Markov Pricing kernel

I'm reading about Markov pricing kernels in the lecture notes of a course I'm following, but I have a big doubt on an application of Ito's lemma. The setting is the following:

We define the pricing kernel as $$\xi_t = \xi(D_t,y_t,t) = e^{\int_0^t \delta(D_s,y_s)ds} T(D_t,y_t), \qquad \xi_0=1$$ where $D$ and $y$ are Ito processes following the dynamics $$dD_t = m(D_t,y_t)dt + \sigma(D_t,y_t) dW_{1t}$$ and $$dy_t = \varphi(y_t)dt + v_1(y_t) dW_{1t} + v_2(y_t) dW_{2t}$$ Moreover, we assume the pricing kernel follows the dynamics $$\frac{d\xi_t}{\xi_t} = -R_t dt -\lambda_{1t} dW_{1t}-\lambda_{2t} dW_{2t}$$

Now, the claim in the lecture notes is that by applying the Ito's lemma to $\xi_t$, one finds $$R(D,y) = \delta(D,y) - \frac{\mathscr{L} T(D,y)}{T(D,y)},$$ where $\mathscr{L}$ is the infinitesimal generator.

Now, I can see that this result can be obtained - rather trivially - in the case where the function $\delta$ in the integral is not dependent on $D$ and $y$. But in the case the dependence is there (as stated in the lecture notes), the drift obtained with Ito takes a much more complex form, and I really don't see any cancelling of the terms.

Do you agree with me - and thus there's a typo in my lecture notes - or am I applying Ito in the wrong way?

• Is the lecture notes publicly available? We need to have more contexts and the definitions for the respective notations. – Gordon Mar 19 '15 at 13:18
• Not really, it should be kept confidential – Abramo Mar 19 '15 at 13:19
• Then you may need to provide more details on the notations employed here. – Gordon Mar 19 '15 at 13:21
• Just ask me about: what is not clear? – Abramo Mar 19 '15 at 14:09
• Why don't you ask your professor to expand more on the topic? Guess he can be happy someone actually reads his lecture notes, and it could lead to a better understanding for the whole class. – Olorun Mar 20 '15 at 0:30

To shorten the notation, let's write $T_t = T(D_t,y_t)$ and $\delta_t = \delta(D_t,y_t)$.

There are two ways to show that, in fact, the dynamics of $$\xi_t = \xi(D_t, y_t,t) = e^{-\int_0^t \delta_s ds}\, T_t$$ is given by $$\frac{d\xi_t}{\xi_t} = \left( -\delta_t + \frac{\mathscr{L} T_t}{T_t} \right)dt \quad+\quad \text{diffusion terms}.$$

First way (more formal)

Write $\xi_t = g_t T_t$, where $g_t = e^{-\int_0^t \delta_s ds}$. It is easy to show that the dynamics of $g_t$ is given by $$\frac{dg_t}{g_t} = -\delta_t \, dt$$ therefore, by applying Ito's product rule we have $$d\xi_t = d(g_t T_t) = T_t dg_t + g_t dT + dg_t dT_t = -\delta_t \xi_t dt + g_t dT_t$$ because the product term vanishes since $dg_t$ has no diffusion. Hence it follows

$$\frac{d\xi_t}{\xi_t} = \frac{d\xi_t}{g_tT_t} = -\delta_t dt + \frac{dT_t}{T_t} = \left( -\delta_t + \frac{\mathscr{L} T_t}{T_t} \right)dt \quad+\quad \text{diffusion terms}$$

Second way (less formal)

The function $g_t = -\int_0^t \delta(D_s,y_s) ds$ is constant w.r.t. both $D_t$ and $y_t$, in the sense that the contribution of the very last realisation at time $t$ of the processes to the integral over $[0,t]$ is almost surely equal to $0$. In other words,

$$\frac{\partial g_t}{\partial D_t} = \frac{\partial g_t}{\partial y_t} = 0 \, .$$ Therefore the result follows by noticing that $$\mathscr{L} \xi_t = g_t \mathscr{L} T_t \implies \frac{\mathscr{L} \xi_t}{\xi_t} = \frac{\mathscr{L} T_t}{T_t}$$

I think you are having trouble differentiating the integral of $\delta$.

You should remember the differential notation is just notation for an integral: $A_td B_t = A'_t dB'_t$ just means $\int_0^T A_td B_t = \int_0^T A'_t dB'_t$.

In particular, $d\int_0^t A'_s dB'_s = A'_t dB'_t$ is a tautology. So $$d ( e^{\int_0^t \delta(s,X_s) ds} )= e^{\int_0^t \delta(s,X_s) ds} d (\int_0^t \delta(s,X_s) ds ) = e^{\int_0^t \delta(s,X_s) ds}\delta(t,X_t)dt$$

Applying Ito formula to the product with $T$ gives the result.

• For $A_t'$, you meant for $A_{t'}$? – Gordon Mar 20 '15 at 19:07
• No I meant $A'$ like another process playing a similar role to $A$. – AFK Mar 21 '15 at 11:06