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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)):
gau012. Substitute y = x - s and you get:gau2More general, if X ~ N(μ*, σ2), then X = μ* + σZ with Z ~N(0,1). It follows thatgau03Of 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

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