= Solution
\b[Pricing kernel and replication.] In the nondegenerate <Black-Scholes model>, put $\kappa=(\mu-r)/\sigma$ and normalize the <state-price density> by $\zeta_0=1$. The process is
$$
\boxed{\zeta_t=\exp\{-rt-\kappa W_t-\tfrac12\kappa^2t\},\qquad
d\zeta_t=-\zeta_t(r\,dt+\kappa\,dW_t).}
$$
The density $e^{rt}\zeta_t$ changes probability to the <risk-neutral measure>. Under that measure $W_t+\kappa t$ is a <Brownian motion> and the stock drift is $r$. Thus an integrable <contingent claim> $H$ has time-$t$ price
$$
\boxed{P_t=\frac{\mathbb E[\zeta_T H\mid\mathcal F_t]}{\zeta_t}
=e^{-r(T-t)}\mathbb E^{\mathbb Q}[H\mid\mathcal F_t].}
$$
In the usual augmented <natural Brownian filtration>, the <Brownian martingale representation theorem> supplies a <replicating strategy>; this is the <complete market> assumption. For a nonnegative <admissible trading strategy> without intermediate <consumption>, the <state-price budget constraint> is $\mathbb E[\zeta_Tw_T]\leq w_0$, with equality for a fully invested replicated claim. In particular
$$
\mathbb E\zeta_T=e^{-rT},\qquad \mathbb E[\zeta_TS_T]=S_0.
$$
\b[Feasibility and the largest slope.] First take the intended regime $r>0$, $T>0$, $w_0,S_0>0$, and the usual nonnegative <portfolio wealth> constraint. The <terminal wealth floor> is
$$
\xi=w_0+\alpha(S_T-S_0).
$$
For any feasible claim the <state-price budget constraint> implies
$$
w_0\geq\mathbb E[\zeta_Tw_T]\geq
\mathbb E[\zeta_T\xi]
=w_0e^{-rT}+\alpha S_0(1-e^{-rT}).
$$
Therefore $\alpha\leq w_0/S_0$. Conversely, when $0\leq\alpha\leq w_0/S_0$, hold $\alpha$ shares and put the remaining $w_0-\alpha S_0$ in the <continuous-time bank account>. Its terminal <portfolio wealth> is
$$
\alpha S_T+(w_0-\alpha S_0)e^{rT}
\geq\alpha S_T+(w_0-\alpha S_0)=\xi.
$$
Hence
$$
\boxed{\bar\alpha=\frac{w_0}{S_0}.}
$$
At equality, $\xi=\bar\alpha S_T$ has cost exactly $w_0$. Positivity of the <state-price density> forces $w_T=\bar\alpha S_T$ almost surely: any strict improvement would cost more. \b[Invest all initial <portfolio wealth> in $w_0/S_0$ shares and hold them until $T$.]
\b[Optimal payoff below the feasibility limit.] For $\alpha<\bar\alpha$ the floor is strictly positive, and its price is strictly less than $w_0$. Assume the <utility function> is increasing, differentiable and strictly <concave>, satisfies the <Inada conditions>, and has the integrability needed for the finite-budget optimization. These are the usual hypotheses implicit in using <inverse marginal utility>. For each positive multiplier $\lambda$, maximize
$$
U(y)-\lambda\zeta_Ty\qquad\text{over }y\geq\xi
$$
separately in every state. Its derivative decreases through zero at $I(\lambda\zeta_T)$, so the <floored marginal utility optimizer> is
$$
\boxed{w_T^*=\max\{w_0+\alpha(S_T-S_0),\,I(\lambda\zeta_T)\}.}
$$
The multiplier is characterized by
$$
\boxed{\mathbb E\!\left[\zeta_T\max\{\xi,I(\lambda\zeta_T)\}\right]=w_0,\qquad \lambda>0.}
$$
Under the stated integrability hypotheses the left side is continuous and decreasing, tends to the floor cost as $\lambda\uparrow\infty$, and tends to infinity as $\lambda\downarrow0$. It is strictly decreasing wherever it exceeds the floor cost: on the event where the inverse-marginal-utility payoff exceeds the floor, a larger multiplier strictly reduces that payoff. Therefore the budget determines a unique finite multiplier. For <CRRA utility> the <inverse marginal utility> is $I(y)=y^{-1/R}$; lognormal moments provide the needed integrability.
For completeness, pointwise maximality gives, for any feasible competing terminal <portfolio wealth> $Y$,
$$
U(Y)-\lambda\zeta_TY
\leq U(w_T^*)-\lambda\zeta_Tw_T^*.
$$
Taking expectations and using $\mathbb E[\zeta_TY]\leq w_0=\mathbb E[\zeta_Tw_T^*]$ proves optimality. Strict <concavity> gives uniqueness of the terminal claim. Its price process
$$
w_t^*=\frac{\mathbb E[\zeta_Tw_T^*\mid\mathcal F_t]}{\zeta_t}
$$
is nonnegative and starts from $w_0$; <claim replication> therefore turns the payoff optimizer into an admissible portfolio.
\b[What the missing interest-rate hypothesis changes.] The PDF does not explicitly assume $r>0$ or give the utility and admissibility hypotheses above. These omissions matter. At $r=0$, under nonnegative admissibility, the largest feasible slope remains $w_0/S_0$: for larger slopes the positive-part floor costs strictly more than $w_0$, since $S_T$ has support $(0,\infty)$. But every $\alpha\leq w_0/S_0$ already gives floor cost exactly $w_0$, so the only feasible terminal claim is $\xi$. There is no spare budget for an inverse-marginal-utility improvement. For example $\alpha=0$, $\kappa\ne0$ and <CRRA utility> make $\mathbb E[\zeta_T\max\{w_0,I(\lambda\zeta_T)\}]>w_0$ for every finite $\lambda$. The prescribed positive finite multiplier then does not exist.
For negative $r$, nonnegative admissibility requires the effective floor $\xi_+=\max\{\xi,0\}$. Define its cost
$$
G(\alpha)=\mathbb E[\zeta_T(w_0+\alpha(S_T-S_0))_+].
$$
This is continuous and convex. For $\alpha\leq w_0/S_0$ it equals $w_0e^{-rT}+\alpha S_0(1-e^{-rT})$, exceeding $w_0$ below the endpoint. At $\alpha_0=w_0/S_0$, $G(\alpha_0)=w_0$ and $G'(\alpha_0)=S_0(1-e^{-rT})<0$. Above that endpoint the lognormal stock gives strict convexity, and $G(\alpha)\to\infty$. Thus there is a unique second root $\alpha_1>\alpha_0$ of $G(\alpha_1)=w_0$, the feasible slopes are $[\alpha_0,\alpha_1]$, and the largest is $\alpha_1$. Equivalently, for $\alpha>\alpha_0$ the floor price is $\alpha$ times the <European call option> price with strike $S_0-w_0/\alpha$. At the upper endpoint replicate $\xi_+$; in the interval with strict budget slack the same <floored marginal utility optimizer> applies with $\xi_+$.
If instead <portfolio wealth> may be negative and utility is defined on all real <portfolio wealth>, the floor itself has its affine replication cost: at $r=0$ every slope is feasible, and at negative $r$ every $\alpha\geq w_0/S_0$ is feasible. There is then no largest finite slope. \b[The intended stock-only endpoint and strict-slack optimizer use positive interest and standard nonnegative admissibility.]
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