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Deriving the Power-Numeraire Measure Change from a Lognormal Expectation

Article Quant Q&A · Author: solid

Summary

The document explains a measure change from the risk-neutral measure to an alpha measure used with a power numeraire in option pricing. It starts from a geometric Brownian motion for the underlying and expresses the Radon–Nikodym derivative in terms of the terminal power of the asset and the corresponding numeraire value.

The key calculation is the time-zero numeraire, obtained as a discounted conditional expectation of the powered asset. Since the Brownian term is normally distributed, its exponential has a lognormal expectation that contributes a variance adjustment. Substituting this numeraire into the density ratio yields an exponential with drift correction proportional to alpha squared and a Brownian term proportional to alpha. The note gives an algebraic derivation for the stated model; it does not discuss broader assumptions or extensions.

Key ideas

  • The alpha-measure density is formed from the powered asset and its numeraire, discounted under the risk-neutral measure.
  • The time-zero numeraire is calculated as a discounted conditional expectation of the terminal powered asset.
  • The exponential of a normally distributed Brownian term contributes a variance adjustment through its lognormal expectation.
  • Substitution produces a density with an alpha-squared drift correction and an alpha-scaled Brownian term.

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# Radon–Nikodym derivative of $\alpha$-measure


# Radon–Nikodym derivative of $\alpha$-measure












I'm struggling a bit to understand how the Radon–Nikodym derivative is computed in pag.3 of this paper written by Mark Joshi titled "THE USE OF POWER NUMERAIRES IN OPTION PRICING". Given a process defined as $dS_t=\mu S_tdt + \sigma S_t dW_t$, the change of measure $\frac{\mathrm{d}\mathbb Q^1}{\mathrm d\mathbb Q^0}$ from bank account numeraire $(\alpha=0)$ to the stock measure $(\alpha=1)$ is clear:

\begin{align*} \frac{\mathrm{d}\mathbb Q^1}{\mathrm d\mathbb Q^0}=\frac{S_T}{S_0}\frac{B_0}{B_T}=\frac{S_T}{S_0}e^{-rT}=\frac{S_0e^{(r-\frac{1}{2}\sigma^2)T+\sigma W_T -rT}}{S_0}=e^{-\frac{1}{2}\sigma^2T+\sigma W_T} \end{align*}

I'm confused by the definition of the change of measure $\frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0}$ from the risk neutral measure $(\alpha=0)$ to the $\alpha$-measure, given that $S^\alpha_T=S^\alpha_0e^{\alpha(r-\frac{1}{2}\sigma^2)T+\alpha \sigma W_T}$:

\begin{align*} \frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0} = \frac{N_{T,T}^\alpha B_0}{N_{0,T}^\alpha B_T}=\frac{N_{T,T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\frac{S_{T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\color{red}{e^{-\frac{1}{2}\alpha^2\sigma^2T+\alpha\sigma W_T}}. \end{align*}

## Answer by solid (score 1, accepted)

https://quant.stackexchange.com/a/79619

\begin{align*} \frac{\mathrm{d}\mathbb Q^\alpha}{\mathrm d\mathbb Q^0} = \frac{N_{T,T}^\alpha B_0}{N_{0,T}^\alpha B_T}=\frac{N_{T,T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\frac{S_{T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\color{red}{e^{-\frac{1}{2}\alpha^2\sigma^2T+\alpha\sigma W_T}}. \end{align*}

I was confused by the definition of $N^\alpha_0=S_0^\alpha \exp\left(rT(a-1)+0.5\sigma^2(a^2-\alpha)T\right)$ given that $S_T^\alpha=S_0^\alpha \exp\left(\alpha(r-0.5\sigma^2)T+\alpha\sigma W_T\right)$. To be specific I did not understood the red step in the definition of $N^\alpha_0$:

\begin{align*} N_0^\alpha &= e^{-rT}\mathbb{E}^{\mathbb Q}[S_T^\alpha|\mathcal{F}_0] \\ &= e^{-rT}\mathbb{E}^{\mathbb Q}\left[S_0^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)T+\alpha\sigma W_T \right)\bigg|\mathcal{F}_t\right] \\ &= e^{-rT}S_0^\alpha\exp\left(\alpha\left(r-\frac{1}{2}\sigma^2\right)T+\frac{1}{2}\alpha^2\sigma^2T\right) \\ &= S_0^\alpha \exp\left(rT(\alpha-1)+0.5\sigma^2(\alpha^2-\alpha)T\right) \end{align*}

Turns out to to be the expectation of a lognormal variable: $$\mathbb EY^{\alpha}=\mathbb Ee^{{\alpha}\mu+{\alpha}\sigma U}=e^{{\alpha}\mu}\mathbb Ee^{{\alpha}\sigma U}=e^{{\alpha}\mu}M_U({\alpha}\sigma)=e^{{\alpha}\mu+\frac12{\alpha}^2\sigma^2}$$

Afterwards it is pretty straightforward:

\begin{align*} \frac{N_{T,T}^\alpha e^{-rT}}{N_{0,T}^\alpha}=\frac{S_0^\alpha \exp\left(\alpha(r-0.5\sigma^2)T+\alpha\sigma W_T\right) e^{-rT}}{S_0^\alpha \exp\left(rT(\alpha-1)+0.5\sigma^2(\alpha^2-\alpha)T\right)}= e^{-\frac{1}{2}\alpha^2\sigma^2T+\alpha\sigma W_T} \end{align*}

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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