Distribution of rates of return Chapter 15 Black Scholes

DenisAmbrosov

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Hello! May someone explain how to get the distribution like below. We cancel out T in the mean by dividing by 1/t but we also should cancel the T out in the variance part but after canceling we still have sigma/T but why not just sigma. So we have (mu - sigma2 / 2) * T, sigma2 * T and all this divided by T then T should cancel out but at the end in the distribution we have mu - sigma2 / 2 , sigma / T. Why is it so? Explain,please
 

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lushukai

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Hi @DenisAmbrosov ,

You really need to provide more context to your questions. Fortunately, I know you are referencing Chapter 15 of OFD by Hull. A simple way to understand it is that multiplicative or divisive terms that affect the represented variable, in this case x (which is also 1/T*ln[S(T)/S(0)]), when transposed, affects the variance component by its square. This is why

ln[S(T)/S(0)] ~ N([mu-sigma^2/2]*T , sigma^2/T) → x = 1/T*ln[S(T)/S(0)] ~ N([mu-sigma^2/2]*T/T , sigma^2*T/T^2) = N(mu-sigma^2/2 , sigma^2/T)

Which just so happens to be (15.2) and (15.7) equations. Hope this is helpful!
 
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