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I have a simple questionsuppose S is a process that follow a lognormal distribution of mean 0 and volatilite sigthen, does S follow this equadiff: dS/S=sigma*dW ??? thank you

yes,where dW is a Wiener process, i.e. dW = sqrt(dt) * epsilon,and epsilon is taken from a normal distribution with mean = 0, std dev = 1.

not accurate.the log normal distribution is defined for random variable. if a process follows lognormal that means that it is so for each t and it is not means that the process can be represented as a SDE solution. On the other hand if you SDE, dS/S=sigma*dW then its solution has lognormal distribution. There exist explicit form of a linear equation and taking log you will arrive at the Gaussian distributed function/

Why cant a random variable be a solution to SDE? SDE's have random variables as solutions right? Why isnt this equivalence true both ways?

You probably meant a random function. If you have a deterministic function then it probably is not a solution a linear deterministic differential equation. Several years ago I tried the Hi square test t to check hypothesis regarding geometric SDE. Note that test regarding with constant coefficients linear SDE easy could be reduced to test Gaussian distribution and independence of increments. Gaussian distribution evidence say was 50%-50% independence was much worse.

QuoteOriginally posted by: pierrelefouI have a simple questionsuppose S is a process that follow a lognormal distribution of mean 0 and volatilite sigthen, does S follow this equadiff: dS/S=sigma*dW ??? thank youSaying that S follows a lognormal distr. is equivalent to say that log(S) follows a normal distr. This doesn't imply that S must satisfy the equation above. Anyway the reverse is true: if S follows dS/S = mu dt + sigma dWt => S(t) = S(0) * exp{(mu - 0.5*sigma^2)*t + sigma*Wt} is lognormal.

<< lognormal distribution of mean 0 and volatility sig >>means dlnS = -0.5*sigma^2*t+sigma*dWor (if you apply ITO with f = exp)dS/S = sigma*dW