Expected Value of a Time-Weighted Geometric Brownian Motion Integral

Summary

The question considers a process defined by integrating time multiplied by the increments of a geometric Brownian motion, and asks for its expectation and variance. The response applies the product rule to the product of time and the asset process, expressing the integral in terms of the terminal asset value and an ordinary time integral. It then computes the expectation by integrating the drift contribution, using the exponential form of the process’s mean.

The displayed result gives an expression for the expectation in terms of the initial value, drift, and time. The treatment does not provide the requested variance: the author explicitly notes that an earlier variance calculation was incorrect and leaves it unresolved. The derivation also assumes the stated GBM dynamics and does not discuss special handling when the drift parameter is zero.

Key ideas

• The product rule relates the time-weighted stochastic integral to the product of time and the asset process.
• The expected increment of the GBM is determined by its drift term because the Brownian increment has zero conditional mean.
• Integrating the drift-weighted expectation yields the stated expectation for the process.
• The variance is not established in the response, so the problem is only partially solved.

Tags

# Evaluating the SDE $dX_t = t\,dS_t$


# Evaluating the SDE $dX_t = t\,dS_t$












The process $S$ is a geometric Brownian motion with an SDE: $dS_t = S_t(\sigma\, dB_t + \mu\, dt)$. I'm stuck evaluating $E(X_t)$ and $V(X_t)$, where $dX_t = t\,dS_t$.

## Answer by Daneel Olivaw (score 4)

https://quant.stackexchange.com/a/49229

Using Itô's Lemma, notice that: $$d(tS_t)=tdS_t+S_tdt=dX_t+S_tdt$$ Hence: $$X_t=tS_t-\int S_udu$$ Using independence of Brownian increments, $E(S_udW_u)=E(S_u)E(dW_u)=0$, and the chain rule for the 4th step: $$\begin{align} E(X_t)&=E\left(\int dX_u\right) \\ &=\int uE(dS_u) \\ &=\int u\mu E(S_u)du \\ &=S_0\int u\mu e^{\mu u}du \\ &=S_0\left(te^{\mu t}-\int e^{\mu u}du\right) \\ &=S_0\left(te^{\mu t}-\frac{1}{\mu}(e^{\mu t}-1)\right) \end{align} $$

[Note: my previous variance calculation was wrong, will fix it when available.]

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