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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

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

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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