# Finding a probability mass function of joint discrete rv

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?

• 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 Nov 4, 2018 at 9:55

$$x$$ 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}

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