Relating Time and Brownian Integrals of a Lognormal Process

Summary

The document considers two integrals of an exponential Brownian process: one integrated over time and another integrated with respect to Brownian motion. It asks how to compute them and whether either integral has a familiar distribution, such as a lognormal one. The main method offered is Itô’s formula applied to the exponential process, which yields an identity involving both integrals and the process value at the terminal time.

That identity links the two quantities but does not by itself evaluate either integral or establish its distribution. In particular, knowing the terminal exponential does not determine the separate time integral without additional information. The excerpt gives no closed-form distribution, numerical method, or assumptions beyond the displayed process. It is therefore a useful starting point for applying stochastic calculus, but it leaves the original distributional question unresolved.

Key ideas

• Applying Itô’s formula to an exponential Brownian process produces drift and Brownian integral terms.
• The identity relates the time integral and the Brownian integral through the process value at the endpoint.
• The relation alone does not determine either integral as a standalone random variable.
• The excerpt does not show that either integral follows a lognormal distribution.

Tags

# How to compute the stochastic integral of log-normal process?


# How to compute the stochastic integral of log-normal process?












How do you compute the following integral:

$$\int_0^t e^{\mu s + \sigma W_s} ds$$

or

$$\int_0^t e^{\mu s + \sigma W_s} dW_s$$

?

Are those integrals stochastic processes of some well-know type (log-normal, for instance)?

Since they are interrelated through Ito's integral, if we compute one of those we will immediately get the other one, so It doesn't matter which integral to consider:

Ito's integral for $e^{\mu t + \sigma W_t}$:

$e^{\mu t + \sigma W_t} = e^{W_0} + \int\limits_0^t (\mu + \frac{\sigma^2}{2})e^{\mu s + \sigma W_s} ds + \int\limits_0^t \sigma e^{\mu s + \sigma W_s} d W_s $

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