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

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

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