Solution: 2016 Fall Final - 23

Author: Michiel Smid

Question

Consider the set $S = \{1,2,3,...,10\}$. You choose a uniformly random element $z$ in $S$. Define the random variables $$ X = \begin{cases} 0\; \ \text{if $z$ is even}, \\ 1\; \ \text{if $z$ is odd} \end{cases} $$ and $$ Y = \begin{cases} 0\; \ \mathrm{if}\ z \in \{1,2\}, \\ 1\; \ \mathrm{if}\ z \in \{3,4,5,6\}, \\ 2\; \ \mathrm{if}\ z \in \{7,8,9,10\}. \end{cases} $$ Which of the following is true?

(a)

None of the above.

(b)

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

(c)

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

Solution

For questions like these, we can answer by checking for a value $i$ if the following is true: $ Pr(X=i \cap Y=i) = Pr(X=i) \cdot Pr(Y=i) $

Let's determine $ Pr(X=1) $
Since half the numbers are odd,
$ Pr(X=1) = \frac{5}{10} $
$ Pr(X=1) = \frac{1}{2} $
Let's determine $ Pr(Y=1) $
Since 4 of the numbers are in the set {3, 4, 5, 6},
$ Pr(Y=1) = \frac{4}{10} $
$ Pr(Y=1) = \frac{2}{5} $
Let's determine $ Pr(X=1 \cap Y=1) $
The values 3 and 5 are the only values that are both odd and in the set {3, 4, 5, 6}
$ Pr(X=1 \cap Y=1) = \frac{2}{10} $
$ Pr(X=1 \cap Y=1) = \frac{1}{5} $

$ Pr(X=1 \cap Y=1) = Pr(X=1) \cdot Pr(Y=1) $

$ \frac{1}{5} = \frac{1}{2} \cdot \frac{2}{5} $

$ \frac{1}{5} = \frac{1}{5} $

Since the equation is true, X and Y are independent.