# Montecarlo pricing

I have some problems with the Montecarlo simulation to price a generic Call option. I want to explain something regarding MC simulation with a simple cases, and after that I am going to talk about my problem.

1. Montecarlo - Simple case: considering a set of parameters: S=1, K=1, sigma=0.5, r=0, T=1, N=1000 (number simulation MC); the montecarlo with this example works in this way:

1.1 $$S$$ = [1,1, ..., 1] I am going to repeat the value of underlying a number of times equal to $$N$$

1.2 $$X$$ = $$S e^{Z}$$ where $$Z$$ = ($$1\times N$$) vector of Browniam motion --> I get a vector ($$1\times N$$) where I multiply each value of $$S$$ with each value of $$e^{Z}$$

1.3 In the vector $$X$$, I take the $$\max(X-K,0)$$ so between each value inside the vector minus K, and zero

1.4 Finally, I find the payoff, that is the average of the vector.

1. Montecarlo - Another case: considering a set of parameters: S=[1.1,1.2], K=1, sigma=0.5, r=0, T=1, N=1000 (number simulation MC); the montecarlo with this example works in this way:

2.1 $$S = \begin{pmatrix} 1.1 & \dots & 1.1\\ 1.2 & \dots & 1.2 \end{pmatrix}$$ ($$2 \times N$$); so we repeat the value of underlying a number of times equal to $$N$$

1.2 $$X$$ = $$S e^{Z}$$ = $$\begin{pmatrix} 1.1 \exp{Z_1} & \dots & 1.1\exp{Z_N}\\ 1.2 \exp{Z_1}& \dots & 1.2\exp{Z_N} \end{pmatrix}$$ where $$Z$$ = ($$1\times N$$) vector of Browniam motion --> I get a vector ($$2\times N$$) where I multiply each value of first row $$S$$ with each value of $$e^{Z}$$, the same for the second row of $$S$$

1.3 In the vector $$X$$, I take the $$\max(X-K,0)$$

1.4 Finally, I find the two payoffs, that is the average of the first row (for first payoff) and average of second payoff (for second row)

Now i can explain my problem: How can I find the payoff, if I have both $$S$$ and $$\sigma$$ that are vectors? for example, $$S$$=[1.1,1.2], $$\sigma$$=[0.5,0.6]

I have tried in this way, but I think it is wrong..

• $$S = \begin{pmatrix} 1.1 & \dots & 1.1\\ 1.2 & \dots & 1.2 \end{pmatrix}$$($$2 \times N$$)

• $$\sigma = \begin{pmatrix} 0.5 & \dots & 0.5\\ 0.6 & \dots & 0.6 \end{pmatrix}$$($$2 \times N$$)

• I generate two Brownian motion (because I have two values of sigma) = $$Z = \begin{pmatrix} BM_{11} & \dots & BM_{1N}\\ BM_{21} & \dots & BM_{2N} \end{pmatrix}$$

• $$X = S e^{Z} = \begin{pmatrix} \begin{pmatrix} 1.1 e^{BM_{11}} & \dots & 1.1 e^{BM_{1N}}\\ 1.2 e^{BM_{11}} & \dots & 1.2 e^{BM_{1N}} \end{pmatrix} \\ \begin{pmatrix} 1.1 e^{BM_{21}} & \dots & 1.1 e^{BM_{2N}}\\ 1.2 e^{BM_{21}} & \dots & 1.2 e^{BM_{2N}} \end{pmatrix} \end{pmatrix}$$

• After that, I take the maximum as before, and the average, but in this case I obtain 4 payoff! And for this reason I am not sure about this method..

• Hi, it would be helpful if you could add the code / software that you are using as well. Vectorisation and matrix operations behave a bit differently between Matlab, Python, R or other software. Also, you put greek in your tag list, but I cannot find anything related to greeks in you question. Commented Aug 23, 2021 at 9:49
• @Kermittfrog I have not already developed the code because I am not sure about the method, before I need if possible some hints for this case.. For the greeks yes, you are right, I should delete it in the tags! Commented Aug 23, 2021 at 9:54
• @Kermittfrog However, the software that I will use is Python Commented Aug 23, 2021 at 10:00

Commonly, you do not use 'pure' matrix algebra when formulating a Monte Carlo valuation setup.

If your options are of the European type, and you truly want to price all options in one go, you could go as follows. Let $$M$$ denote the number of simulations $$m=1\ldots M$$, let $$N$$ denote the number of underlyings $$n=1\ldots N$$, and let $$K_m$$ denote each strike.

Let $$\Sigma$$ denote the covariance matrix of your asset returns, i.e. $$\Sigma_{i,j}=\sigma_i\sigma_j\rho_{i,j}$$ and $$\Sigma_{ii}=\sigma_i^2$$. Then, $$C$$ denotes the Cholesky decomposition of $$\Sigma$$, i.e. $$CC^T=\Sigma$$. Note that $$\Sigma$$ is $$N \times N$$ and $$C$$ is $$N \times N$$ as well.

To stick with your example, let $$\sigma\equiv\begin{pmatrix}\sigma_1\\\sigma_2\\\ldots\\\sigma_N\end{pmatrix}$$

If you have no correlation between assets, then \begin{align} \Sigma&=\begin{pmatrix}\sigma_1^2&0&\ldots&0\\ 0&\sigma_2^2&\ldots&0\\0&0&\ldots&0\\0&0&\ldots&\sigma_N^2\\ \end{pmatrix}=\mathrm{Diag}\left(\sigma_1^2,\sigma_2^2,\ldots,\sigma_N^2\right)=\mathrm{Diag}\left(\sigma\right)\mathrm{Diag}\left(\sigma\right) \end{align} and, of course, $$C=\begin{pmatrix}\sigma_1&0&\ldots&0\\ 0&\sigma_2&\ldots&0\\0&0&\ldots&0\\0&0&\ldots&\sigma_N\\ \end{pmatrix}=\mathrm{Diag}(\sigma)$$

Disregarding interest rates and dividend yields, let's introduce the $$1\times M$$ vector $$e$$ which consists of ones, only. Given an $$N\times M$$ matrix of standard normal variates $$U$$, we can now set $$X=-\frac{1}{2}\sigma\otimes\sigma e T+ \sqrt{T}CU$$ where $$\otimes$$ denotes element-wise multiplication, and $$T$$ denotes the time to expiry of your option. The dimension of $$X$$ is then $$N \times M$$ as well.

Ultimately, we find

$$S_T=S_0\otimes e^{X}$$

and $$P=\max(S_T-Ke,0)$$ with $$S_0$$ the $$N\times 1$$ vector of initial prices and the exponential operator $$e$$ to be understood in an element-by-element way. Furthermore, $$\max$$ is to be understood in an element-by-element way as well, and $$K$$ is the vector of strikes. You can then average over each row of the $$N\times M$$ option payoff matrix $$P$$.

Again, your programming language of choice should have some helpers and common interpretation of scalar-by-vector, scalar-by-matrix and vector-by-matrix multiplication; and you will most probably not follow this setup in totality.

• thanks for the answer! In your notation, what is the dimension of X? I have written all the history in a matrix form, because is easier to me for the implementation in Python Commented Aug 23, 2021 at 10:53
• I've updated accordingly. $X$ is $N\times M$. Commented Aug 23, 2021 at 11:59