Joint probability mass function

Definition taught in Joint distributions, part of Probability

Joint probability mass function

Type Definition
Taught in Joint distributions
Part Probability
Properties 1
Worked examples 3
Questions using it 2

Discrete random variables X and Y have joint probability mass function f_{X,Y}(x,y)=\mathbb{P}(X=x,Y=y). The marginal mass functions are the row and column sums,

f_X(x)=\sum_y f_{X,Y}(x,y), \qquad f_Y(y)=\sum_x f_{X,Y}(x,y).

Properties

Tower property

Main article: Tower property

If the expectations exist, then \mathbb{E}\bigl(\mathbb{E}(Y\mid X)\bigr)=\mathbb{E} Y. More generally, \mathbb{E}\bigl(\mathbb{E}(Y\mid X,Z)\mid X\bigr)=\mathbb{E}(Y\mid X).

Examples

A box of three dice

Worked example from Joint distributions

A box holds three dice. Die 1 is ordinary. Die 2 has no six but two fives. Die 3 has no five but two sixes. A die is chosen at random and thrown. Let X be the number of the die and Y the face shown.

Question. Find the joint mass function and both marginals. Are X and Y independent?

Solution. Each die is chosen with probability 1/3, and each of its six faces then shows with probability 1/6, so each face position has probability 1/18. A value printed on two faces gets 2/18=1/9.

y=1 y=2 y=3 y=4 y=5 y=6 f_X(x)
x=1 1/18 1/18 1/18 1/18 1/18 1/18 1/3
x=2 1/18 1/18 1/18 1/18 1/9 0 1/3
x=3 1/18 1/18 1/18 1/18 0 1/9 1/3
f_Y(y) 1/6 1/6 1/6 1/6 1/6 1/6 1

The marginals are the row and column sums. Remarkably, Y is uniform on \{1,\ldots,6\}, exactly as for a single fair die. Yet X and Y are not independent: \mathbb{P}(X=2,Y=6)=0, while f_X(2)f_Y(6)=1/18. Seeing a six rules out die 2. The marginals alone cannot reveal the dependence; only the joint table can.

A joint table

Question page: A joint table

The joint probability mass function of X and Y is

x\backslash y 1 3 6 8
2 0 0.1 0.1 0
5 0.2 0 0 0
6 0 0.2 0.1 0.3

  1. Find the marginal mass functions of X and of Y.

  2. Are X and Y independent?

  3. Calculate \mathbb{E}(X^2 Y).

  1. Row sums: \mathbb{P}(X=2)=0.2, \mathbb{P}(X=5)=0.2, \mathbb{P}(X=6)=0.6. Column sums: \mathbb{P}(Y=1)=0.2, \mathbb{P}(Y=3)=0.3, \mathbb{P}(Y=6)=0.2, \mathbb{P}(Y=8)=0.3.

  2. They are not independent. For example \mathbb{P}(X=2,Y=1)=0, while \mathbb{P}(X=2)\mathbb{P}(Y=1)=0.04.

  3. Only the six positive cells contribute:

\begin{aligned} \mathbb{E}(X^2 Y) &=0.1\cdot 4\cdot 3+0.1\cdot 4\cdot 6+0.2\cdot 25\cdot 1\\ &\quad+0.2\cdot 36\cdot 3+0.1\cdot 36\cdot 6+0.3\cdot 36\cdot 8\\ &=138.2. \end{aligned}

Signs and their sum

Question page: Signs and their sum

Let U and V be independent, with \mathbb{P}(U=1)=\mathbb{P}(V=1)=1/4 and \mathbb{P}(U=-1)=\mathbb{P}(V=-1)=3/4. Set X=U/V and Y=U+V.

  1. Give the joint probability mass function of (X,Y).

  2. Calculate \operatorname{Cov}(X,Y).

  1. Since V=\pm1, 1/V=V and X=UV. The four sign pairs give
(U,V) Probability X Y
(1,1) 1/16 1 2
(1,-1) 3/16 -1 0
(-1,1) 3/16 -1 0
(-1,-1) 9/16 1 -2

so \mathbb{P}(X=1,Y=2)=1/16, \mathbb{P}(X=-1,Y=0)=6/16 and \mathbb{P}(X=1,Y=-2)=9/16.

  1. \mathbb{E} X=10/16-6/16=1/4, \mathbb{E} Y=\mathbb{E} U+\mathbb{E} V=2(1/4-3/4)=-1 and \mathbb{E}(XY)=2\times1/16-2\times9/16=-1. So

\operatorname{Cov}(X,Y)=-1-\tfrac14\times(-1)=-\tfrac34.

Used in

See also

Random variable · Discrete distribution · Independent random variables · Joint probability density · Covariance · Expectation vector and covariance matrix · Conditional distribution · Tower property

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