Question

ciule jolh! PMF and the marginal PMFs? 6.14 Let X and Y be discrete random variables. Show that the function p: R2 R defined by p(r, y) px(x)pr(y) is a joint PMF by verifying that it satisfies properties (a)-(c) of Proposition 6.1 on page 262. Hint: A subset of a countable set is countable CHAPTER SIX Joindy Discrete Random Variables 6.2 Joint and marginal PMFs of the discrete random variables x numher of bedrooms and momber of bwthrooms of a randomly selected home 262 X and Y, the Table Batthrooms, y 0.0 0. 2 006 000 3028/ 024 004000.56 4 004 0.22 0.10 0.02 038 Pv(y) | 0.38 | 046 | 0.14 | 0.02 1.00 Proposition 6.1 Basic Properties of a Joint PMF: Bivariate Case The joint probability mass function px.y of two discrete random variables X and y satisties the following three properties a) px ra. y)2 0 for all (. y) e R2: that is, a joint PMF is a nonnegative function b) [(r. y)ER:Px.r(t. y)0 is countable: that is, the set of pairs of real n I numbers for which a joint PMF is nonzero is countable. c) ΣΣ(xmPx.r(x, y) 1; that is, the sum of the values of a joint PMF equals 1. In Exercise 6.17, we ask you to show that a function p:R R that satisfies prop- erties (a), (b). and (c) of Proposition 6.1 is the joint PMF of some pair of discrete random variables. Therefore, for any such function, we can say,. "Let X and Y be discrete ran- dom variables with joint PMF p." This statement makes sense regardless of whether we explicitly give X and Y and the sample space on which they are defined. Marginal Probability Mass Functions In Example 6 l, as we already noted, te values of the joint PMF, pXY (xJ), fall inside the heavy lines in Table 6.2. For instance, PX.y(4,3) 0.22 Let's now examine the values that fall outside those heavy lines. The row and column eads Total" in Table 6.1 were changed in Table 6.2 to py(y) and px(x), respeo- tively. Thus, in Table 6.2, the last row gives the PMF of Y, and the last column gives the PMF of X. In this context, for purposes of clarity and because these two probability mass functions are in the margins of the table, the PMF of each is called a margi probablity mass function. Technically, however, the adjective "marginal" is redundant. Note that the sum of the values of the joint PMF in a row or column equals the value ot he marginal PMF at the end (right or bottom) of that row or column. For instance, the sum of the values of the joint PMF in the column labeled "3" is 0.00+0.24 +0.22-0.46. the valuc of the marginal PMP at the bottom of that column The openicn ean tnning over all pints(x, y) in R

Answer 1

Property (a)

Now, since both p_X(x) and p_Y(y) are pmfs, we have:

$\\ p_X(x) \ge 0 \text{ for all } x \in R \text{ ; and} \\ p_Y(y) \ge 0 \text{ for all } y \in R$

Thus, PX.y(x,y) PX(z)PV(y) > 0 for all (x,y) E R2

Property (b)

Now, for p_X,Y(x,y) to be non-zero we need to have both p_X(x) and p_Y(y) not equal to zero.

Thus, the set $\{ (x,y) \in R^2 : p_{X,Y}(x,y) = p_X(x)*p_Y(y) \ne 0 \}$ is countable if both

$\{ x \in R : p_X(x) \ne 0\} \text{ and } \{ y \in R : p_Y(y) \ne 0\} \text{ are countable}$.

Now, since both p_X(x) and p_Y(y) are pmfs, we have:

$\\ \{ x \in R : p_X(x) \ne 0\} \text{ is countable} \text{ ; and} \\ \{ y \in R : p_Y(y) \ne 0\} \text{ is countable}$

Thus, the set

$\{ (x,y) \in R^2 : p_{X,Y}(x,y) = p_X(x)*p_Y(y) \ne 0 \} \text{ is countable}.$

Property (c)

Now,consider:

$\begin{align*} {\sum \sum}_{(x,y)} p_{X,Y}(x,y) &= \sum_x \sum_y p_{X,Y}(x,y) \\ &= \sum_x \sum_y p_X(x)*p_Y(y) \\ &= \sum_x p_X(x) * \sum_y p_Y(y) \\ &= 1*1 \\ &= 1 \end{align*}$

Thus, is a Joint PMF.

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