Take the 2-minute tour ×
Quantitative Finance Stack Exchange is a question and answer site for finance professionals and academics. It's 100% free, no registration required.

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

How can you prove that?

share|improve this question
    
Like SolitonK I'm not sure about the intended question @quinlai can you please confirm that the edits are correct? –  Bob Jansen yesterday

2 Answers 2

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.

share|improve this answer

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: $$ \\ d\bigg(\frac{1}{S_t}\bigg) = \frac{1}{S_t}(\sigma^2 - r)dt - \frac{1}{S_t}\sigma dW_t$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.