# Is the volatility for these two SDEs the same

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

I’m not sure about the question. Say you have two processes $$(1) \ \ d\left(\frac{1}{X_t}\right) =\frac{1}{X_t}\left(\sigma^2-r\right)dt +\frac{1}{X_t}\sigma dW_t$$ and $$(2) \ \ dY_t = Y_t rdt + Y_t \sigma dW_t$$ If I interpret it correctly then you should apply Ito’s formula to $f(Y,t) = \frac{1}{Y}$ to obtain $$(2)-bis \ \ \ \ d\left(\frac{1}{Y_t}\right) =\frac{1}{Y_t}\left[\left(\frac{\sigma}{Y_t}\right)^2-\frac{r}{Y_t}\right]dt +\frac{1}{Y_t}\sigma dW_t$$ So the two processes (1) and (2)-bis have different drifts but same volatilities.
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: $$\\ d\bigg(\frac{1}{S_t}\bigg) = \frac{1}{S_t}(\sigma^2 - r)dt - \frac{1}{S_t}\sigma dW_t$$