The Itô Quadratic Variation Term in Incentive-Compatible Consumption Dynamics

# Ito's Lemma for this problem


# Ito's Lemma for this problem












I'm attempting to prove a lemma from a paper, in the context of optimal contracts.

$r,\rho,\gamma,\alpha,\sigma$ are all known constants.

$dR_t = (\alpha + r)dt + \sigma dZ_t$ where $Z_t$ is a standard Brownian motion.

Lemma 1

Given an incentive compatible contract, the agent's consumption must satisfy $$\frac{dc_t}{c_t} = \left( \frac{r - \rho}{\gamma} + \frac{1+\gamma}{2} (\sigma^c_t)^2 \right) dt + \sigma^c_t \frac{1}{\sigma} \left( dR_t - (\alpha + r) dt \right) + dL_t$$ for some stochastic process $\sigma^c$ and a weakly increasing stochastic process $L$.

Proof

The authors provide the following steps:

- $e^{-(\rho - r)t}c_t^{\gamma}$ is a supermartingale, thus we can express it as $$ e^{-(\rho - r)t}c_t^{\gamma} = M_t - A_t$$ where $M_t$ is a martingale and $A_t$ is a weakly increasing process.

- Applying the martingale representation theorem to $M_t$, there exists a stochastic process $\sigma^M_t$ such that $$M_t = \int_0^{t} \sigma^M_t dZ_t$$ where $Z_t$ is a standard Brownian motion.

- They then apply Ito's Lemma to get the first equation by setting $\sigma^M_t = -\gamma \sigma^c_t e^{-(\rho - r)t}c_t^{\gamma}$.

I'm struggling at step 3, as I am not sure how the Ito differential looks like for $M_t$.

This is what I've done: $$- (\rho - r) e^{-(\rho - r) t}c_t^{-\gamma} dt - \gamma e^{-(\rho - r)t} c_t^{\gamma - 1} dc_t = dM_t - dA_t $$ Substituting in $dM_t$ and dividing by $K = e^{-(\rho - r) t} c_t^{-\gamma}$,

$$(r - \rho) dt - \gamma \frac{dc_t}{c_t} = K^{-1} \sigma^M_t dZ_t - K^{-1} dA_t$$ Define $\sigma^c_t = (-\gamma K)^{-1} \sigma^M_t$ and $dL_t = (\gamma K)^{-1} dA_t $, and thus $$ \frac{dc_t}{c_t} = \frac{r - \rho}{\gamma} dt + \sigma^c_t dZ_t + dL_t $$ Plug in $dZ_t = \frac{1}{\sigma} \left( dR_t - (r + \alpha) dt \right)$ (a previous result) and the result follows.

Where does the $\frac{1+\gamma}{2} (\sigma^c_t)^2$ term come from?

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/44537

You missed the quadratic term, specifically, rather than \begin{align*} - (\rho - r) e^{-(\rho - r) t}c_t^{-\gamma} dt - \gamma e^{-(\rho - r)t} c_t^{-\gamma - 1} dc_t &= dM_t - dA_t, \end{align*} we have \begin{align*} & - (\rho - r) e^{-(\rho - r) t}c_t^{-\gamma} dt - \gamma e^{-(\rho - r)t} c_t^{-\gamma - 1} dc_t + \frac{1}{2} \gamma (\gamma+1)e^{-(\rho - r)t} c_t^{-\gamma - 2} d\langle c, c\rangle_t \\ &=\ dM_t - dA_t. \end{align*} That is, \begin{align*} \frac{dc_t}{c_t} = \frac{r - \rho}{\gamma} dt +\frac{1}{2}(\gamma+1)c_t^{-2}d\langle c, c\rangle_t + \sigma^c_t dZ_t + dL_t. \end{align*} Moreover, \begin{align*} c_t^{-2}d\langle c, c\rangle_t = \big(\sigma_t^c\big)^2 dt. \end{align*} Therefore, \begin{align*} \frac{dc_t}{c_t} = \frac{r - \rho}{\gamma} dt +\frac{1}{2}(\gamma+1)\big(\sigma_t^c\big)^2 dt + \sigma^c_t dZ_t + dL_t. \end{align*}