{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Variance than the average of two independent random

This preview shows pages 140–145. Sign up to view the full content.

variance than the average of two independent random variable distributed according to N. Firstly, if 1 2 U U , N = 2 with probability !/2 and other is N = 2 + N a with probability ½. So, [ ] [ ] a N 2E 2 N E + = [ ] ( ) [ ] [ ] [ ] 2 a a 2 2 N E 1/2 N E 2 4 N 2 E 1/2 (4) 1/2 N E + + = + + = a [ ] [ ] ( ) [ ] [ ] ( ) 0.4997 8 - 4e - 14e N E - N E N Var Thus, 2e - 8 N E and 4 - 2e N E 2 2 a 2 a a 2 a a = = = = Now consider the random variable N and M, ( ) ( ) ( ) a a N Var N 4 Var M N Var = + = +

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
So, ( ) ( ) 3.065 0.4997 1.5316 M N Var N N Var + + Therefore, the use of antithetic variable reduces the variance of the estimator by a factor of slightly more than 3. 21. Suppose we were interested in using simulation to compute θ =E [e u ]. Here, a natural variant to use as a control is the random number U. Discuss what sort of improvement over the raw estimator is possible. (20 marks) Solution θ = E[e U ] By using raw estimator, ( ) [ ] [ ] ( ) ( ) 0.2420 1 - e - dx e x e E - U E e Var 1 0 2 x 2 U 2U U = = = By using antithetic variable U, ( ) [ ] [ ] [ ] ( ) ( ) 0.14086 2 1 - e - 1 2 1 - e - dx e x e E U E - e U E U , e Cov 1 0 x U U U = = = = And then, ( ) [ ] [ ] ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( ) ( ) 0.0039 0.2380 - 0.2420 0.14086 12 - e Var 1/2 - U c e Var Y Var Y X, Cov - X Var μ - Y c X Var have we 1/12 U E - U E U Var 2 U * U 2 y * 2 2 = = = + = + = = Therefore, the use of control variate U can lead to a variance reduction of up to 98.4 percent.
Computer Integrated Manufacturing IT 4036 for Second Semester Sample Question

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document