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Let $F : T \mapsto S e^{\int_0^T{\mu \left(t\right)\mathrm{d}t}}$, $S \in \mathbb{R}_+^*$, and define $y \left(T\right) = \ln \left(\frac{K}{F_T}\right)$, where again $K \in \mathbb{R}_+^*$. Note that this implies $K = F_T e^y = S e^{\int_0^T{\mu \left(t\right)\mathrm{d}t}+y}$.

I want to differentiate $y$ with respect to $T$. Doing so directly yields $$ \partial_T y = \partial_T \left[ \ln \left(K\right) - \ln \left(S\right) - \int_0^T{\mu \left(t\right)\mathrm{d}t} \right] = - \mu \left(T\right) $$

However, in the paper I am reading, the author uses the chain rule and gets $$ \partial_T y = \partial_K y \times \partial_T K = \frac{1}{K} \times \mu \left(T\right) S e^{\int_0^T{\mu \left(t\right)\mathrm{d}t}+y} = \frac{1}{K} \times \mu \left(T\right) \times K = \mu \left(T\right) $$

I have to be missing something since I do not see why I would obtain two different results for the derivative. Any idea?

siou0107
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  • Nowhere in that paper do I see the thing you claim the author is deriving. He doesn't use $\frac{\partial y}{\partial T}$ anywhere that I can see. – MPW Apr 23 '20 at 20:56
  • The author does NOT differentiate $y$ against $T$. He does so indirectly by differentiating $K$ against $T$ in equation 29. It is I that have tried differentiating $y$ directly; since $K$ is an input of the problem, I don’t like the fact of differentiating it and prefer differentiating $y = f \left(K\right)$. Sorry if that was not clear. – siou0107 Apr 23 '20 at 21:18
  • The first line of equation (29) seems to be nonsense. It says that LHS = LHS + TERM, which cannot be true unless TERM is identically zero. So it looks like hand-waving to me. – MPW Apr 23 '20 at 21:45

1 Answers1

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You indicate that $S$ and $K$ are positive real constants. In this case, it makes no sense to differentiate with respect to $K$. So either you are mistaken about that, or the author is just wrong.

Assuming these are indeed constants, you would have

If $y(x) = \ln(K/F(x)) = \ln K - \ln F(x)$, then $$\frac{dy}{dx} = -\frac{F'(x)}{F(x)}$$ Now $$F(x) = Se^{\int_0^x\mu(t)\;dt}$$ so $$F'(x) = Se^{\int_0^x\mu(t)\;dt}\cdot \frac{d}{dx}\left(\int_0^x\mu(t)\;dt\right)$$ $$=Se^{\int_0^x\mu(t)\;dt}\cdot \mu(x)$$ and therefore $$\frac{dy}{dx} = -\frac{F'(x)}{F(x)} = -\frac{Se^{\int_0^x\mu(t)\;dt}\cdot \mu(x)}{Se^{\int_0^x\mu(t)\;dt}}=\boxed{-\mu(x)}.$$

Question: Should there be a minus sign in the exponent in the definition of $F$?

This would produce the desired result.

MPW
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  • $K$ is an input of the problem. But the author, for some reasons, transforms it as a function of $T$ (, $S$) and $y$, which is an approach I don’t like so much. That is why I wanted to differentiate $y$ directly wrt $T$. However, I still do not get how come the two approaches do not yield the same result – siou0107 Apr 23 '20 at 19:01
  • The paper, page 8: https://www.frouah.com/finance%20notes/Dupire%20Local%20Volatility.pdf – siou0107 Apr 23 '20 at 19:05
  • I don't see that particular computation, not on page 8. Can you be specific? Which equation number? – MPW Apr 23 '20 at 20:38