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RE-WRITTEN QUESTION: I have a set of inequalities which forms a region in the 2D space. My main goal is to test whether the area of this region is larger or less than a threshold. Since the inequalities are difficult to solve by analytical method. I sample a set of points in the 2D space, and to justify how many points fall into that region formed by inequalities (it's easier to do so). Consequently, I can get an estimate of the region area. Now to satisfy accuracy (maybe a ratio of the region area, or an absolute value), I need to mathematically calculate out how many sample points in total I have to generate. So this is the main problem.

Now there is a whole space (whose area is 1), and a target region (whose area is $\theta$). I want to know whether the area of target region is larger than $\theta_0$ or less than $\theta_0$. Now I use hit-or-loss Monte Carlo method, which samples $N$ points from the whole space, and suppose we can get $M$ of them are in the target region. By which we obtain an estimate of $\theta$, i.e., $\hat{\theta} = \frac{M}{N}$.

Obviously, $\hat{\theta}$ is an unbiased estimate of $\theta$ with variance $\frac{\theta(1-\theta)}{N}$. Suppose given a confidence level $(1-\gamma)$, the estimate error satisfies $|\theta-\hat{\theta}|\leq z_{\gamma/2} \sqrt{\frac{\theta(1-\theta)}{N}} $. Now I want to determine a required sample size with an error bound $E$.

If the error bound is absolute error, i.e., $|\theta-\hat{\theta}|\leq z_{\gamma/2} \sqrt{\frac{\theta(1-\theta)}{N}} \leq E$, I can get \begin{align} N_1 \geq \frac{(z_{\gamma/2})^2 \theta (1-\theta)}{E^2}. \end{align}

And if the error bound is relative error, i.e., we need to make sure $\frac{|\theta - \hat{\theta}|}{\theta} \leq E$, then it equals to $|\theta-\hat{\theta}|\leq \theta E$, which means \begin{align} N_2 \geq \frac{(1-\theta)}{\theta}\frac{(z_{\gamma/2})^2}{E^2}. \end{align}

MY QUESTION IS: Since $\theta$ is unknown before, can I use $\theta_0$ (the threshold I want to judge) to substitute $\theta$ to get $N_1$ and $N_2$? Is this choice reasonable enough? What is the drawback of doing so? And what is the difference between using absolute error and relative error on the result of sample size?

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It seems that you are using the standard Wald CI for an unknown binomial proportion $\theta$. This is an asymptotic CI, which is a poor choice for small $n.$ However, if you are doing a simulation with very large $n$ you should get reasonable results.

This style of CI assumes that $Z = \frac{\hat \theta - \theta}{\sqrt{\theta(1-\theta)/n}} \stackrel{aprx}{\sim} \mathsf{Norm}(0,1)$ (by the Central Limit Theorem) and then, because unknown $\theta$ is estimated by $\hat \theta = M/n$, that it is OK to use $\hat \theta$ for $\theta.$ Notice that you have two approximations: (a) assuming that $Z$ is normal and (b) assuming that $\hat \theta$ is near $\theta.$ Both assumptions are made more reasonable by large $n.$

If you want the worst-case scenario, you can estimate the margin of error using $\theta = 1/2$ because that gives the largest estimated margin of error (and so for you, the largest estimated sample size $n$ for a desired maximum margin of error).

If $\theta$ is very near $0,$ then the relative error can be very large. I have never tried getting a CI for relative error in such circumstances, but I suspect that for $\theta \approx 0$ there would be serious problems using $\hat \theta$ for $\theta.$ Notice the $\theta$ in the denominator of your last displayed equation.


Example: Simulating $\pi/4$ as the fraction of the unit circle within the square with vertices at $(0,1)$ and $(1,1)$ with R code. We simulate $B = 1,000,000$ points placed at random in the square, of which 785,161 within a unit of the origin are accepted, giving $\hat \theta = .7852$ compared with $\theta = \pi/4 = .7854.$ The absolute error is about .0002, which is well within the Wald 95% margin of error .0008.

