最近伊藤過程から関数の期待値と変動率の計算を考えています。It is well-documented (eg Hull) that, starting with the diffusion process for a stock price S as geometric Brownian motion, one can find the diffusion process for ln(S) via Ito's lemma. The diffusion process for S is dS = mu*S*dt+ sigma*S*dW. Ito's lemma is used to derive the diffusion process for dln(S). This value is normally distributed with mean (mu-sigma^2/2)t and variance sigma^2*t. However, it does not seem possible to work in the opposite direction, starting with dx = dW and then deriving the diffusion process for f(x) = exp(alpha+beta*x). This seems to be a contradiction in Ito's lemma. Firstly, consider the first and second derivatives of f with respect to x, df/dx = beta*f, and d^2f/dx^2=beta^2*f. df/dt = 0. Ito's lemma indicates the following (given a diffusion process dx = mu*dt+sigma*dz): df = (df/dt + mu*(df/dx)+0.5*sigma^2*d^2f/dx^2)dt+sigma*(df/dx)*dW. Thus, for f(x) = exp(alpha+beta*x), we set sigma=1, mu=0, df = (0+0+0.5*(beta^2f))dt+beta*f*dW. This would imply a mean of (0.5*beta^2f)T, which differs markedly from the lognormal mean value of exp(alpha+0.5*beta^2). However, some other sources seem to show comparable derivations of a diffusion using Ito's lemma (cf NYU stochastic calculus (2013) lecture notes by Goodman) without comparing the mean or variance with that of the lognormal distribution which uses similar first and second order derivatives of the exponential. It may be that it is necessary to write the alpha value above in terms of the time variable t. This would enable the time derivative to be non-zero also. However, it is unclear how the beta multiplier outside of the exponential would be nullified and an additive sigma^2/2 term be moved into the exponential. In fact, setting alpha=(r-beta^2/2)*t ,which is the expected rate of return over time t for a lognormal random variable with growth rate r, means that df/dt = (-0.5*beta^2)*f. In addition, rather than use the form of Ito's lemma above, which assumes a geometric Brownian motion for x, we use the Taylor series expansion (2nd order in x, 1st order in time) for df = (df/dt)dt+(df/dx)dx+(1/2)(d^2f/(dx)^2)(dx)^2, where dx=dW and (dx)^2=dt. Thus, the (0.5*beta^2)f terms drop out. We are left with df = r*f*dt+beta*f*dW. Note that "f" appears in both terms, making this a geometric Brownian motion. Differential equation theory would indicate that this yields E(f) = E(S)= exp(rt), which is equivalent to exp(alpha+0.5*beta^2). The variance would be based on the notion that (E(f))^2 = exp(2*alpha+beta^2) and var(f) = E(f^2)-(E(f))^2. Assuming that the probability distribution can be used to show E(f^2) = exp(2*alpha+2*beta^2), we have solved for the lognormal variance of exp(2*alpha+beta^2)*(exp(beta^2)-1).
Thus, there does seem to be a way to derive the lognormal diffusion, mean, and variance via Ito's lemma starting from a standard normal arithmetic Brownian motion diffusion dx = dW. However, the intuition to set f(x) = exp((r-0.5*beta^2)t+beta*x) is not easy to see.
