Distribution of Brownian Motion Integrated Over a Time Interval

Summary

The document asks for the distribution of the time integral of standard Brownian motion over an interval from t to T, and challenges an intuitive variance based only on the interval length. The answer rewrites the integral as a stochastic integral with deterministic weight T minus s. Since a stochastic integral of a deterministic function against Brownian motion is normally distributed, its variance is the integral of the squared weight over the interval.

Evaluating that variance gives one third of the cube of T minus t, so the result is a zero-mean normal distribution with that variance. This derivation directly addresses the requested interval and establishes why the same cubic scaling appears as in the integral from zero over a matching duration. The document gives a concise mathematical argument and a textbook reference, but does not discuss extensions such as nonstandard Brownian motion, drift, or conditional distributions.

Key ideas

• The integral of Brownian motion over an interval can be expressed as a weighted stochastic integral.
• A deterministic-integrand Brownian stochastic integral is normally distributed with variance equal to the integrated squared weight.
• For the interval from t to T, the variance is one third of the interval length cubed.
• The resulting integral has zero mean under standard Brownian motion.

Tags

# Integral of brownian motion wrt. time over [t;T]


# Integral of brownian motion wrt. time over [t;T]












From the post Integral of Brownian motion w.r.t. time we have an argument for

$$\int_0^t W_sds \sim N\left(0,\frac{1}{3}t^3\right).$$

However, how does this generalise for the interval $[t;T]$? I.e. what is the distribution of

$$\int_t^T W_sds.$$

I would expect it to be $$\int_t^T W_sds \sim N\left(0,\frac{1}{3}(T-t)^3\right),$$

but I cannot see why.

## Answer by Landscape (score 4)

https://quant.stackexchange.com/a/70160

The last integral is correct as

$$\int_t^T W_s ds = \int_t^T (T-s) dW_s \sim N\left(0, \int_t^T(T-s)^2ds\right) = N\left(0,\frac{1}{3}(T-t)^3\right).$$

Ref. Arbitrage Theory in Continuos Time (Björk, 4th edition)

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