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Home/ Questions/Q 8655011
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T14:55:40+00:00 2026-06-12T14:55:40+00:00

I was trying to execute this functions, but I kept on getting an error

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I was trying to execute this functions, but I kept on getting an error with my if statment: Error in if (value[1][i] < 0) { : missing value where TRUE/FALSE needed:

Monte_Carlo <- function(trial)
{
    S_T <- S_o*exp((r - q - (1/2)*sigma^2)*period + sigma*rnorm(trial, mean = 0, sd = 1))
    K <- matrix(100, nrow = 1, ncol = 20)   
    value <- K - S_T

    for(i in 1:trial)
    {
        if(value[1][i] < 0)
        {
            value[1][i] = 0;
        }
    }
    return (mean(value)*exp(-r))
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-12T14:55:41+00:00Added an answer on June 12, 2026 at 2:55 pm

    You are indexing your matrix incorrectly. value[1] will return a single value, which you are then trying to assign the ith element of, for i up to trial

    If you assign to the ith element in the 1st row (which looks as if you are trying to do), then it will work

    Monte_Carlo <- function(trial)
    {
      S_T <- S_o*exp((r - q - (1/2)*sigma^2)*period + sigma*rnorm(trial, mean = 0, sd = 1))
      K <- matrix(100, nrow = 1, ncol = 20)   
      value <- K - S_T
    
      for(i in 1:trial)
      {
        if(value[1, i] < 0)
        {
          value[1,i] = 0;
        }
      }
      return (mean(value)*exp(-r))
    }
    

    you can remove the for loop and if statement with pmax which is vectorized and will return elementwise the maximum of value and 0.

    Monte_Carlo <- function(trial)
    {
      S_T <- S_o*exp((r - q - (1/2)*sigma^2)*period + sigma*rnorm(trial, mean = 0, sd = 1))
      K <- matrix(100, nrow = 1, ncol = trial)   
      value <- K - S_T
    
      .value <- pmax(value,0)
      return (mean(.value)*exp(-r)
    }
    

    As a matter of good programming, I would add r,S_o, q, sigma and period as arguments to your function, so that it doesn’t depend on global variables

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