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$$ (1) \ \ d\left(\frac{1}{S_t}\right) =\frac{1}{S_t}\left(\sigma^2-r\right)dt +\frac{1}{S_t}\sigma dW_t $$ and $$ (2) \ \ dS_t = S_t rdt + \sigma S_t dW_t $$

How can you prove that?

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Like SolitonK I'm not sure about the intended question @quinlai can you please confirm that the edits are correct? –  Bob Jansen Nov 22 at 13:43

2 Answers 2

Let's first rewrite the tow processes and let $X_t = 1/S_t$ Then we have $$ dX_t/X_t = (\sigma^2-r)dt + \sigma dW_t, $$ with the solution (apply Ito) $$ X_t = X_0 \exp((\sigma^2/2-r) t + \sigma W_t), $$ and $$ dS_t/S_t = r dt + \sigma dW_t, $$ with the solution (apply Ito) $$ S_t = S_0 \exp((r-\sigma^2/2) t + \sigma W_t). $$ If we look at the two processes at point $t$ then their volatilities defined as the coefficient of $dW_t$ in the SDEs is different. It is $X_t \sigma = \sigma/S_t$ for the first and $\sigma S_t$ for the second - these are very different numbers for any path of $S_t$.

If you look at the variance of the processes then we know from results about log-normal distributions (http://en.wikipedia.org/wiki/Log-normal_distribution) that $$ Var[X_t] = X_0^2 \exp(-2rt+2\sigma^2t)(\exp(\sigma^2 t)-1), $$ and $$ Var[S_t] = S_0^2 \exp(2rt)(\exp(\sigma^2t)-1). $$ Also very different: so the answer is no.

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We need to start from $(2)$. We start from Ito's Lemma which stipulates for the single variable case that: $$ \\ df(S_t) = f'(S_t)dS_t + \frac{1}{2}df''(S_t)Var[dS_t] $$ Setting $f=1/S_t$ yields: $$ \\ d\bigg(\frac{1}{S_t}\bigg) = -\frac{1}{(S_t)^2}dS_t+\frac{1}{2}\frac{2}{(S_t)^3}\sigma^2(S_t)^2Var[dW_t] \Rightarrow $$ $$\\d\bigg(\frac{1}{S_t}\bigg) = -\frac{1}{(S_t)^2}(S_trdt+\sigma S_tsW_t)+\frac{1}{2}\frac{2}{(S_t)^3}\sigma^2(S_t)^2Var[dW_t] $$ As $Var[dW_t] = dt$, after some algebra and common factoring we end up with: $$ \\ d\bigg(\frac{1}{S_t}\bigg) = \frac{1}{S_t}(\sigma^2 - r)dt - \frac{1}{S_t}\sigma dW_t$$

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