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I am tossing a fair coin 6 times. I define $X$ to be the number of heads in first 3 tosses and $Y$ to be the number of heads in all 6 tosses. $X$ and $Y$ are random variables.

I want to create the joint and marginal distribution tables for $p(X,Y)$ and $p(X)$ and $p(Y)$. For creating the joint distribution table, I have the following (drawing from this post):

$X = B(3, 0.5)$ and $Y = B(6, 0.5)$

Then are the table columns and rows something like this $P(X = i, Y=j)$ with all the values filled in?

\begin{array}{|c|c|c|c|} \hline & i=1 & i=2 & i=3 \\ \hline j=1& & &\\ \hline j=2& & &\\ \hline j=3& & &\\ \hline j=4& & &\\ \hline j=5& & &\\ \hline j=6& & &\\ \hline \end{array}

For example, if I have $P(X = 2, Y= 5)$, I'm interpreting the statement as probability of getting at "exactly 2 heads in the first three tosses and exactly 5 heads from the 6 tosses." So then I calculated that $P(X=2) = 0.375$ and $P(X=6) = 0.09375$. So would I multiply these two values and enter them as such:

\begin{array}{|c|c|c|c|} \hline & i=1 & i=2 & i=3 \\ \hline j=1& & &\\ \hline j=2& & &\\ \hline j=3& & &\\ \hline j=4& & &\\ \hline j=5& & 0.375\times 0.09375 = 0.035 &\\ \hline j=6& & &\\ \hline \end{array}

I'm not sure if this the proper way to represent this and if assumptions are correct or if I'm erring in my concept.

StubbornAtom
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Jonathan
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  • No, you need Pr(first 3) times Pr((second 3). Note pr(i=2, j=6) must be zero. – Peter Oct 12 '20 at 22:13
  • @Peter sorry I'm a bit confused by your statement. Why "second 3"? And the pr must be zero? – Jonathan Oct 12 '20 at 22:15
  • To get 6 heads in the 6 throws you must have had 3 heads in the first 3 throws, so the two probabilities i and j are not indeoendent. This means you can't multiply them to get the probability of both occurring. However the first 3 throws and the last 3 throws are actually independent results so you can multiply them. For i=2, j=5 you would multiply Pr(2 heads from 3) times Pr(3 heads from 3). – Peter Oct 12 '20 at 22:23
  • I see. But thinking of it in that perspective makes it considerably more complicated when calculating P(X=2, Y=1), no? – Jonathan Oct 12 '20 at 22:26
  • If X=0, Ycan be 0,1,2,3. If X=2 Y can be 2,3,4,5. You might want to extend your table to include zero, and you can fairly easily put in the entries that must be zero. – Peter Oct 12 '20 at 22:32

1 Answers1

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First point to note is that $X$ and $Y$ can be $0$ with positive probability, so you might want to extend your table

Second point to note is that $Y \ge X$, so $P(Y < X)=0$

Third point to note is that $Y$ is not independent of $X$, and the higher $X$ is, the higher $Y$ can be. But if $Z$ is the number of heads in the last three tosses then you can say

(a) $Z \sim B(3,0.5)$,

(b) $X$ and $Z$ are independent,

(c) $Y=X+Z$ so $Z=Y-X$, and

(d) $P(X=i,Y=j) =P(X=i, Z=j-i)= P(X=i) \times P(Z=j-i)$ so for example $P(X=2,Y=5)=P(X=2)\times P(Z=3)$

So the start of your table could be

\begin{array}{|c|c|c|c|c|} \hline & i=0 & i=1 & i=2 & i=3 \\ \hline j=0& & 0& 0&0\\ \hline j=1& & & 0&0\\ \hline j=2& & & &0\\ \hline j=3& & & &\\ \hline j=4& & & &\\ \hline j=5& & & 0.375\times 0.125 = 0.046875 &\\ \hline j=6& & & &\\ \hline \end{array}

Henry
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