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

Question

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) = λe^−λx . Show that the mean number of iterations needed in the rejection scheme is minimized when λ = 1.

Accepted Answer

Let

[latex]\begin{aligned}& c(\lambda)=\max _{x}\left\{\frac{f(x)}{\lambda e^{-\lambda x}} ight\}=\frac{2}{\lambda \sqrt{2 \pi}} \max _{x}\left[\exp \left\{\frac{-x^{2}}{2}+\lambda x ight\} ight] \&\qquad =\frac{2}{\lambda \sqrt{2 \pi}} \exp \left\{\frac{\lambda^{2}}{2} ight\}\Hence,\&\frac{d}{d \lambda} c(\lambda)=\sqrt{2 / \pi} \exp \left\{\frac{\lambda^{2}}{2} ight\}\left[1-\frac{1}{\lambda^{2}} ight]\end{aligned}[/latex]

Hence (d/dλ)c(λ) = 0 when λ = 1 and it is easy to check that this yields the minimal value of c(λ).

Question 11.EX.6: In Example 11.4 we simulated the absolute value of a standar......

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