Drill - P1.t2.323, page24
- Thread starter Tipo
- Start date
Hi @Tipo
The conventional way to represent the normal distribution is X ~ N(μ, σ^2); e.g., http://en.wikipedia.org/wiki/Normal_distribution.
So the (given) assumption that volatility = 36% tells us that sigma is 0.36 and the variance is 0.36^2 = 0.12960
In the displayed lnS(t) ~ Φ[lnS(0) + (μ - σ^2/2)T, σ^2T], the term σ^2T is a variance such that the standard deviation of this approximately normal distribution is sqrt(σ^2T) = σ*sqrt(T). You are correct to be looking for that but as T = 1.0 --> σ*sqrt(T) = σ*sqrt(1) = σ = 0.36, the standard deviation of this approximately normal distribution is sigma.
So the step the calculation of N[(4.09434 - 3.98640)/0.36] is just the standard normal cumulative lookup of a typical Z value: N[Z] = N[(X - μ)/σ] where μ is the expected value lnS(0) + (μ - σ^2/2)T. I hope that explains!
The conventional way to represent the normal distribution is X ~ N(μ, σ^2); e.g., http://en.wikipedia.org/wiki/Normal_distribution.
So the (given) assumption that volatility = 36% tells us that sigma is 0.36 and the variance is 0.36^2 = 0.12960
In the displayed lnS(t) ~ Φ[lnS(0) + (μ - σ^2/2)T, σ^2T], the term σ^2T is a variance such that the standard deviation of this approximately normal distribution is sqrt(σ^2T) = σ*sqrt(T). You are correct to be looking for that but as T = 1.0 --> σ*sqrt(T) = σ*sqrt(1) = σ = 0.36, the standard deviation of this approximately normal distribution is sigma.
So the step the calculation of N[(4.09434 - 3.98640)/0.36] is just the standard normal cumulative lookup of a typical Z value: N[Z] = N[(X - μ)/σ] where μ is the expected value lnS(0) + (μ - σ^2/2)T. I hope that explains!
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