Monte Carlo Pricing of Down-and-Out Calls and Discretization Bias

# Black scholes model for down and out European call option using Monte Carlo


# Black scholes model for down and out European call option using Monte Carlo












I tried to implement Matlab program computing the price of the European down and out call option using Monte Carlo and Euler discretization scheme. I have initial price S0=50, strike K=50, barrier level B=45 and time of expiration 6 months. The final price I obtain is very small(0.005). Even when I increase T to 1 or when I decrease the barrier, the price doesn't increase. I don't know what is the problem. I also have one additional question - how can I find Greeks(Delta,Vega,Gamma,Theta,Rho) with Monte Carlo simulation on this model? Here is my code:

```
function [Price]= BlackScholes (n,m,r,T,Var,S0,K,B)
Price=1:50;

for i=1:n
I=1;
for j = 0:(m-1);
Z(j+1)= randn (1 ,1); 
dW=sqrt (T/m)*Z(j+1);
if j==0 
S(j+1) = S0*exp((r-Var/2)*(T/m)+sqrt(Var)*dW);
if (I==1) & (S(j+1) <= B)
I = 0;
end
else 
S(j +1) = S(j)* exp ((r - Var /2) *(T/m) + sqrt ( Var )* dW); 
if all([ I==1 , S(j+1) <=B])
I = 0;
end
end
end 
C=zeros(n,1);
C(i)= exp(-r*T)* max ((S(m-1)-K), 0)*I;
Price = sum (C (1:n))/n; 
end
```

Thanks a lot!

## Answer by Quantuple (score 1, accepted)

https://quant.stackexchange.com/a/28038

There are many things wrong with your code. I'll leave aside the manner in which it is implemented, but note that it is: (1) not Matlab friendly with all the for loops (you should vectorise), (2) the fact that you have splitted the case j==0 in the main loop is a poor coding practice.

```
for i=1:n

I=1;
for j = 0:(m-1);
    Z(j+1)= randn (1 ,1);
    dW=sqrt (T/m)*Z(j+1);
    if j==0
        S(j+1) = S0*exp((r-Var/2)*(T/m)+sqrt(Var)*dW);
        if (I==1) & (S(j+1) <= B)
            I = 0;
        end
    else
        S(j +1) = S(j)* exp ((r - Var /2) *(T/m) + sqrt ( Var )* dW);
        if all([ I==1 , S(j+1) <=B])
            I = 0;
        end
    end
end
C=zeros(n,1);                                           %%% [1] THIS IS WRONG
C(i)= exp(-r*T)* max ((S(m-1)-K), 0)*I;                 
Price = sum (C (1:n))/n;                                %%% [2] THIS IS USELESS

end
```

The reason you observe very small prices, is that only the last path of your MC simulation contributes to the option price. This is because at the line marked with [1] above, you always reinitialise the price vector as a vector full of zeros. Also note that the line [2] is pretty useless and should lie outside of the for loop.

I would recommend an implementation along the lines of:

```
function price = mc_pricer(n,m,r,T,var,S0,K,B)

rng(0);            
S = zeros(n,m);
S(:,1)=S0;
Z = randn(n,m-1);
t = linspace(0,T,m);
dt = diff(t);
for i=2:m
    S(:,i)=S(:,i-1).*exp( (r-var/2)*dt(i-1) + sqrt(var*dt(i-1))*Z(:,i-1) );
end
S_T = S(:,end);
payoff = max(S_T-K, 0).*(1-any(S<=B,2));
price = exp(-r*T)*mean(payoff);

end
```

## Answer by zyy2016 (score 2)

https://quant.stackexchange.com/a/28048

Besides the code's problem, I highly recommend the Brownian Bridge correction method which can compensate the pricing error resulting from discretization of the continuous path.