Discounted Expectations for Functions Under Geometric Brownian Motion

Summary

The document asks how to value the infinite-horizon discounted expectation of a twice differentiable function of a geometric Brownian motion when that function satisfies a differential equation involving the process generator. It proposes that the expectation depends on the relation between the discount rate and the function’s generator eigenvalue: above that rate difference, the value is the function divided by the gap, with a constant term; at or below the threshold, it diverges.

The accepted response derives the result by applying Itô’s lemma to the function and using its defining differential equation to obtain an ordinary differential equation for its expectation. Solving that equation and integrating yields the proposed formula and divergence cases. The argument invokes dominated convergence on the assumption that the function is bounded. The source gives no detailed proof of divergence or separate derivation for the boundary case, and its conclusion should therefore be read with its stated regularity and boundedness assumptions in mind.

Key ideas

• Applying Itô’s lemma converts the function’s generator equation into an evolution equation for its expectation.
• The discounted integral is finite when the discount rate exceeds the generator eigenvalue, under the stated assumptions.
• At and below the threshold, the accepted answer says the infinite-horizon expectation diverges.
• The derivation relies on boundedness of the function to justify dominated convergence.

Tags

# Discounted expectation of generic $\mathbb{C}^2$ function


# Discounted expectation of generic $\mathbb{C}^2$ function












Consider a standard geometric Brownian motion $V_t$ with drift $\mu<r$ and standard deviation $1$.

It holds that the discounted expectation is $$E\left[\int_t^\infty e^{-r(s-t)} V_s ds | V_t \right] = \frac{V_t}{r-\mu}$$ provided that $r > \mu$ and diverges for $r < \mu$.

Does a similar result hold for any $\mathbb{C}^2$ function? That is consider a function that satisfies $$R f(V) = c + \mu V f'(V) + V^2 f''(V)$$ What is the discounted expectation of it for possible values of $R-r$: $$E\left[\int_t^\infty e^{-r(s-t)} f(V_s) ds | V_t \right] = ?$$

My current guess is that the solution is:

- For $r > R$, the solution is $\frac{f(V_t)}{r-R} + const$.

- For $r < R$, the expectation diverges.

- Not sure for $r = R$.

To calculate 1, I have derived an ODE for the expectation and used $f$ coming from the solution of its ODE as a non-homogenous term. Then solved with variation of coefficients. But I am not sure how to prove point 2 and how to proceed with 3.

## Answer by Pollo Gi (score 1, accepted)

https://quant.stackexchange.com/a/73455

The guessed solution is correct. Also the case $r=R$ diverges. To find the solution solve the ODE for the expectation of $f(V)$:

$$d\mathbb{E}(f) = R \mathbb{E}(f) - c$$ which is obtained from applying ito's lemma to $f$ and replacing the ODE for the drift term. The solution can be replaced into the expectation (DCT applies because $f$ is bounded).

The rest is integration.

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