Applying Itô’s Lemma to a Discounted Risky Asset

Summary

The note derives the dynamics of a risky asset after discounting by a riskless account that grows continuously at rate r. Under the geometric Brownian motion specification used in the answer, the asset has proportional return dynamics with drift μ and volatility σ, while the money-market account satisfies dM = rM dt. Applying Itô’s lemma to the ratio S/M gives a discounted asset whose proportional drift is μ − r and whose diffusion term remains σ dZ.

The subtraction arises because the denominator grows at rate r, so discounting removes that growth from the asset’s drift. The answer writes the resulting equation in proportional form, dS*/S* = (μ − r)dt + σ dZ. It also flags ambiguity in the question’s original notation: the displayed asset equation omits a factor of S, whereas the derivation assumes geometric Brownian motion. The derivation is therefore conditional on that intended proportional-process model, and should not be read as applying unchanged to an arithmetic process.

Key ideas

• When a risky asset follows geometric Brownian motion, its returns have drift μ and volatility σ.
• A riskless account growing at rate r accumulates value in the denominator of the discounted asset.
• Applying Itô’s lemma to S/M subtracts r from the asset’s proportional drift.
• The discounted asset retains the same Brownian shock coefficient under the stated model.
• The derivation assumes geometric Brownian motion; the question’s displayed equation is ambiguous without the asset-price factor.

Tags

# Discounted risky asset stochastic process problem


# Discounted risky asset stochastic process problem












$S_t$ is the random variable representing the risky asset price at time $t$.

M_t is the riskless asset. They are governed by the equations

$\frac{dS_t}{dt}=\mu dt + \sigma dZ_t$ and

$dM_t = rM_t dt$

where $Z_t$ is Brownian motion. If we define the discounted risky asset by $S_t^{*}=S_t/M_t$. How does the process $S_t^{*}$ become governed by

$\frac{dS_t^{*}}{dt}=(\mu-r) dt + \sigma dZ_t$ ?

I cannot see why you subtract $rdt$.

## Answer by Chinny84 (score 5, accepted)

https://quant.stackexchange.com/a/14416

$$ \textbf{Preface} $$ I am assuming log normal asset but this is not clear from the question? Or rather I have misinterpreted the question!

Well as I see it from a a purely mathematical exercise $$ d\left(\dfrac{S_t}{M_t}\right) =\frac{1}{M_t}dS_t - \frac{S_t}{M_t^2}dM_t +O(dt^2) $$

using Ito's lemma.

Then we can sub in the original processes yields

\begin{align} d\left(\dfrac{S_t}{M_t}\right)&=&\frac{1}{M_t}S_t\left(\mu dt + \sigma dZ_t\right) - \frac{S_t}{M_t}\frac{1}{M_t}\left(M_t r dt\right)\\ &=& S^{*}_t\left(\mu dt +\sigma dZ_t\right) - S^{*}_trdt \\ &=& S^*_t\left[(\mu-r)dt+\sigma dZ_t\right] \end{align}

or finally $$ \frac{dS^*_t}{S^*_t} = (\mu-r)dt+\sigma dZ_t $$

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.