= Solution
Use column vectors for dollar holdings $\theta_t$ and put $b=\mu-r\mathbf1$, $h=\sigma^{-1}b$ and $V=\sigma\sigma^T$. The normalized <state-price density> is
$$
\boxed{\zeta_t=\exp(-rt-h^TW_t-\tfrac12|h|^2t),\qquad d\zeta_t=-\zeta_t(rdt+h^TdW_t).}
$$
The vector $h$ is the <market price of risk>. The density $e^{rt}\zeta_t$ defines an equivalent <risk-neutral measure>, under which $W_t^{\mathbb Q}=W_t+ht$ is <Brownian motion> and every stock has drift $r$. A terminal claim $H$ with finite state-price cost is priced at $\zeta_t^{-1}\mathbb E[\zeta_TH\mid\mathcal F_t]$. Nonsingularity of $\sigma$ and the natural Brownian market information make this a <complete market>; in particular the optimal nonnegative payoff below can be replicated.
Let $q=n^{-1}\mathbf1$ and $v=\sigma^Tq$. Apply the <Itô formula> to each log price, whose return variance is $V_{ii}$:
$$
d\log S_t^i=\sum_j\sigma_{ij}dW_t^j+(\mu_i-\tfrac12V_{ii})dt.
$$
Summing and dividing by $n$ proves the <geometric stock index> formula
$$
\boxed{\log(J_t/J_0)=\frac1n\left[\mathbf1^T\sigma W_t+\left(\mathbf1^T\mu-\frac12\operatorname{tr}V\right)t\right].}
$$
Thus $J_t=J_0\exp(v^TW_t+a_Jt)$, where $a_J=q^T\mu-\operatorname{tr}(V)/(2n)$. The factor $1/n$ applies to the whole bracket. The geometric average need not itself be a <self-financing portfolio>; its instantaneous drift is $a_J+|v|^2/2$.
If $\theta_t^i$ is the dollar amount invested in stock $i$, the <bank account> holds $w_t-\mathbf1^T\theta_t$. The <self-financing portfolio> equation is
$$
\boxed{dw_t=[rw_t+\theta_t^Tb]dt+\theta_t^T\sigma dW_t.}
$$
Equivalently, $\pi_t=\theta_t/w_t$ gives $dw_t/w_t=(r+\pi_t^Tb)dt+\pi_t^T\sigma dW_t$. For share-holding notation replace $\theta_t^i$ by $S_t^i$ times the number of shares.
Static <expected utility maximization> uses the terminal budget $\mathbb E[\zeta_Tw_T]\leq w_0$. Since $J_T$ is an exogenous positive benchmark, differentiating $U(x/J_T)$ in $x$ gives $J_T^{-1}U'(x/J_T)$. An interior <Lagrange multiplier> $\lambda>0$ for the budget therefore gives
$$
\frac1{J_T}U'(w_T^*/J_T)=\lambda\zeta_T.
$$
For <CRRA utility>, write $p=1/R$. Since $U'(y)=y^{-R}$, solving this marginal equation gives
$$
\boxed{w_T^*=\lambda^{-p}\zeta_T^{-p}J_T^{1-p}.}
$$
Every positive lognormal moment is finite here, so the budget determines the constant uniquely. Put
$$
K_T=\mathbb E[(\zeta_TJ_T)^{1-p}].
$$
Then
$$
\boxed{\lambda^{-p}=w_0/K_T,\qquad\lambda=(K_T/w_0)^R.}
$$
This is also a sufficiency condition: <concavity> gives $U(X/J_T)-U(w_T^*/J_T)\leq\lambda\zeta_T(X-w_T^*)$, and taking expectations and using the budget inequality proves global optimality.
To find the whole optimal process, set $c=a_J-r-|h|^2/2$ and
$$
A=(1-p)c+\frac12(1-p)^2|v-h|^2.
$$
Since $\zeta_tJ_t=J_0\exp((v-h)^TW_t+ct)$, the <moment-generating function of a normal distribution> gives
$$
K_T=J_0^{1-p}e^{AT},\qquad
\mathbb E[(\zeta_TJ_T)^{1-p}\mid\mathcal F_t]=(\zeta_tJ_t)^{1-p}e^{A(T-t)}.
$$
Pricing the terminal payoff yields the explicit <benchmark-relative power-utility portfolio> wealth
$$
\boxed{w_t^*=\lambda^{-p}\zeta_t^{-p}J_t^{1-p}e^{A(T-t)}
=w_0\zeta_t^{-p}(J_t/J_0)^{1-p}e^{-At}.}
$$
Its proportional diffusion coefficient is $\ell=ph+(1-p)v$. Since the wealth diffusion coefficient for dollar fractions is $\sigma^T\pi$, replication requires
$$
\sigma^T\pi^*=ph+(1-p)\sigma^Tq.
$$
Using $\sigma^{-T}h=V^{-1}b$ gives the requested fixed proportions:
$$
\boxed{\pi^*=R^{-1}V^{-1}(\mu-r\mathbf1)+(1-R^{-1})\frac1n\mathbf1,\qquad\theta_t^*=w_t^*\pi^*.}
$$
The drift matches the <self-financing portfolio> equation as well. Indeed, $\ell=h+(1-p)(v-h)$, so the drift of $\log w^*$ from the explicit formula is
$$
r+\frac12|h|^2-\frac12(1-p)^2|v-h|^2
=r+\ell^Th-\frac12|\ell|^2
=r+(\pi^*)^Tb-\frac12(\pi^*)^TV\pi^*.
$$
Consequently an equivalent form of the solution is
$$
\boxed{w_t^*=w_0\exp\left(\left[r+(\pi^*)^Tb-\tfrac12(\pi^*)^TV\pi^*\right]t+(\sigma^T\pi^*)^TW_t\right).}
$$
The <bank account> fraction is $1-\mathbf1^T\pi^*$, with negative values representing borrowing. The first stock term is ordinary power-utility risk exposure, while the second hedges the random benchmark. At the logarithmic limit $R=1$ the benchmark term disappears, consistently with $\log(w_T/J_T)=\log w_T-\log J_T$.
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