Intertwined Conditional Probability | ISI MStat 2016 PSB Problem 4

This is an interesting problem from intertwined conditional probability and Bernoulli random variable mixture, which gives a sweet and sour taste to Problem 4 of ISI MStat 2016 PSB.

Problem

Let $X, Y,$ and $Z$ be three Bernoulli $\left(\frac{1}{2}\right)$ random variables such that $X$ and $Y$ are independent, $Y$ and $Z$ are independent, and $Z$ and $X$ are independent.
(a) Show that $\mathrm{P}(X Y Z=0) \geq \frac{3}{4}$.
(b) Show that if equality holds in (a), then $$
Z=
\begin{cases}
1 & \text { if } X=Y, \\
0 & \text { if } X \neq Y\\
\end{cases}
$$

Prerequisites

Principle of Inclusion and Exclusion $|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|$
Basic Probability Theory
Conditional Probability
$abc = 0$ iff $ a= 0$ or $ b= 0$ or $c = 0$.
$ \cup$ = or; $ \cap$ = and

Solution

(a)

$ P(XYZ = 0) \iff P( { X = 0} \cup {Y = 0} \cup {Z = 0}) $

$$= P(X = 0) + P(Y = 0) + P(Z= 0) - P({ X = 0} \cap {Y = 0}) - P({Y = 0} \cap {Z= 0}) - P({X = 0} \cap {Z= 0}) + P({X = 0} \cap {Y = 0} \cap {Z= 0}). $$

We use the fact that $X$ and $Y$ are independent, $Y$ and $Z$ are independent, and $Z$ and $X$ are independent.

$$= P(X = 0) + P(Y = 0) + P(Z= 0) - P({ X = 0})P({Y = 0}) - P({Y = 0})P({Z= 0}) - P({X = 0})P({Z= 0}) + P({X = 0},{Y = 0},{Z= 0})$$.

$X, Y,$ and $Z$ be three Bernoulli $\left(\frac{1}{2}\right)$ random variables. Hence,

$ P(XYZ = 0) = \frac{3}{4} + P({X = 0},{Y = 0},{Z= 0}) \geq \frac{3}{4}$.

(b)

$ P(XYZ = 0) = \frac{3}{4} \iff P({X = 0},{Y = 0},{Z= 0}) = 0 $.

Now, this is just a logical game with conditional probability.

$ P({X = 0} |{Y = 0},{Z= 0}) = 0 \Rightarrow P({Z= 0} |{Y = 0},{X = 1}) = 1$.

$ P({Y = 0} |{X = 0},{Z= 0}) = 0 \Rightarrow P({Z= 0} |{X = 0},{Y = 1}) = 1$.

$ P({Z = 0} |{X = 0},{Y= 0}) = 0 \Rightarrow P({Z = 1} |{X = 0},{Y= 0}) = 1$.

$ P( Z = 0) = P({X = 1},{Y = 0},{Z= 0}) + P({X = 0},{Y = 1},{Z= 0}) + P({X = 1},{Y = 1},{Z= 0}) + P({X = 0},{Y = 0},{Z= 0})$

$ = \frac{1}{4} + \frac{1}{4} + P({X = 1},{Y = 1},{Z= 0}) $.

Now, $Z$ is a Bernoulli $\left(\frac{1}{2}\right)$ random variable. So, $P(Z = 0) =\frac{1}{2}$ $ \Rightarrow P({X = 1},{Y = 1},{Z= 0}) = 0 \Rightarrow P({Z = 0} | {Y = 1},{X= 1}) = 0 $.

$ P({Z= 0} |{Y = 0},{X = 1}) = 1$.

$P({Z= 0} |{X = 0},{Y = 1}) = 1$.

$P({Z = 1} |{X = 0},{Y= 0}) = 1$.

$ P({Z = 1} | {Y = 1},{X= 1}) = 1$.

Hence, $$
Z=
\begin{cases}
1 & \text { if } X=Y, \\
0 & \text { if } X \neq Y\\
\end{cases}
$$.

Problem

Let $X, Y,$ and $Z$ be three Bernoulli $\left(\frac{1}{2}\right)$ random variables such that $X$ and $Y$ are independent, $Y$ and $Z$ are independent, and $Z$ and $X$ are independent.
(a) Show that $\mathrm{P}(X Y Z=0) \geq \frac{3}{4}$.
(b) Show that if equality holds in (a), then $$
Z=
\begin{cases}
1 & \text { if } X=Y, \\
0 & \text { if } X \neq Y\\
\end{cases}
$$

Prerequisites

Principle of Inclusion and Exclusion $|A \cup B \cup C|=|A|+|B|+|C|-|A \cap B|-|A \cap C|-|B \cap C|+|A \cap B \cap C|$
Basic Probability Theory
Conditional Probability
$abc = 0$ iff $ a= 0$ or $ b= 0$ or $c = 0$.
$ \cup$ = or; $ \cap$ = and