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?

  • $\begingroup$ Is the lecture notes publicly available? We need to have more contexts and the definitions for the respective notations. $\endgroup$ – Gordon Mar 19 '15 at 13:18
  • $\begingroup$ Not really, it should be kept confidential $\endgroup$ – Abramo Mar 19 '15 at 13:19
  • $\begingroup$ Then you may need to provide more details on the notations employed here. $\endgroup$ – Gordon Mar 19 '15 at 13:21
  • $\begingroup$ Just ask me about: what is not clear? $\endgroup$ – Abramo Mar 19 '15 at 14:09
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    $\begingroup$ 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. $\endgroup$ – 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 $$ \mathscr{L} \xi_t = g_t \mathscr{L} T_t \implies \frac{\mathscr{L} \xi_t}{\xi_t} = \frac{\mathscr{L} T_t}{T_t} $$

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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.

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

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