Applying the Itô Product Rule to the Vasicek Integrating Factor

Summary

The note explains why multiplying the Vasicek short-rate SDE by the deterministic factor e^{at} produces the differential d(e^{at}r_t). Using the Itô product rule for e^{at} and r_t, the differential is the sum of e^{at}dr_t and r_t d(e^{at}). Since the factor has finite variation, its covariation with the stochastic rate process is zero, and its differential is a e^{at}dt. This gives the left-hand side in the rearranged equation.

The explanation addresses the calculus step used to solve the linear SDE; it does not derive the full solution or discuss calibration, bond pricing, or model assumptions. Its result applies to this deterministic integrating factor and stochastic rate process, and is a focused clarification rather than a broader treatment of stochastic integration.

Key ideas

• The Itô product rule expands the differential of e^{at}r_t into two terms.
• The factor e^{at} has differential a e^{at}dt.
• The covariation between a deterministic finite-variation factor and the stochastic rate process is zero.
• The expansion matches the left-hand side of the transformed Vasicek equation.

Tags

# Differential of integrating factor $d(e^{at}r_t)$ in Vasicek model


# Differential of integrating factor $d(e^{at}r_t)$ in Vasicek model












I am attempting to solve the Vasicek model SDE (using Wikipedia parametrisation):

$$ dr_t = a(b-r_t)dt + \sigma dW_t $$

Every solution is proceeding to multiply both sides of the equation by the integrating factor $e^{at}$ (akin to solving linear ODEs). After multiplication and rearrangement we get the following equation:

$$ e^{at}dr_t + e^{at}ar_tdt= e^{at}(abdt + \sigma dW_t) $$ Now the left hand side is apparently equal to $d(e^{at}r_t)$. How is that exactly the case?

Is it by Ito product rule? If so what is $X(t)$ and $Y(t)$?

Is it by Ito's lemma but then what is the $f(x,t)$

## Answer by Magic is in the chain (score 5, accepted)

https://quant.stackexchange.com/a/42355

Apply the Ito product rule, noting the cov of a deterministic and stochastic term is zero:

$$\begin{align} d\left(e^{at}r_t\right)&=e^{at} dr_t+r_t de^{at} \\[6pt] &=e^{at} dr_t+r_t e^{at} d(at) \\[6pt] &=e^{at} dr_t+r_t e^{at} a dt \end{align}$$

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