Back

Solution: 2018 Winter Final - 21

Author: Michiel Smid

Question

You are given two independent random variables $X$ and $Y$, where
$\Pr(X = 0)$ $= \Pr(X = 1) = \Pr(Y = 0)$ $= \Pr(Y = 1) = 1/2.$
Define the random variable $Z = X \cdot Y$. Which of the following is correct?
(a)
The random variables $X$ and $Z$ are independent.
(b)
The random variables $X$ and $Z$ are not independent.
(c)
None of the above.
(d)
All of the above.

Solution

For $X$ and $Z$ to be independent, $Pr(X = x \cap Z = z) = Pr(X = x) \cdot Pr(Z = z)$, for all real numbers $x$ and $z$, by definition of independent random variables.

If we can find a counterexample, aka where $Pr(X = x \cap Z = z) \neq Pr(X = x) \cdot Pr(Z = z)$, then we know the two are not independent.

Let $z = 0$, $x = 0$ and $y = 0$:

  • $Pr(X = 0) = \frac{1}{2}$, from expression in question
  • $Pr(Y = 0) = \frac{1}{2}$, from expression in question
  • $Pr(Z = 0) = Pr(X \cdot Y = 0) = Pr(X = 0 \cap Y = 0) + Pr(X = 0 \cap Y = 1) + Pr(X = 1 \cap Y = 0) = ( \frac{1}{2} \cdot \frac{1}{2}) + ( \frac{1}{2} \cdot \frac{1}{2}) + ( \frac{1}{2} \cdot \frac{1}{2}) = \frac{3}{4}$
  • $Pr(X = 0 \cap Z = 0) = Pr(X = 0 \cap Y = 0) + Pr(X = 0 \cap Y = 1) = \frac{1}{2}$

Now we check if LHS = RHS for the expression:

  • $Pr(X = x \cap Z = z) = Pr(X = x) \cdot Pr(Z = z)$
  • $Pr(X = 0 \cap Z = 0) = Pr(X = 0) \cdot Pr(Z = 0)$
  • $\frac{1}{2} = \frac{1}{2} \cdot \frac{3}{4}$
  • $\frac{1}{2} \neq \frac{3}{8}$

Since LHS is not equal to RHS, then the events $X$ and $Z$ are not independent.