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Chapter 11: Problem 32

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

Expert verified
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

01

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: $$ \operatorname{Var}(Z) \leq \operatorname{Var}(X) $$
02

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: $$ \operatorname{Var}(Z) = \frac{1}{4}\cdot\operatorname{Var}(X + Y) $$
03

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: $$ \operatorname{Var}(X + Y) = \operatorname{Var}(X) + \operatorname{Var}(X) - 2\cdot\operatorname{Var}(X) = 0 $$
04

Calculate Var(Z)

Now we can substitute the result for \(\operatorname{Var}(X+Y)\) back into the expression for \(\operatorname{Var}(Z)\): $$ \operatorname{Var}(Z) = \frac{1}{4}\cdot\operatorname{Var}(X + Y) = \frac{1}{4}\cdot 0 = 0 $$
05

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.

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Most popular questions from this chapter

The Discrete Hazard Rate Method: Let \(X\) denote a nonnegative integer valued random variable. The function \(\lambda(n)=P\\{X=n \mid X \geqslant n\\}, n \geqslant 0\), is called the discrete hazard rate function. (a) Show that \(P\\{X=n\\}=\lambda(n) \prod_{i=0}^{n-1}(1-\lambda(i))\). (b) Show that we can simulate \(X\) by generating random numbers \(U_{1}, U_{2}, \ldots\) stopping at $$ X=\min \left\\{n: U_{n} \leqslant \lambda(n)\right\\} $$ (c) Apply this method to simulating a geometric random variable. Explain, intuitively, why it works. (d) Suppose that \(\lambda(n) \leqslant p<1\) for all \(n\). Consider the following algorithm for simulating \(X\) and explain why it works: Simulate \(X_{i}, U_{i}, i \geqslant 1\) where \(X_{i}\) is geometric with mean \(1 / p\) and \(U_{i}\) is a random number. Set \(S_{k}=X_{1}+\cdots+X_{k}\) and let $$ X=\min \left\\{S_{k}: U_{k} \leqslant \lambda\left(S_{k}\right) / p\right\\} $$

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In Example \(11.4\) we simulated the absolute value of a standard normal by using the Von Neumann rejection procedure on exponential random variables with rate \(1 .\) This raises the question of whether we could obtain a more efficient algorithm by using a different exponential density-that is, we could use the density \(g(x)=\lambda e^{-\lambda x}\). Show that the mean number of iterations needed in the rejection scheme is minimized when \(\lambda=1\).

Consider a queueing system in which each service time, independent of the past, has mean \(\mu .\) Let \(W_{n}\) and \(D_{n}\) denote, respectively, the amounts of time customer \(n\) spends in the system and in queue. Hence, \(D_{n}=W_{n}-S_{n}\) where \(S_{n}\) is the service time of customer \(n\). Therefore, $$ E\left[D_{n}\right]=E\left[W_{n}\right]-\mu $$ If we use simulation to estimate \(E\left[D_{n}\right]\), should we (a) use the simulated data to determine \(D_{n}\), which is then used as an estimate of \(E\left[D_{n}\right] ;\) or (b) use the simulated data to determine \(W_{n}\) and then use this quantity minus \(\mu\) as an estimate of \(E\left[D_{n}\right]\) ? Repeat for when we want to estimate \(E\left[W_{n}\right]\).

If \(f\) is the density function of a normal random variable with mean \(\mu\) and variance \(\sigma^{2}\), show that the tilted density \(f_{t}\) is the density of a normal random variable with mean \(\mu+\sigma^{2} t\) and variance \(\sigma^{2}\).
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