# Tag Info

## Hot answers tagged stochastic-calculus

2

if we forget about $S_0$, you are just trying to price a power option, i.e. an option on $S^\alpha$. By Ito $$d \log S^\alpha = \alpha d\log S = \alpha (r- q - \frac{1}{2}\sigma^2 ) dt + \alpha\sigma dW_t$$ This can be rewritten $$d \log S^\alpha = (r-q'-\frac{1}{2}\sigma'^2 ) dt + \sigma' dW_t$$ If you set $\sigma' = \alpha \sigma$ $q' = r ... 2 this is not the way to do it. The Black-_Scholes argument requires the underlying to be tradable.$S_{t}^{0.5}$is not tradable. Instead, recognize that the underlying is still$S_t$but the pay-off has changed to $$(\alpha S_{t}^{1/2} - \beta)_+$$ for appropriate constants$\alpha,\beta.\$ So the derivation of the BS equation still holds and the ...

1

As you have guessed correctly, these type of questions can be answered using Ito's Lemma.We have: $$d(M_t)= d(Z_t e^{\int_0^tF(Z_u)du})=d(Z_t) e^{\int_0^tF(Z_u)du}+Z_t d(e^{\int_0^tF(Z_u)du})+d(Z_t)d(e^{\int_0^tF(Z_u)du})$$ For the first two terms on R.H.S, we have: d(Z_t) e^{\int_0^tF(Z_u)du} = (f(W_t)dW_t + ...

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