Question

Question: Let\(A\left( X \right) = E\left( {\left| {X - E\left( X \right)} \right|} \right)\), the expected value of the absolute value of the deviation of\(X\), where\(X\)is a random variable. Prove or disprove that\(A\left( {X + Y} \right) = A\left( X \right) + A\left( Y \right)\) for all random variables\(X\)and\(Y\).

Short answer

Answer

For all random variables \(X\) and \(Y\), \(A\left( {X + Y} \right) \ne A\left( X \right) + A\left( Y \right)\).

Step-by-step solution

Given information

Let \(A\left( X \right) = E\left( {\left| {X - E\left( X \right)} \right|} \right)\), the expected value of the absolute value of the deviation of \(X\), where \(X\) is a random variable.

Definition and formula used

A random variable is a mathematical formalization of a quantity or object which depends on random events.

Proof

This is certainly not true if\(X\)and\(Y\)are not independent.

For example, if \(Y = - X\),

then\(X + Y = 0\), so it has expected absolute deviation\(0\), whereas\(X\)and\(Y\)can have non zero expected absolute deviation.

For a more concrete example, let\(X\)be the number of heads in one flip of a fair coin, and let\(Y\)be the number of tails for that same flip.

Each takes values\(0\)and\(1\)with probability\(0.5\).

Then,\(E\left( X \right) = 0.5\),

\(\left| {x - E\left( X \right)} \right| = 0.5\)for each outcome\(x\),

And therefore,\(A\left( X \right) = 0.5\).

Similarly,\(A\left( Y \right) = 0.5\).

So,\(A\left( X \right) + A\left( Y \right) = 1\)

On the other hand,\(X + Y\)has constant value\(1\), so\(\left| {\left( {x + y} \right) - E\left( {X + Y} \right)} \right| = 0\)for each pair of outcomes\(\left( {x,y} \right)\), and\(A\left( {X + Y} \right) = 0\).

Thus,\(A\left( {X + Y} \right) \ne A\left( X \right) + A\left( Y \right)\)for all random variables\(X\)and\(Y\).