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Why the Scaled Brownian Integral Has Dependent Increments

Article Quant Q&A · Author: ben tenyson
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Summary

The document examines whether the process formed by dividing a time-integrated Brownian motion by time has independent increments. Although the unscaled stochastic integral over a new interval is independent of its earlier history, the time-dependent scaling changes the increment structure.

For times t greater than s, the process difference can be decomposed into a new stochastic integral over the interval from s to t and a term proportional to the earlier process value. The new integral is independent of the earlier value, but the proportional term creates a nonzero covariance between the process increment and its value at time s. The argument therefore shows that these quantities are not independent. It addresses this pair of times and does not claim that all other dependence properties of the process have been characterized.

Key ideas

  • The process scales a Brownian stochastic integral by the reciprocal of time.
  • Its increment separates into a new integral and a term involving the earlier process value.
  • The new integral is independent of the process value at the earlier time.
  • A nonzero covariance shows that the increment and earlier value are dependent.

Tags

StatisticsVolatility
Full text
# Independence of increments of the stochastic process $\frac{1}{t}\int_0^t u dW_u $


# Independence of increments of the stochastic process $\frac{1}{t}\int_0^t u dW_u $












Let $X_t$ be a stochastic process such that

$$X_{t} =\frac{1}{t}\int_0^t u dW_u $$

I know that for

$$Y_{t} =\int_0^t u dW_u$$ $Y_t-Y_s$ is independent of $Y_s$ where $t>s$. But is this also true for $X_t$ which has explicit time dependence in it? Edit The covariance is $$E[X_tX_s] - E[X_s^2]$$

$$E[X_t X_s] =\frac{1}{ts} \cdot E\biggl[\int_t^s u dW_u \int_0^s u dW_u\biggr] +E\biggl[\int_0^s u dW_u \int_0^s dW_u\biggr] $$ The first integral in the first expectation is limit of sequence of normal random variables which are independent of the second one and thus first expectation can be split and using wiener process properties it vanishes.

## Answer by Gordon (score 3, accepted)

https://quant.stackexchange.com/a/49847

Note that, for $t>s>0$, \begin{align*} X_t-X_s &= \frac{1}{t}\int_0^t udW_u - \frac{1}{s}\int_0^s udW_u\\ &=\frac{1}{t}\bigg(\int_s^t u dW_u + \int_0^s udW_u \bigg)- \frac{1}{s}\int_0^s udW_u\\ &=\frac{1}{t} \int_s^t u dW_u + \Big(\frac{1}{t} -\frac{1}{s}\Big)\int_0^s udW_u\\ &=\frac{1}{t} \int_s^t u dW_u - \frac{t-s}{t} X_s. \end{align*} Here, $\int_s^t u dW_u$ is independent of $X_s$. Then \begin{align*} E\big((X_t-X_s) X_s \big) &= -\frac{t-s}{t} E\big(X_s^2\big) \ne 0. \end{align*} That is, $X_t-X_s$ is not independent of $X_s$.

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