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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,

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

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,

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

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