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Problem: a fair die is rolled at the same time as a fair coin is tossed. Let A be the number on the upper surface of the die and let B describe the outcome of the coin toss, where B is equal to 1 if the result is “head” and it is equal to 0 if the result if “tail”. The random variables X and Y are given by X = A + B and Y = A − B, respectively. Calculate P(X,Y)?

Answer: since both events A and B are independent, P(A=a,B=b) can be found; P(A=a,B=b) = 1/6*1/2 = 1/12. With the provided information, P(X) and P(Y) can also be calculated. I am not sure how to calculate P(X,Y)? Any help will be appreciated?

X = {1,2,3,4,5,6,7} -> pX(x) = {1/12, 1/6, 1/6, 1/6, 1/6, 1/12}


Y = {0,1,2,3,4,5,6} -> pY(y) = {1/12, 1/6, 1/6, 1/6, 1/6, 1/12}
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    $\begingroup$ Well, what are the possible values of $X,Y$? What is the probability that, say, $(X,Y)=(1,1)$? $\endgroup$
    – lulu
    Commented Nov 2, 2020 at 13:11
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    $\begingroup$ If you know $X$ and $Y$ you can use that to find $A$ and $B$. Notice that $A = \dfrac{X+Y}{2}$ and $B=\dfrac{X-Y}{2}$ $\endgroup$
    – JMoravitz
    Commented Nov 2, 2020 at 13:14
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    $\begingroup$ Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to. $\endgroup$ Commented Nov 2, 2020 at 13:16
  • $\begingroup$ @JMoravitz you mean $p(a_i)p(b_i) = P(\frac{x_i+y_i}{2})P(\frac{x_i-y_i}{2})$ ? $\endgroup$
    – GPrathap
    Commented Nov 2, 2020 at 13:21
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    $\begingroup$ Your notation is lacking, but yes I do mean to imply that $P(X=x,Y=y)=P(A=\frac{x+y}{2},B=\frac{x-y}{2})=P(A=\frac{x+y}{2})P(B=\frac{x-y}{2})$ $\endgroup$
    – JMoravitz
    Commented Nov 2, 2020 at 13:23

1 Answer 1

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It is difficult to get to $P(X = x, Y = y)$ directly from $P(X = x)$ and $P(Y = y)$. It is simpler to get there from $P(A = a, B = b)$.

Take an example,

$P(X = 1) = \frac{1}{12}$, $P(Y = 1) = \frac{1}{6}$.

There is only one possible way for $X = 1$ which is die shows $1$ and you get a tail on the coin. So the probability of $\frac{1}{12}$.

But when it comes to $P(Y = 1)$, you have two possible ways: $1$ on the die and tail on the coin OR $2$ on the die and head on the coin. So the probability of $\frac{1}{6}$.

Now when it comes to finding $P(X = 1, Y = 1)$, please note that you are looking for probability where the sum and difference are both $1$ which is only possible if die showed $1$ and you got a tail on the coin.

So to find $P(X = 1, Y = 1)$, we focus on $1$ on the die and tail on the coin which is nothing but $P(A = 1, B = 0) = \frac{1}{12}$.

If $A = 1, B = 1$, given $X = A + B, Y = A - B \implies A = \frac{X + Y}{2} = 1, B = \frac{X - Y}{2} = 0$ and that is why we found $P(A = 1, B = 0)$.

Valid combinations of X, Y are

i) $\{1, 1\}, \{2, 2\}, \{3, 3\}, \{4, 4\}, \{5, 5\}, \{6, 6\}$ that correspond to $A = a, B = 0$.

ii) $\{7, 5\}, \{6, 4\}, \{5, 3\}, \{4, 2\}, \{3, 1\}, \{2, 0\}$ that correspond to $A = a, B = 1$.

All of the above combinations will have probability of $\frac{1}{12}$. Rest of them will have probability of zero. For example , $P(X = 1, Y = 0) = 0$.

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  • $\begingroup$ Thank you for your nice explanation :) $\endgroup$
    – GPrathap
    Commented Nov 3, 2020 at 6:56
  • $\begingroup$ @GPrathap you are welcome. $\endgroup$
    – Math Lover
    Commented Nov 3, 2020 at 7:04

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