Solution: 2017 Fall Final - 21

Author: Michiel Smid

Question

You are given a fair red die and a fair blue die. You roll each die once, independently of each other. Let $(i,j)$ be the outcome, where $i$ is the result of the red die and $j$ is the result of the blue die. Define the random variables $$ X = i + j $$ and $$ Y = i - j. $$ Which of the following is true?

(a)

The random variables $X$ and $Y$ are not independent.

(b)

None of the above.

(c)

The random variables $X$ and $Y$ are independent.

Solution

Counterexample to Show $ X $ and $ Y $ Are Not Independent

We want to demonstrate mathematically that the random variables:

$ X = i + j $ (sum of two six-sided dice rolls)
$ Y = i - j $ (difference between two six-sided dice rolls)

are not independent by showing an example where the independence condition $P(X = x \land Y = y) = P(X = x) \cdot P(Y = y)$ fails for certain values of $ x $ and $ y $.

Choose Specific Values

Let:

$ X = 4 $ (sum of the two dice is 4)
$ Y = 2 $ (difference between the two dice is 2)

We will compute:

$ P(X = 4 \land Y = 2) $
$ P(X = 4) $
$ P(Y = 2) $
Check if the independence formula holds.

Joint Probability $ P(X = 4 \land Y = 2) $

For $ X = 4 $ and $ Y = 2 $:

The sum $ i + j = 4 $
The difference $ i - j = 2 $

Possible pairs $(i, j)$ that satisfy both conditions: $ (3,1) $

There is 1 valid outcome out of $ 36 $ total outcomes, so $P(X = 4 \land Y = 2) = \frac{1}{36}$

Marginal Probability $ P(X = 4) $

Possible pairs where the sum is 4: $ (1,3), (2,2), (3,1) $

There are 3 valid outcomes out of $36$ total outcomes, so $P(X = 4) = \frac{3}{36} = \frac{1}{12}$

Marginal Probability $ P(Y = 2) $

Possible pairs where the difference is 2: $ (3,1), (4,2), (5,3), (6,4) $

There are 4 outcomes out of $36$ total outcomes, so $P(Y = 2) = \frac{4}{36} = \frac{1}{9}$

Check Independence Condition

If $ X $ and $ Y $ were independent, we would expect: $P(X = 4 \land Y = 2) = P(X = 4) \cdot P(Y = 2)$

Substituting the probabilities: $\frac{1}{36} \stackrel{?}{=} \frac{1}{12} \cdot \frac{1}{9}$

Simplify the right-hand side: $\frac{1}{12} \cdot \frac{1}{9} = \frac{1}{108}$

Comparing the two probabilities: $\frac{1}{36} \neq \frac{1}{108}$

Clearly, the probabilities do not match, which means that the random variables $ X $ and $ Y $ are not independent.