1  PP: Joint, Conditional, and Marginal Distributions

Author

Rachel Roggenkemper

Published

January 8, 2025

The latest series of collectible Lego Minifigures contains 3 different Minifigure prizes (labeled 1, 2, 3). Each package contains a single unknown prize. Suppose we only buy 3 packages and we consider as our sample space outcome the results of just these 3 packages (prize in package 1, prize in package 2, prize in package 3). For example, 323 (or (3, 2, 3)) represents prize 3 in the first package, prize 2 in the second package, prize 3 in the third package. Let \(X\) be the number of distinct prizes obtained in these 3 packages. Let \(Y\) be the number of these 3 packages that contain prize 1. Suppose that each package is equally likely to contain any of the 3 prizes, regardless of the contents of other packages; let \(P\) denote the corresponding probability measure.

It can be shown that the joint distribution of \(X\) and \(Y\) can be represented by the following table.

\(P(X=x,\;Y=x)\)

\(y\)
\(x\) 0 1 2 3
1 2/27 0 0 1/27
2 6/27 6/27 6/27 0
3 0 6/27 0 0

1. Briefly explain why there are 27 possible outcomes.

There are 27 possible outcomes because in each of the 3 packages, any of the 3 prizes could occur, and \(3*3*3 = 27\).

  1. 1, 1, 1 -> X=1, Y=3
  2. 1, 1, 2 -> X=2, Y=2
  3. 1, 1, 3 -> X=2, Y=2
  4. 1, 2, 1 -> X=2, Y=2
  5. 1, 3, 1 -> X=2, Y=2
  6. 1, 2, 3 -> X=3, Y=1
  7. 1, 3, 2 -> X=3, Y=1
  8. 1, 2, 2 -> X=2, Y=1
  9. 1, 3, 3 -> X=2, Y=1
  10. 2, 1, 3 -> X=3, Y=1
  11. 2, 3, 1 -> X=3, Y=1
  12. 2, 2, 2 -> X=1, Y=0
  13. 2, 1, 1 -> X=2, Y=2
  14. 2, 3, 3 -> X=2, Y=0
  15. 2, 1, 2 -> X=2, Y=1
  16. 2, 3, 2 -> X=2, Y=0
  17. 2, 2, 1 -> X=2, Y=1
  18. 2, 2, 3 -> X=2, Y=0
  19. 3, 1, 2 -> X=3, Y=1
  20. 3, 2, 1 -> X=3, Y=1
  21. 3, 3, 3 -> X=1, Y=0
  22. 3, 2, 2 -> X=2, Y=0
  23. 3, 1, 1 -> X=2, Y=2
  24. 3, 3, 1 -> X=2, Y=1
  25. 3, 3, 2 -> X=2, Y=0
  26. 3, 1, 3 -> X=2, Y=1
  27. 3, 2, 3 -> X=2, Y=0

2. Show that \(P(X=1,\;Y=0) = 2/27\) by listing the outcomes that comprise the event {\(X=1,Y=0\)}.

(2, 2, 2) -> X=1, Y=0

(3, 3, 3) -> X=1, Y=0

3. Show that \(P(X=1,\;Y=3) = 1/27\) by listing the outcomes that comprise the event {\(X=1,Y=3\)}.

(1, 1, 1) -> X=1, Y=3

4. Show that \(P(X=2,\;Y=0) = 6/27\) by listing the outcomes that comprise the event {\(X=2,Y=0\)}.

(2, 3, 3) -> X=2, Y=0

(2, 3, 2) -> X=2, Y=0

(2, 2, 3) -> X=2, Y=0

(3, 2, 2) -> X=2, Y=0

(3, 3, 2) -> X=2, Y=0

(3, 2, 3) -> X=2, Y=0

5. Make a table representing the marginal distribution of \(X\) and compute \(E(X)\).

x 1 2 3
p(x) 3/27 18/27 6/27

\(E(X) = (1)(3/27) + (2)(18/27) + (3)(6/27) = 2.111\)

