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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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