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.
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 $$
Z=
\begin{cases}
1 & \text { if } X=Y, \\
0 & \text { if } X \neq Y\\
\end{cases}
$$
\( P(XYZ = 0) \iff P( { X = 0} \cup {Y = 0} \cup {Z = 0}) \)
$$= P(X = 0) + P(Y = 0) + P(Z= 0) - P({ X = 0} \cap {Y = 0}) - P({Y = 0} \cap {Z= 0}) - P({X = 0} \cap {Z= 0}) + P({X = 0} \cap {Y = 0} \cap {Z= 0}). $$
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,
\( P(XYZ = 0) = \frac{3}{4} + P({X = 0},{Y = 0},{Z= 0}) \geq \frac{3}{4}\).
\( P(XYZ = 0) = \frac{3}{4} \iff P({X = 0},{Y = 0},{Z= 0}) = 0 \).
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.
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 $$
Z=
\begin{cases}
1 & \text { if } X=Y, \\
0 & \text { if } X \neq Y\\
\end{cases}
$$
\( P(XYZ = 0) \iff P( { X = 0} \cup {Y = 0} \cup {Z = 0}) \)
$$= P(X = 0) + P(Y = 0) + P(Z= 0) - P({ X = 0} \cap {Y = 0}) - P({Y = 0} \cap {Z= 0}) - P({X = 0} \cap {Z= 0}) + P({X = 0} \cap {Y = 0} \cap {Z= 0}). $$
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,
\( P(XYZ = 0) = \frac{3}{4} + P({X = 0},{Y = 0},{Z= 0}) \geq \frac{3}{4}\).
\( P(XYZ = 0) = \frac{3}{4} \iff P({X = 0},{Y = 0},{Z= 0}) = 0 \).
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, $$
Z=
\begin{cases}
1 & \text { if } X=Y, \\
0 & \text { if } X \neq Y\\
\end{cases}
$$.