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

Venny Venny AMy GMy | ISI MStat 2016 PSB Problem 3

This problem is a very basic and cute application of set theory, Venn diagram and AM GM inequality to solve the ISI MStat 2016 PSB Problem 3.

Problem - Venn diagram and AM GM inequality

For any two events \(A\) and \(B\), show that
$$
(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}
$$

Prerequisites

Solution

Draw the Venn Diagram

venn diagram and am gm inequality problem

P(region Red) = \(Y\)

P(region Blue) = \(Z\)

P(region Grey) = \(W\)

P(region Brown) = \(X\)

Observe that \( W + X + Y + Z = 1\). \( W, X, Y, Z \geq 0\).

Now, Calculate Given Probability of Sets in terms of \( W, X, Y, Z \).

\({P}(A \cap B) = Z\).

\({P}\left(A \cap B^{c}\right) = Y\).

\({P}\left(A^{c} \cap B\right) = W\).

\( {P}\left(A^{c} \cap B^{c}\right) = X\).

The Final Inequality

\( W, X, Y, Z \geq 0\).

\( W + X + Y + Z = 1\).

Observe that \( 3(W^2 + X^2 + Y^2 + Z^2) = (W^2+X^2) + (W^2+Y^2) + (W^2+Z^2) + (X^2+Y^2) + (X^2+Z^2) + (Y^2+Z^2)\).

\( 3(W^2 + X^2 + Y^2 + Z^2) \geq 2WX + 2WY + 2WZ + 2XY + 2XZ + 2YZ \) by AM - GM Inequality.

\( \Rightarrow 4(W^2 + X^2 + Y^2 + Z^2) \geq (W + X + Y + Z)^2 = 1\).

\( \Rightarrow (W^2 + X^2 + Y^2 + Z^2) \geq \frac{1}{4} \).

Hence,

$$
(\mathrm{P}(A \cap B))^{2}+\left(\mathrm{P}\left(A \cap B^{c}\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B\right)\right)^{2}+\left(\mathrm{P}\left(A^{c} \cap B^{c}\right)\right)^{2} \geq \frac{1}{4}
$$