There's a problem in my actex manual:
For a stochastic process Y(t),
i) dY(t) = mu*Y(t)dt + sigma*Y(t)dZ(t)
ii) P(Y(3)> E[Y(3)] = .1736
find sigma.
In the solution, it states that E[Y(t)] = Y(o)exp(mu*t) and I guess this is where I am confused. The average for the lognormal is Y(0)exp(mu+.5*sigma^2), but they seem to be plugging in 3 for t, and ignoring sigma, which I thought that was how you found the median.
An aside if anyone wants to help me with this also: I feel like I understand the concept that the lognormal distribution is found by taking normal random variables raising these values to the base e. We generate a random walk by keeping a running count of normal variables, and if we want to add a trend to that, we add in an alpha*t. Geometric brownian motion is a cumulative random term + trend term, raised to the power e.
For a stochastic process Y(t),
i) dY(t) = mu*Y(t)dt + sigma*Y(t)dZ(t)
ii) P(Y(3)> E[Y(3)] = .1736
find sigma.
In the solution, it states that E[Y(t)] = Y(o)exp(mu*t) and I guess this is where I am confused. The average for the lognormal is Y(0)exp(mu+.5*sigma^2), but they seem to be plugging in 3 for t, and ignoring sigma, which I thought that was how you found the median.
An aside if anyone wants to help me with this also: I feel like I understand the concept that the lognormal distribution is found by taking normal random variables raising these values to the base e. We generate a random walk by keeping a running count of normal variables, and if we want to add a trend to that, we add in an alpha*t. Geometric brownian motion is a cumulative random term + trend term, raised to the power e.
Expectations.. Lognormal