B = 10^6;  u1 = runif(B);  u2 = runif(B)
acc = (u1^2+u2^2<1);  th.hat= mean(acc);  th.hat; pi/4;  abs(th.hat - pi/4)
## 0.785161                  # th.hat
## 0.7853982                 # th = pi/4
## 0.0002371634              # absolute error
1.96*sd(acc)/sqrt(B)      
## 0.0008049932              # approx margin of error from simulation
1.96*sqrt(th.hat*(1-th.hat)/B)
## 0.0008049928              # Wald margin of error

Addendum (responding to comments and revision): In theory, you are asking something impossible. But maybe there is a solution that will work in practice.

Take my simulation as an example. If you want to be 95% confident that $|\hat \theta - \theta| < \Delta$, then you want the 95% CI $\hat \theta \pm 1.96\sqrt{\hat\theta(1-\hat\theta)/N}$ to have $1.96\sqrt{\hat\theta(1-\hat\theta)/N} < \Delta.$ If $\Delta = 0.001$ and $\theta \approx 0.785$ then you can solve to get $N \approx 800,000$. And in my simulation with $N = B = 1,000,000$ I got a margin of error about 0.0008. As it turned out my simulated value $\hat \theta$ was within 0.0002 of $\theta_0 = \pi/4.$ However, knowing the exact value of $\pi$ is a famous mathematical quest. So if I had gotten $\hat \theta = 0.7853982,$ then I would have to know the value of $\pi/4$ to many places to know whether $\hat\theta > \pi/4$ or $\hat\theta < \pi/4.$

For planning purposes, you should pick a small $\Delta$ that seems good enough, and then find $N$ such $z_{\gamma/2}\sqrt{\theta_0(1-\theta_0)/N} < \Delta.$ So you can very likely come "within $\Delta$" of $\theta_0$ with your simulation. (For 'insurance' you might use $\theta_0 = 1/2$ to get the largest $N$ according to this method.) But from simulation and statistical analysis of simulation results, you cannot know for sure whether your simulated $\hat \theta$ is a little above or a little below the true $\theta_0.$ (Otherwise, some clever person could have used simulation to find $\pi$ to infinitely many digits.)

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  • $\begingroup$ Thanks for your answer. However, I have two questions. The first is, what I want to test is whether $\hat{\theta}>\theta_0$ or $\hat{\theta}<\theta_0$, and I do not want to use the upper bound (i.e., using 1/2 to substitute $\theta$), and prefer to use $\theta_0$ (not $\hat{\theta}$) to calculate the sample size which is a little less. I wonder is there any mathematical guarantee to use it. The second is, in your simulation, the number of samples does not satisfies $N_1$ in the equation. And I cannot know $\pi/4$ in advance in practice, because I am estimating an unknown region. $\endgroup$ – WYC May 11 '17 at 21:53
  • $\begingroup$ Where does $\theta_0$ come from? (Strange you mentioned only margins of error for CIs in your question if your purpose is testing.) In the usual statistical sense of 'testing', it does not make sense to 'test' whether $\hat \theta > \theta_0.$ For large $n,$ the test statistic for $H_0: \theta = \theta_0$ against $H_a: \theta > \theta_0$ is $Z = \frac{\hat \theta - \theta_0}{\sqrt{\theta_)(1-\theta_0)/n}}.$ If you really are testing, then a 'power and sample size' procedure will give the required $n$ to achieve a given power (prob of rejecting $H_0$ when it is false in a specified way). $\endgroup$ – BruceET May 11 '17 at 22:06
  • $\begingroup$ My simulation example was intended to illustrate the the Wald margin of error works as intended for the particular case simulated. $\endgroup$ – BruceET May 11 '17 at 22:11
  • $\begingroup$ I am not good at statistics. The main purpose is to test whether $\theta$ is larger or less than a threshold, say $\theta_0$. However, I do not know $\theta$ in advance, thus my think is to estimate $\theta$ by $\hat{\theta}$, and then judge according to $\hat{\theta}$. I am not sure whether this is a typical hypothesis testing problem. If it is a testing problem, how can I get a required sample size? $\endgroup$ – WYC May 11 '17 at 22:11
  • $\begingroup$ It might help if you give an example of what you are simulating and how, explain what you mean by 'threshold' $\theta_0$ and give a value for it. Try plain English, not 'pseudo' stat-speak. Then maybe one of us can help. I'm on my way to teach a class. Will check back later. $\endgroup$ – BruceET May 11 '17 at 22:17

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