# Ito's Lemma for a Brownian motion

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:

1. $$e^{-(\rho - r)t}c_t^{\gamma}$$ 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.

2. 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.

3. 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?

• Your question is impossible to answer unless you explain the framework, i.e. what are $L_t$, $R_t$, $\sigma_t^c$ and so on... your problem involves plenty of objects and you explain neither of them. – saz Mar 12 '19 at 7:38
• You haven't applied Ito lemma correctly to calculate $d c_t^\gamma$ you are missing the second term – clark Mar 12 '19 at 7:43
• @saz Thanks for the comment. I've tried to improve my question by adding all details relevant for this question. – Walrasian Auctioneer Mar 12 '19 at 16:40
• @clark Could you expand on your comment? Am I missing the second order derivative? – Walrasian Auctioneer Mar 12 '19 at 16:41
• @WalrasianAuctioneer Yes, you are applying Ito to $f(c_t)$ where $f(x)=x^\gamma$. So, $$d f(c_t) = f'(c_t) dt + 1/2f''(c_t) d<c_t,c_t>,$$ and here $d<c_t,c_t>= (\sigma_t^M)^2 dt$, but here you have included only the first term. – clark Mar 12 '19 at 17:33

I 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*}