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Supposing $g(x)=\sqrt[3]{x}$, I want to calculate the expected value of g, $E(\sqrt[3]{x})$, using Monte Carlo method, by generating $x_i$ from a Weibull distribution with parameters $(1,5)$.

After that, I want to use the control variates method and the antithetic method in order to to reduce the variance of my estimator, which I found with the simple Monte Carlo. And here is my problem, I do not know how to do these methods.

I would appreciate if someone could help me do that or give any tip/help.Thank you very much for your concern, in advance.

What I have done so far

Supposing $S$ is our estimator, then we know that $S=(\sum \limits_{i=1}^{N} g(x_i))/N$.

Using Matlab, I found the expected value $S$ by generating 1000 random numbers from the weibull(1,5) distribution and calculate the sum. Here is my algorithm:

N=1000
sum=0;
for i=1:N;
  X = wblrnd(1,5);  
  res(i)=X.^(1/3); 
  sum=sum+res(i);
end
S=sum/N
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    $\begingroup$ Why are you being asked to do homework without having gone over what you need to know in order to do it in the class? Or is this something extra you're doing? $\endgroup$ – jbowman Dec 10 '11 at 15:23
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    $\begingroup$ @jbowman the part i was given is only the simple monte carlo, but i read that it's not the best method to estimate the expected value. For myself, i want to do the other two methods, in order to reduce the variance of my estimator as much as i can. $\endgroup$ – johan Dec 10 '11 at 15:27
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There is no one way to implement either control or antithetic variates, however, a couple of examples may help.

Antithetic variables: Imagine instead of the random number generator you actually used, you generated a $U(0,1)$ variate, call it $u$, and ran it through the inverse CDF of the Weibull(1,5) distribution, thereby generating a Weibull(1,5) variate. For the next random number, use $1-u$ instead of generating a new $u$. For subsequent random numbers, alternate generating a new $u$ and using $1-u$. This helps to "balance" high and low values from your random number stream, thus reducing variability of your final estimates.

Control variates: These are "extra" variables that are correlated with the result, enabling you to do something like run a regression on your results against the control variates to get a more accurate estimate. In your case, for example, you know the true mean of the Weibull dist'n (5), so you could use the $x_i$ as a control variate. You would calculate the improved estimate:

$S^* = S - \frac{\widehat{cov}(x,res)}{\widehat{var}(x)} * (\bar{x} - 5)$

where the covariance and variance terms are estimated from the data. This helps correct the estimate for random number streams that are not, in some relevant way, totally representative of the underlying distribution.

Both methods, esp. control variates, are more general than these two examples might lead you to believe. The wikipedia links are at best rough introductions; plenty of books and papers covering both techniques are out there if you want to go more into depth.

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  • $\begingroup$ Pretty much i found the same things in different pages, but i would be grateful if you could be more specific (or anyone who knows) concerning my exercise. $\endgroup$ – johan Dec 10 '11 at 18:48
  • $\begingroup$ @Kostas - I'm not sure what more you need; I've tried to put the explanation in terms of your problem, but I can't tell without more detail from you where the weak spots are. $\endgroup$ – jbowman Dec 10 '11 at 23:16
  • $\begingroup$ nevermind your help was quite helpful, I would like to thank you again for your concern. $\endgroup$ – johan Dec 12 '11 at 19:25
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As correctly explained by jbowman, you have to create a negative correlation between your simulations to implement antithetic variables: with a generic wblrndfunction this is not possible. You thus have to get back to the definition of a Weibull $\mathcal{W}(\lambda,k)$, which is a scale transform of the $k$th power of an exponential variate, i.e. $$ X\sim \mathcal{W}(\lambda,k) $$ is equivalent to $$ (X/\lambda)^k \sim \mathcal{E}(1) $$ This can be reworded in terms of simulation as $$ X=\lambda(-\log U)^{1/k}\,,\qquad U\sim\mathcal{U}(0,1)\,. $$ Therefore, you can implement the antithetic method by using a sample of uniforms, $U_1,\ldots,U_n$ and its complent $1-U_1,\ldots,1-U_n$ and compare the variance of the estimator of $\mathbb{E}[X^{1/3}]$ with the corresponding estimator based on $U_1,\ldots,U_{2n}$. To exhibit the improvement, you have to run a Monte Carlo experiment repeating the computation of those variances on many samples of size $n$. (You cannot see the impact of antithetic simulation on a single run.)

The control variate is implemented in your case by choosing a known moment of the Weibull, for instance as suggested by jbowman, $$ \mathbb{E}[X] = \lambda \Gamma(1+1/k) $$ and using $X$ as the control variate. This means you compute the average of the simulated $X_i$'s along the average of the $X_i^{1/3}$ and the empirical covariance between the $X_i$'s and the $X_i^{1/3}$ as well as the empirical variance of the $X_i$'s to use jbowman formula. Again, checking the improvement brought by the control variate requires a Monte Carlo experiment with several runs.

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  • $\begingroup$ Xi'an I would like to thank you too, very much, for your help. $\endgroup$ – johan Dec 12 '11 at 19:25

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