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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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