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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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