# Deriving eq. 5 in Carr & Madan 1998

I don't understand this derivation from Carr & Madan (1998), specifically the derivation of the third term on the left (left-most term on the bottom line).

### My attempt

Let $$h(t, F_t) := V(F_t, t; \sigma_h) e^{r(T-t)}$$. Applying Ito's lemma, specifically Shreve's theorem 4.4.6, I get:

$$h(T, F_T) = h(0, F_0) = \overbrace{\int_0^T \frac{\partial h}{\partial t} \text{d}t}^{A} + \overbrace{\int_0^T \frac{\partial h}{\partial t} \sigma F_t \text{d}W_t}^{B} + \overbrace{\int_0^T \frac{1}{2} \frac{\partial^2 h}{\partial F_t^2} \sigma F_t^2 \text{d}t}^{C}.$$

I can convince myself of integrals $$B$$ and $$C$$. These are:

$$B := \int_0^T \frac{\partial h}{\partial F_t} \sigma_t F_t \text{d}W_t = \int_0^T \frac{\partial V}{\partial F_t} e^{r(T-t)} \sigma_t F_t \text{d}W_t = \int_0^T \frac{\partial V}{\partial F_t} e^{r(T-t)} \text{d}F_t.$$

The last equality falls out of the definition of the Black-76 model with volatility indexed by time, i.e.:

$$\text{d}F_t = F_t \sigma_t \text{d}W_t.$$

And $$C$$ is:

$$C := \int_0^T \frac{1}{2} \frac{\partial^2 h}{\partial F_t^2} \sigma F_t^2 \text{d}t = \frac{1}{2} \int_0^T e^{r(T-t)} \frac{\partial^2 V}{\partial F_t} \sigma_t^2 F_t^2 \text{d}t.$$

My only issue is with integral $$A$$. I get:

$$A := \int_0^T \frac{\partial h}{\partial t} \text{d}t = \int_0^T e^{r(T-t)} \frac{\partial V}{\partial t} \text{d}t.$$

But now what? I could apply the Black-76 differential equation, but then I'd get:

$$\frac{\partial V}{\partial t} = V r - \frac{1}{2} \frac{\partial^2 V}{\partial F_t} \sigma_t^2 F_t \text{d}t.$$

This is not what they have. Where am I going wrong? It feels like I'm missing a financial rather than mathematical argument, where they're showing that

$$\frac{\partial}{\partial t} V(t, F_t; \sigma_t) \stackrel{CM98}{=} -rV(F_t, t; \sigma_h) + \frac{\partial}{\partial t} V(F_t, t; \sigma_h)$$

somehow.

Your problem is in the $$A$$ term. Using your notation, you should have

$$d h(t, F_t) = \dfrac{\partial h}{\partial t} dt + \dfrac{\partial h}{\partial F_t} dF_t + \dfrac{1}{2} \dfrac{\partial^2 h}{\partial F_t^2} .$$ Integrating this from $$T$$ to $$T'$$, we get

$$h(T', F_{T'}) = h(T, F_{T}) + \int_T^{T'}\dfrac{\partial h}{\partial t} dt + \int_T^{T'}\dfrac{\partial h}{\partial F_t} dF_t + \int_T^{T'} \dfrac{1}{2} \dfrac{\partial^2 h}{\partial F_t^2} ,$$

where the second and third terms are just $$\int_T^{T'}\dfrac{\partial h}{\partial F_t} dF_t = \int_T^{T'} e^{r(T' - t)}\dfrac{\partial V(t, F_t)}{\partial F_t} dF_t,$$ and $$\int_T^{T'} \dfrac{1}{2} \dfrac{\partial^2 h}{\partial F_t^2} = \int_T^{T'} e^{r(T' - t)} \dfrac{\partial^2 V(t, F_t)}{\partial F_t^2} \dfrac{F_t^2 \sigma_t^2}{2} dt .$$ However, the first term is $$\int_T^{T'}\dfrac{\partial h}{\partial t} dt = \int_T^{T'}\dfrac{\partial (e^{r(T' - t)}V(t, F_t))}{\partial t} dt = \int_T^{T'} e^{r(T' - t)} \left( \dfrac{\partial V(t, F_t)}{\partial t} - r V(t, F_t) \right) dt,$$ note the $$t$$ in the exponential, that "adds" the extra term you were missing.

Hope this helps.

• Ah, the product rule, of course. Thank you.
– jds
Feb 19 at 13:43