Applying Itô’s Product Rule to a Discounted Stochastic Value

Summary

The document describes a question about deriving the differential of a discounted contract value whose state variables follow a chain of stochastic differential equations. Applying Itô’s formula to the value function produces drift terms from time evolution, state drift, and the quadratic variation of the diffusive state, plus a Brownian increment term. The states with deterministic dynamics contribute through their stated drift relationships.

The remaining step is the product rule for the discount factor and the value function. Since the discount factor is deterministic, its differential contributes a term equal to the negative rate times the discounted value; its cross variation with the value process is zero. Combining this with the Itô differential gives the displayed expression for the change in contract value. The source contains the question and its attempted derivation, but no answer or independent validation, so the explanation follows the mathematical rule implicit in the expressions rather than reporting a supplied resolution.

Key ideas

• Itô’s formula adds drift, diffusion, and quadratic-variation terms for the stochastic value function.
• The deterministic state equations contribute their drift values to the chain rule.
• The discount factor contributes a negative-rate term under the product rule.
• A deterministic discount factor has zero quadratic covariation with the value process.
• The document presents the derivation question but does not include a response or validation.

Tags

# Ito formula and confusion with the differential operator $d$


# Ito formula and confusion with the differential operator $d$












Thanks for visiting my question.

Im am currently working on this paper (https://arxiv.org/abs/2305.02523) and I am stuck at page 21 (Theorem 14 proof). First these SDE's were defined: \begin{align*} dZ_t&=Y_tdt\\ dY_t&=X_tdt\\ dX_t&=A(t,X_t)dt+B(t,X_t)dW_t \end{align*} And we define "the value of the contract" by: $P=de^{-rt} v(t,x,y,z)$ Then the authors use the general Ito formula to derive: ($v_x$ is the derivative with respect to x) $dv(t,x,y,z)=v_tdt+v_xdx+\frac{1}{2}v_{xx}d[x,x]+v_ydy+v_zdz$ up to this point I understand the proof. Then they want to calculate $dP$ which is given without further explanation by: $dP=e^{-rt}(-rv+v_t+\frac{1}{2} v_{xx}B(t,X_t)^2+v_xA(t,X_t)+xv_y+yv_z )dt+ v_xB(t,X_t)e^{-rt}dW_t$ I dont unterstand how to get to this formula

My attempt: starting from $dv(t,x,y,z)=v_tdt+v_xdx+\frac{1}{2}v_{xx}d[x,x]+v_ydy+v_zdz$ i have plugged in the SDE's: e.g. $dz=ydt$, $dy=xdt$, $dx=A(t,X_t)dt+B(t,X_t)dW_t$ and $d[x,x]=B(t,X_t)^2dt$ $\Rightarrow dv(t,x,y,z)=(v_t+\frac{1}{2} v_{xx}B(t,X_t)^2+v_xA(t,X_t)+xv_y+yv_z )dt+v_xB(t,X_t)dW_t$

Now i used: $P=de^{-rt} v(t,x,y,z)=(de^{-rt})v(t,x,y,z)+e^{-rt} (dv(t,x,y,z))$ Is this correct? assuming this is correct and with our results from bevor we get: $P=e^{-rt}(-rv+v_t+\frac{1}{2} v_{xx}B(t,X_t)^2+v_xA(t,X_t)+xv_y+yv_z )dt+ v_xB(t,X_t)e^{-rt}dW_t$ But that is the same formula that the authors get for $dP$ So there must be a mistake

I dont realy understand how to handle $P=de^{-rt} v(t,x,y,z)$ and $dP=d(de^{-rt} v(t,x,y,z))$

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