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You choose three different numbers at random from the numbers $1,2,...,10$. Let $X$ be the smallest of these three numbers and $Y$ be the largest. What is the joint density mass of $X$ and $Y$? What are the marginal distributions of $X$ and $Y$? What is the probability mass functions of $Y-X$?

Try.

We need to find $P(X=x,Y=y)$, to pick the largest and smallest number we do it in one way only, but to pick the third number, it has to be between $x$ and $y$ no inclusive, so all posible values for the third number is ${y-x-1 \choose 1}$. Threfore,

$$ P(X=x,Y=y) = \frac{ {y-x-1 \choose 1} }{ {10 \choose 3 } } = \frac{y-x-1 }{120} $$

Now, $x$ can't be 10 so $x=1,2,3,4,5,6,7,8,9$ and $y=x+1,x+2,...,10$. So,

$$ p_X(x) = \sum_{y=x+1}^{10} \frac{y-x-1 }{120} $$

$$ p_Y(y) = \sum_{x=1}^{y-1}\frac{y-x-1 }{120} $$

$$ p_{Y-X}(n) = P(X=x, Y-X=n) = P(X=x, Y=x+n) = \frac{n-1}{120} $$

is this correct?

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  • $\begingroup$ You can simplify the sums to $p_x(x) = \dfrac{(9-x)(10-x)}{2}$ and $p_y(y) = \dfrac{(y-1)(y-2)}{2}$ on these ten values $\endgroup$
    – Henry
    Nov 4, 2018 at 9:55

1 Answer 1

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$x$ can't be $10$ or $9$ and $y=x+2 , \ldots, 10$

$$p_X(x) =\sum_{y=x+2}^{10}\frac{y-x-1}{120}$$

$$p_Y(y) =\sum_{x=1}^{y-2}\frac{y-x-1}{120}$$

If $n=2, \ldots, 8$, \begin{align} p_{Y-X}(n) &= \sum_{x=1}^{10-n} Pr(X=x,Y=x+n)\\ &= \sum_{x=1}^{10-n} \frac{n-1}{120}\\ &= \frac{(10-n)(n-1)}{120} \end{align}

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  • $\begingroup$ for the mass fn $Y-X$, why do we need to add the probabilities of the joint of $X$ and $Y$? $\endgroup$
    – James
    Nov 4, 2018 at 8:03
  • $\begingroup$ $Pr(Y-X=n) = \sum_{x=1}^{10-n}Pr(Y-X=n|X=x)Pr(X=x)$ is the law of total probability. $\endgroup$ Nov 4, 2018 at 8:09
  • $\begingroup$ but you havent multiplied the left hand side by $P(X=x)$ $\endgroup$
    – James
    Nov 4, 2018 at 8:13
  • $\begingroup$ $Pr(Y-X=n) = \sum_{x=1}^{10-n}Pr(Y-X=n|X=x)Pr(X=x) = \sum_{x=1}^{10-n}Pr(Y-X=n,X=x)$ $\endgroup$ Nov 4, 2018 at 8:14
  • $\begingroup$ In law of total probability, we condition on a variable and sum things up $\endgroup$ Nov 4, 2018 at 8:15

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