Recently, I asked myself the question: "Why is the expected value of e^sX (X ~ N(0,1)) equal to e^(0.5*s^2)?". You know that compounding a sum of money (A) at a continously compounded rate X for a period involves multiplying it by e^X. In case X is non-random, the expected value is A*e^X. In case X is random, calculating the expected value becomes quite difficult. The solution for a standard normally distributed X can be derived as follows:

1. Write down the equation (μ = E[e^sX] where X ~ N(0,1)):

2. Substitute y = x - s and you get:More general, if X ~ N(μ*, σ

s <- 0.3

n <- 10000

x <- rnorm(n)

answer <- mean(exp(s*x))

Or a bit more sophisticated and less arbitrary:

s <- 0.3

n <- 10000

i <- (1:n)/(n+1)

x <- qnorm(i)

answer <- mean(exp(s*x))

NB: The expected value can be very misleading.

Source: Stochastische Grundlagen der Finanzmathematik, Klaus Pötzelberger

1. Write down the equation (μ = E[e^sX] where X ~ N(0,1)):

2. Substitute y = x - s and you get:More general, if X ~ N(μ*, σ

^{2}), then X = μ* + σZ with Z ~N(0,1). It follows thatOf course, practitioners often know the parameters they are interested in (e.g. s = 0.3), so they can take a shortcut and run a monte carlo analysis:s <- 0.3

n <- 10000

x <- rnorm(n)

answer <- mean(exp(s*x))

Or a bit more sophisticated and less arbitrary:

s <- 0.3

n <- 10000

i <- (1:n)/(n+1)

x <- qnorm(i)

answer <- mean(exp(s*x))

NB: The expected value can be very misleading.

Source: Stochastische Grundlagen der Finanzmathematik, Klaus Pötzelberger

Mahalanobis - am 2008-12-27 22:58 - Rubrik: mathstat