6. Make a table representing the marginal distribution of \(Y\) and compute \(E(Y)\).

y 0 1 2 3
p(y) 8/27 12/27 6/27 1/27

\(E(Y) = (0)(8/27) + (1)(12/27) + (2)(6/27) + (3)(1/27) = 1\)

7. Find the conditional distribution of \(Y\) given \(X=x\) for each possible value of \(x\).

Y 0 1 2 3
P(Y | X = 1) 2/3 0 0 1/3
Y 0 1 2 3
P(Y | X = 2) 6/18 6/18 6/18 0
Y 0 1 2 3
P(Y | X = 3) 0 1 0 0

8. Make a table representing the distribution of \(E(Y|X)\).

X E(Y | X = x)
1 1
2 1
3 1

9. Find the conditional distribution of \(X\) given \(Y=y\) for each possible value of \(y\).

X 1 2 3
P(X | Y = 0) 2/8 6/8 0
X 1 2 3
P(X | Y = 1) 0 6/12 6/12
X 1 2 3
P(X | Y = 2) 0 1 0
X 1 2 3
P(X | Y = 3) 1 0 0

10. Make a table representing the distribution of \(E(X|Y)\).

Y E(X | Y = y)
0 1.75
1 2.5
2 2
3 1

11. Describe three methods for how you could use physical objects (e.g., cards, dice, spinners) to simulate an \((X,Y)\) pair with the joint distribution given by the table above.

a) Method 1: simulate outcomes from the probability space (i.e., prizes in the packages)

Assign equal probabilities (1/27) to all 27 outcomes, simulate by randomly selecting, compute X and Y to get an (X, Y) pair.

b) Method 2: simulate an \((X,Y)\) pair directly from the joint distribution (without simulating outcomes from the probability space)

Using the joint distribution, we have the probabilities for each (X, Y) pair, so randomly select an (X, Y) pair according to each of the pairs probabilities.

c) Method 3: simulate an \((X,Y)\) pair by first simulating \(X\) from directly from its marginal distribution (without simulating outcomes from the probability space).

First simulate X using its marginal distribution, once you have an X, use the appropriate Y given X = x distribution giving you the Y.

12. Suppose you have simulated many \((X,Y)\) pairs. Explain how you could use the simulation results to approximate each of the following. You should not do any of the calculations; rather, explain in words how you would use the simulation results and simple operations like counting and averaging.

a) \(P(X=3)\)

Number of pairs where X = 3 / number of total pairs

b) the marginal distribution of \(X\)

For each value of X, count how many times that X appears in a pair

For each X, number of pairs where X is that x / number of total pairs

c) \(E(X)\)

Take an average of X from all pairs

d) \(Var(X)\)

\(E(X^2) \approx\) average of all X’s squared

\(Var(X) = E(X^2) - [E(X)]^2\)

e) \(P(X=2,Y=1)\)

Number of pairs where X = 2 and Y =1 / total number of pairs

f) \(E(XY)\)

For all pairs, multiply X*Y, then take average

g) \(Cov(X,Y)\)

\(Cov(X,Y) = E(XY) - E(X)E(Y)\)

h) \(P(X=1|Y=0)\)

Number of pairs where X = 1 and Y = 0 / number of pairs where Y = 0

i) the conditional distribution of \(X\) given \(Y=0\)

Filter out to only look at pairs where Y = 0

For each value of X, count how many times that X appears in a pair where Y = 0

For each X, number of pairs where X is that x and Y = 0 / number of total pairs where Y = 0

j) \(E(X|Y=0)\)

Filter out to only look at pairs where Y = 0

Take an average of all X values

k) \(P(Y=0|X=1)\)

Number of pairs where Y = 0 and X =1 / number of pairs where X = 1

l) the conditional distribution of \(Y\) given \(X=1\)

Filter out to only look at pairs where X = 1

For each value of Y, count how many times that Y appears in a pair where X = 1

For each Y, number of pairs where Y is that y and X = 1 / number of total pairs where X = 1

m) \(E(Y|X=1)\)

Filter out to only look at pairs where X = 1

Take an average of all Y values