Vickson

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    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:
    [tex] \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}
    [/tex]
    Right click on the image to see the tex commands used to produce that.
     
  13. Apr 10, 2016 #12
    Cool thanks.
     
  14. Apr 10, 2016 #13
    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##
    $$\begin{split}
    \text{Pr}(X_2=4|X_1=2) &= \frac{\text{Pr}(X_2=4\cap X_1=2)}{\text{Pr}(X_1=2)} \\
    &= \frac{f(2,4)}{\frac{1}{4}} \\
    &= \frac{0}{\frac{1}{4}} \\
    &= 0.
    \end{split}$$
     
  15. Apr 11, 2016 #14
    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???
     
  16. Apr 11, 2016 #15
    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)##???
     
  17. Apr 12, 2016 #16

    andrewkirk

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    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)##.
     
  18. Apr 12, 2016 #17
    Okay so is post 15 the way to go???
    Also what is ##n##??? is it ##n=2,3,4##???
    Thanks.
     
  19. Apr 12, 2016 #18

    andrewkirk

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    ##n## is a positive integer (not excluding 1).
     
  20. Apr 12, 2016 #19
    Oh right. So say, for example, if we have ##\{1,1,1\}##, then ##X_1=X_2=2##. Thus, ##X_1X_2=4##.
     
  21. Apr 12, 2016 #20
    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???
     
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