Show that if X and Y have the same distribution then Var((X + Y)/2) ≤ Var(X) Hence,

Question

Show that if X and Y have the same distribution then

Var((X + Y)/2) ≤ 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).

Accepted Answer

[latex]\begin{aligned}\operatorname{Var}[(X+Y) / 2]&=\frac{1}{4}[\operatorname{Var}(X)+\operatorname{Var}(Y)+2 \operatorname{Cov}(X, Y)]\&=\frac{\operatorname{Var}(X)+\operatorname{Cov}(X, Y)}{2}\end{aligned}[/latex]

Now it is always true that

[latex]\frac{\operatorname{Cov}(V, W)}{\sqrt{\operatorname{Var}(V) \operatorname{Var}(W)}} \leqslant 1[/latex]

and so when X and Y have the same distribution Cov(X,Y) ≤ Var(X).

Question 11.EX.32: Show that if X and Y have the same distribution then Var((X ......

This Question has been Answered.

Already have an account?

Login

Related Answered Questions

For a nonhomogeneous Poisson process with intensity function λ(t), t ≥ 0, where ∫0^∞ ...

Verified Answer:

Suppose n balls having weights w1, w2, . . . , wn are in an urn. These balls are ...

Verified Answer:

In Example 11.4 we simulated the absolute value of a standard normal by using the Von ...

Verified Answer:

Give a method for simulating a hypergeometric random variable. ...

Verified Answer:

Suppose it is relatively easy to simulate from the distributions Fi, i = 1, 2, . . . , n. ...

Verified Answer: