我需要多少次蒙地卡羅執行來為通話定價？

我必須使用 Monte Carlo 對幾個電話進行定價。顯然，在執行次數和看漲期權的公平價格之間存在巨大的權衡。我知道我可以檢查近似值如何改變我減少了模擬的數量，但我想知道是否有一個大概值。

目前我正在執行 1000 次模擬。但也許500就足夠了？

它真的，真的，真的取決於你的參數，即 $ r $ , $ \sigma $ , $ K $ , $ T $ , $ S_0 $ . 例如，以下是實施我在此處回答中解釋的停止標準的一些結果。這些是迭代次數，以使 MC 呼叫價格與確切呼叫價格相差小於 0.01（例如一美分）的機率約為 0.95：

$ r = 0.02 $ , $ \sigma = 0.2 $ , $ K = 10 $ , $ T = 1 $ , $ S_0 = 10 $ : 72,382 次迭代

$ r = 0.02 $ , $ \sigma = 0.2 $ , $ K = 10 $ , $ T = 10 $ , $ S_0 = 10 $ : 1,391,379 次迭代

$ r = 0.02 $ , $ \sigma = 0.5 $ , $ K = 10 $ , $ T = 1 $ , $ S_0 = 10 $ : 625,365 次迭代

這是我的停止標準的 MATLAB 實現，如果您想自己嘗試：

%%% Estimates the price of a European call option using MC.
% Continues MC estimation until we reach a 0.05 probability that
% MC call price differs by more than "tol" units from exact 
% call price.

function C = CallPricePointEstimate()
%%% BS parameters  - Try varying these to see how number of iters (n)
% changes %%%
K = 10;
S = 10;
r = 0.02;
T = 1;
sigma = 0.5;

%%% MC parameters %%%
% number of stock prices to generate, may need more, which are
% generated in the MC loop below if idx > N_MC
N_MC = 1000000;
tol = 0.01;  % price within $tol
bound = 5;  % something larger than tol
n = 1;  % current iteration number

%%% generate normal RVs and price paths %%%
Z = normrnd((r-sigma^2/2)*T, sigma*sqrt(T), N_MC, 1);
ST = S*exp(Z);

%%% init MC accumulators %%%
sampleMean = 0;
sampleVar = 0;
oldMean = 0;

idx = 1;  % will reset to 1 if idx > N_MC
%%% MC loop %%%
while ((bound > tol) || (n < 30))
   % if idx > N_MC, need to generate more normals
   if (idx > N_MC)
       Z = normrnd((r-sigma^2/2)*T, sigma*sqrt(T), N_MC, 1);
       ST = S*exp(Z);
       idx = 1;
   end
   payoff = exp(-r*T)*max((ST(idx) - K),0);

   % update mean and var
   sampleMean = oldMean + (payoff - oldMean)/n;
   if (n>1)
       sampleVar = (1-(1/(n-1)))*sampleVar + n*(sampleMean - oldMean)^2;
   end
   oldMean = sampleMean;

   bound = 1.96*sqrt(sampleVar/n);  %%% updated this line %%%
   n = n+1;
   idx = idx+1;
end

%%% Display num samples, statistics and vfy bound = tol %%%
n
sampleMean;
sampleVar;
bound;

C = sampleMean;  % call price

%%% exact BS price using BS formula %%%
d1 = 1/(sigma*sqrt(T)) * (log(S/K) + (r + sigma^2/2)*T);
d2 = d1 - sigma*sqrt(T);
exact = S*normcdf(d1) - K*exp(-r*T)*normcdf(d2)

**更新：**我意識到我的程式碼中有一個錯誤：bound更新的地方sampleVar/sqrt(n)應該是sqrt(sampleVar/n). 這導致給定參數的更快收斂。所需的迭代次數已更新。

引用自：https://quant.stackexchange.com/questions/22153