Monte Carlo Present Value: Expected Value and the Law of Large Numbers

# Basic Monte Carlo Present value calculation in R question


# Basic Monte Carlo Present value calculation in R question












I'm self studying monte carlo applications with the application towards present values.

However the values that I am using are of the uniform distribution variety with a pre defined minimum and maximum value.

I'm trying to apply the central limit theorem to it so it should approximate a normal distribution giving me the values that are more likely than others.

Now i'm having some difficulties applying the CLT to my current set up.

```
#Empty variable to store the list in
results = NULL 

#loop 
for(i in 1:1000){
  #rate  between 1% and 20%
  r <- runif(1, 0.01, 0.2)

  #expected capital return
  k <- runif(1, 500, 900)

  #periods
  p <- 2

  #do the actual calculation
  pv <- k * (1 + r)^p 

  #Store the calculation
  results <- rbind(results, pv)
}

histogram <- hist(results)
plot(histogram)
summary <- summary(results)
print(summary)
standarddeviation <- sd(results)
print(standarddeviation)

165.108

       V1        
 Min.   : 512.2  
 1st Qu.: 727.8  
 Median : 848.8  
 Mean   : 856.8  
 3rd Qu.: 974.8  
 Max.   :1292.6
```

So what I take from this is that the mean is 856.8 and this is what the project currently would be worth investing for to me more or less.

I am wondering if my methodology and reasoning is correct because I feel I might have gone off the deep end.

## Answer by Cettt (score 2, accepted)

https://quant.stackexchange.com/a/49776

what you are doing is not really an application of the central limit theorem (CLT) but rather an application of the law of large numbers. If I understood your problem correctly you start with the following information:

- The future discounting rate has a uniform distribution: $r \sim U(1\%, 20\%)$.

- The future capital has a uniform distribution: $k \sim U(500, 900)$.

- $r$ and $k$ are independent.

From this you can easy calculate the "true" expected present value: \begin{align} \mathbb{E}[pv] &= \mathbb{E}[k \cdot (1+r)^p] = \mathbb{E}[k] \cdot \mathbb{E}[(1+r)^p] \\[2mm] &= 700 \cdot \int_{0.01}^{0.2} (1+x)^p \frac {1}{0.2 - 0.01} \; dx \\[2mm] &= 700 \cdot \frac{1.2^{p+1} - 1.01^{p+1}}{0.19 \cdot (p+1)}. \end{align}

For $p = 2$ you get that $\mathbb{E}[pv] = 856.8233$.

Your code shows that the law of large numbers works: if you repeat an experiment often enough than the average results of that experiment will be close to its "true" expected value.

One last comment about the code itself. It is advisable to use random seeds in R when working with random numbers. This can be achieved using the `set.seed` function. By setting a random seed you make your results reproducible. Also you can use `replicate` instead of a for loop to make your code more readable:

```
my_pv <- function(p){

  r <- runif(1, 0.01, 0.2)
  k <- runif(1, 500, 900)

  return(k * (1 + r)^p) 
}

set.seed(1234)
result <- replicate(1000, my_pv(2))
mean(result)
[1] 854.8702
```