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For a maturity-differentiable Heath-Jarrow-Morton model forward field with the integrability needed for stochastic differentiation, the short rate $r_t=f(t,t)$ has the displayed dynamics. Indeed changing observation time uses the forward-rate stochastic differential equation, while changing maturity adds ${\partial }_{T}f(t,t)dt$. Under a risk-neutral measure, $\alpha (t,T)=\sigma (t,T){\int }_t^T\sigma (t,u)du$, so $\alpha (t,t)=0$. The remaining drift depends on the diagonal slope of the whole forward curve, which need not be determined by $r_t$; a Markov process for the short rate is an additional modelling restriction. Without sufficient maturity regularity, the diagonal need not admit this classical Itô process formula.

For an Itô process $d{X}_t=a_{t}dt+b_{t}d{W}_t$ and $F\in C^{1,2}$, the differential Itô formula iswhere the derivatives are evaluated at $(t,X_t)$. A multidimensional version replaces $b_t^2{F}_{xx}$ by the contraction of the Itô diffusion covariance matrix with the spatial Hessian.

For a heuristic derivation, partition time and use the Euler increment $\Delta X=a\Delta t+b\sqrt{\Delta t}Z$, where $Z$ is standard normal. A second-order Taylor expansion givesThe term $b^2(\Delta W{)}^2$ has order $\Delta t$, so it survives summation; ordinary first-order calculus would incorrectly discard it. For independent Brownian motion increments,This gives the quadratic variation $[W{]}_t=t$. The summed mixed $dtdW$ terms and $d{t}^2$ terms vanish, while the linear random increments converge to an Itô integral. After localizing bounded coefficients and derivatives, Taylor remainders are negligible and the surviving terms give the formula. Thus $(dW{)}^2=dt$ is shorthand for a statement about summed quadratic variation, not an equality of individual random increments.

Suppose the forward field is differentiable in maturity with the stochastic integrability needed for differentiation. As time increases, both arguments of $r_t=f(t,t)$ increase. The diagonal short-rate dynamics in the Heath-Jarrow-Morton model are thereforeUnder the HJM drift restriction from part (a), ${\alpha }^Q(t,t)=0$, givingMore explicitly the slope isIt depends on the forward curve and potentially the whole history, not necessarily on $r_t$ alone. Without the stated maturity regularity, the original forward equations do not by themselves guarantee this classical Itô process formula for the diagonal.

The short rate need not be Markov, even with one Brownian factor and deterministic volatility. An explicit example of One Brownian factor does not imply a Markov short rate takes $f_0(T)=0$ and $\sigma (t,T)=T-t$. Its no-arbitrage forward drift is ${\alpha }^Q(t,T)=(T-t{)}^3/2$, soThe continuous rate history determines $W_t$ from the left derivative of $I_t$. Hence for $h>0$,But $W_t$ is not determined by $I_t$: joint normality, $\mathrm{Var}{I}_t=t^3/3$, and $\mathrm{Cov}(W_t,I_t)=t^2/2$ giveThus the conditional mean of the future given the whole rate history cannot be a function of the current rate alone, contradicting the Markov property. Enlarging the state to $(I_t,W_t)$ gives a two-dimensional Markov process; special HJM specifications can instead close on a single short rate.

Concavity gives the supporting-tangent inequalityMultiply by the discount factor $e^{-bt}$ and use the first-order condition $e^{-bt}{U}^{\prime }(c_t)=Y_t$. This gives the pointwise marginal-utility verification of optimal consumption inequalityApply the budget inequality from part (d) to the competing admissible strategy, and use equality for the proposed one:The two weighted consumption integrals are finite, so their difference is integrable. Under the usual positive discount-rate assumption $b>0$, the utility integrals are also integrable: $U(0)\le U(x)\le L$ for a finite upper bound $L$, and ${\int }_0^{\infty }{e}^{-bt}dt=1/b$. Integrating the tangent inequality and taking expectations therefore givesEconomically, both consumers face the same state-price budget, and the candidate spends it exactly where its discounted marginal utility equals the state price. This is utility duality with martingale deflators in its consumption form.

A positive $b$, or another hypothesis making the infinite-horizon objectives well defined and permitting this integration, is needed. The printed question does not specify the sign of $b$. Bounded utility alone does not ensure that an undiscounted infinite time integral exists: a bounded integrand can have both infinite positive and negative parts. Thus the conclusion is established under the standard discount convention $b>0$, and also whenever the displayed objectives satisfy the stated integrability conditions; without either convention the literal infinite-horizon comparison need not be a defined mathematical expression.

This can occur within an admissible financial model, not just for an abstract bounded integrand. Take $U(x)=1-2e^{-x}$ and $b=0$. Choose a deterministic smooth nonnegative $c$ with successive unit-length plateaus alternating between zero and the integer $n$, and connect them over intervals whose lengths have finite sum. Then ${\int }_0^{\infty }2c_t{e}^{-c_t}dt<\infty$: the high-plateau contributions sum to ${\sum }_{n\ge 1}2n{e}^{-n}$ and the transition integrand is bounded by $2/e$. SetThe bank account is a positive deterministic Itô process and $Y_t{B}_t=Y_0$, so $Y$ is a state-price density. Holding $X_t/B_t$ bank units finances consumption with nonnegative wealth and exact budget equality. Also $U^{\prime }(c_t)=Y_t$. Nevertheless every zero plateau contributes $-1$ to the utility integral, while each sufficiently high plateau contributes a fixed positive amount. Both its negative and positive parts are infinite, so the undiscounted objective is undefined. This establishes why the missing discount or objective-integrability hypothesis is substantive.