Evaluating an Integrated Stochastic Differential with a Deterministic Factor

Summary

The question checks how to integrate the differential of a stochastic process multiplied by a time-dependent deterministic exponential over an interval. The response confirms the endpoint expression by rewriting the product with a transformed process: let the transformed process equal the original process multiplied by the inverse exponential in its running time. The original integral can then be expressed as the deterministic factor at the terminal time times the integral of the transformed process differential.

Integrating a process differential over the interval gives the transformed process at the upper endpoint minus its value at the lower endpoint. Substituting back yields the terminal process value minus the initial value multiplied by the exponential evaluated across the interval. The answer says the proposed endpoint result is correct and flags a likely notation typo in the initial equation. The explanation is a basic sanity check and does not specify assumptions on the process or address more general stochastic product rules where additional terms may arise.

Key ideas

• A differential integrated over an interval gives the change in the underlying process between its endpoints.
• A deterministic exponential factor can be used to define a transformed process for simplifying the integral.
• Evaluating the transformed process at the interval endpoints recovers the proposed expression.
• The running variable in the differential notation should be consistent with the process argument.
• The answer does not cover more general stochastic product rules or their assumptions.

Tags

# Integration of stochastic total derivative


# Integration of stochastic total derivative












Super basic question. I think I am doing this correctly, but just want a sanity check.

Say I have a stochastic process $r(t)$.

Say I have an equation

$$d(e^{\beta (t-s)}r(s))=\dots$$

where the $e^{\beta (t-s)}$ term is deterministic and $t\geq s > 0$.

Then say we want to integrate both sides from $s$ to $t$

$$\int_s^t{d(e^{\beta (t-u)}r(u))}=\dots$$

we then have

$$e^{\beta (t-t)}r(t)-e^{\beta (t-s)}r(s)=\dots$$

$$r(t)-e^{\beta (t-s)}r(s)=\dots$$

Please set me straight if I have this wrong. Thanks.

## Answer by wsw (score 4, accepted)

https://quant.stackexchange.com/a/9087

I think there is a typo in your first equation. The running variable should be $s$, as in $d\left( e^{\beta(t-s)} r(s) \right)$.

Let's start with your integral. Let $R_u = e^{-\beta u} r_u$. Your integral becomes $$ e^{\beta t} \int_s^t d R_u \, . $$ Recall that $dR_u = R_{u+du} - R_u$. The integral evaluates to $e^{\beta t}(R_t -R_s)$, which simplifies to $r_t - e^{\beta(t-s)} r_s$, which is what you got.

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