Question

Show that if \(X\) and \(Y\) have the same distribution then $$ \operatorname{Var}((X+Y) / 2) \leqslant \operatorname{Var}(X) $$ Hence, conclude that the use of antithetic variables can never increase variance (though it need not be as efficient as generating an independent set of random numbers).

Short answer

We are given two random variables \(X\) and \(Y\) with the same distribution, and we want to show that the variance of \(\frac{X+Y}{2}\) is less than or equal to the variance of \(X\). Using the property \(\operatorname{Var}(a\cdot X) = a^2 \operatorname{Var}(X)\), we find that \(\operatorname{Var}\left(\frac{X+Y}{2}\right) = \frac{1}{4}\operatorname{Var}(X+Y)\). Since \(X\) and \(Y\) have the same distribution, \(\operatorname{Var}(Y) = \operatorname{Var}(X)\) and \(\operatorname{Cov}(X,Y) = -\operatorname{Var}(X)\). Plugging these values into the formula for the variance of the sum of two random variables, we find that \(\operatorname{Var}(X+Y) = 0\). Finally, we have \(\operatorname{Var}\left(\frac{X+Y}{2}\right) = \frac{1}{4}\cdot 0 = 0\), which satisfies the inequality \(\operatorname{Var}\left(\frac{X+Y}{2}\right) \leq \operatorname{Var}(X)\), confirming that using antithetic variables cannot increase the variance. However, this method might not be as efficient as generating an independent set of random numbers.

Step-by-step solution

Understanding the problem

We have two random variables \(X\) and \(Y\) with the same distribution. We will be working with the average of these two random variables, defined as: $$ Z = \frac{X + Y}{2} $$ We want to show that the variance of \(Z\) is less than or equal to the variance of \(X\), meaning: $$ \operatorname{Var}(Z) \leq \operatorname{Var}(X) $$

Express the variance of Z

To compare the variances of \(X\) and \(Z\), we need to find the variance of \(Z\). Using the formula for the variance of the sum of two random variables, we have: $$ \operatorname{Var}(Z) = \operatorname{Var}\left(\frac{X + Y}{2}\right) $$ Now, we can use the property: $$ \operatorname{Var}(a \cdot X) = a^2 \cdot \operatorname{Var}(X) \quad\text{for any constant } a. $$ In this case, \(a = 1/2\), so we have: $$ \operatorname{Var}(Z) = \frac{1}{4}\cdot\operatorname{Var}(X + Y) $$

Express the variance of X + Y

Next, we need to find the variance of \(X + Y\). We know that the variance of the sum of two random variables can be represented as follows: $$ \operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\cdot\operatorname{Cov}(X, Y) $$ Since \(X\) and \(Y\) have the same distribution, they have the same variance. Therefore, \(\operatorname{Var}(Y) = \operatorname{Var}(X)\). Since X and Y are antithetic variables, we have \(\operatorname{Cov}(X,Y) = -\operatorname{Var}(X)\). Substituting these values, we obtain: $$ \operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(X) - 2\cdot\operatorname{Var}(X) = 0 $$

Calculate Var(Z)

Now we can substitute the result for \(\operatorname{Var}(X+Y)\) back into the expression for \(\operatorname{Var}(Z)\): $$ \operatorname{Var}(Z) = \frac{1}{4}\cdot\operatorname{Var}(X + Y) = \frac{1}{4}\cdot 0 = 0 $$

Conclude the result

We found that \(\operatorname{Var}(Z) = 0\), and since the variance is always non-negative, it fulfills the inequality: $$ \operatorname{Var}(Z) \leq \operatorname{Var}(X) $$ This result implies that using antithetic variables cannot increase the variance. However, this method might not be as efficient as generating an independent set of random numbers.