Deriving a Stochastic Discount Factor with Itô’s Lemma

Summary

The document explains how to obtain an exponential expression for a stochastic discount factor from its stochastic differential equation. The process has Brownian shocks from two independent sources, with one diffusion coefficient depending on time. The derivation applies Itô’s lemma to the logarithm of the discount factor, turning the multiplicative differential into an additive expression and introducing a drift correction from quadratic variation.

Integrating the resulting equation gives the log process as Brownian integrals minus one half of the accumulated squared diffusion coefficients; exponentiating recovers the level of the discount factor. The two independent Brownian motions imply that their cross variation is zero, so the instantaneous variance contribution is the sum of their squared coefficients. The derivation shown assumes a specified initial value, which appears as a multiplicative factor in the final expression; the displayed target formula implicitly sets that value to one. No numerical example or broader conditions on the time-varying coefficient are provided.

Key ideas

• Apply Itô’s lemma to the logarithm of a positive stochastic process to derive its exponential form.
• The quadratic variation creates a negative one-half variance correction in the logarithmic drift.
• Independent Brownian drivers have zero cross variation, so their variance contributions add.
• Integrating the log differential and exponentiating yields the level process.
• The general solution includes the initial value as a multiplicative factor.

Tags

# stochastic discount factor transformation


# stochastic discount factor transformation












I have $$\frac{dM_t}{M_t}=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t, \tag{1}$$ where $B_t$ and $W_t$ are two independent Brownian Motions, which was further presented as $$ M_t=\exp \left( -\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s -\frac{1}{2} \int_0^t \gamma_s^2 ds \right) \tag{2}$$

can anybody explain how $(1)$ was transitioned into $(2)$?

## Answer by Gordon (score 2, accepted)

https://quant.stackexchange.com/a/25986

Apply Ito's lemma to $\ln M_t$, we obtain that \begin{align*} d\ln M_t &= \frac{1}{M_t} dM_t -\frac{1}{2} \frac{1}{M_t^2} d\langle M, M\rangle_t\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \frac{1}{M_t^2}\left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt\\ &=-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right) dt.\tag{Eq. 1} \end{align*} Here, since $dM_t=M_t\big[-\frac{\mu}{\sigma} dW_t + \gamma_t dB_t\big]$, \begin{align*} d\langle M, M\rangle_t = \left(\frac{\mu^2}{\sigma^2} + \gamma_t^2\right)M_t^2dt. \end{align*} From $(\textrm{Eq.} 1)$, \begin{align*} \ln M_t - \ln M_0 &= \int_0^t\left[-\frac{\mu}{\sigma} dW_s + \gamma_s dB_s -\frac{1}{2} \left(\frac{\mu^2}{\sigma^2} + \gamma_s^2\right) ds\right]\\ &=-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds. \end{align*} That is, \begin{align*} M_t = M_0 \exp\left(-\frac{\mu}{\sigma}W_t - \frac{1}{2} \frac{\mu^2}{\sigma^2}t + \int_0^t \gamma_s dB_s - \frac{1}{2} \int_0^t \gamma_s^2 ds\right). \end{align*}

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