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

## Solution

#### Draw the Venn Diagram

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

## Solution

#### Draw the Venn Diagram

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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