Toy example of lognormal simulations

Math

Let \(X\) follow normal distribution \(X \sim Normal(\mu_x, \sigma_x^2)\) with mean \(\mu_x\) and variance \(\sigma_x^2\) and moment generating function \(M_x(t) = exp(\mu_x t + \sigma^2 t^2 / 2)\).

Let \(Y = e^{X}\), then \(Y\) follows lognormal distribution with \(Y \sim Lognormal(\mu_x, \sigma_x^2)\). Then we have:

\[\mu_y = E(Y) = E(e^{X}) = M_x(1) = e^{\mu_x + \sigma_x^2/2}\] \[\sigma_y^2 = E(Y^2) - (E(Y))^2 = E(e^{2X}) - (E(e^{X}))^2 = M_x(2) - M_x(1)^2\] \[=(exp(\sigma_x^2) -1)exp(2\mu_x + \sigma_x^2)\]

In simulation, given that we know our target \(\mu_y\) and \(\sigma_y^2\) from emperical observation, solving for \(\mu_x\) and \(\sigma_x^2\), we have: \[\mu_x = \ln{\frac{\mu_y^2}{\sqrt{\mu_y^2 + \sigma_y^2}}}\] \[\sigma_x^2 = \ln{(1+\frac{\sigma_y^2}{\mu_y^2})}\]

Simulation

Assuming \(0.10\) interest rate, 0.4 annualized volatility and stock price of \(100\) in year one. In five years we have \(\mu_y = 146.41\) and \(vol_y = 0.89\). Therefore, \(\sigma_y = \mu_yvol_y = 130.30\). Therefore, we can easily simulate 10000 data points with \(Lognormal(4.69, 0.58)\).

Hovering over the boxplot, we can get 15% and 75% percentile on Q1 and Q3, respectively. We can also obtain upper fence and min corresponding to the max and min estimates with outliers removed.

Raw data output from simulation