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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.

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  • $\begingroup$ is this a duplicate of your recently asked question? $\endgroup$
    – Attack68 ♦
    Nov 21, 2019 at 6:39
  • $\begingroup$ @Attack68 no it's not a duplicate that one is about the probability distribution of X_t this one is about the properties of X_t itself $\endgroup$
    – ben tenyson
    Nov 21, 2019 at 6:41
  • $\begingroup$ Since each increment is normally distributed with mean $0$, you can check if the increments are bivariate normal (check if linear combinations of them are also going to be normal) and then check that the covariance of the increments is $0$. This would imply that the increments are independent since uncorrelated bivariate normal random variables are independent. $\endgroup$
    – Slade
    Nov 21, 2019 at 11:41
  • $\begingroup$ @Slade Thanks I checked the covariance it doesn't vanish unless s=t or s=0 ( trivial cases). Thus X_t doesn't have independent increments unlike Y_t $\endgroup$
    – ben tenyson
    Nov 21, 2019 at 14:03
  • $\begingroup$ When you multiply the two stochastic integrals together they don't have any overlap in their integral limits. Did you account for this? If you edit your post with your attempt someone can check it out easier. $\endgroup$
    – Slade
    Nov 21, 2019 at 14:05

1 Answer 1

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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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  • $\begingroup$ @is there a formal way to show that $\int_s^t u dW_u$ is independent of X_s? I mean if both integrals were replaced by finite sums then yes but what about their limit? $\endgroup$
    – ben tenyson
    Nov 21, 2019 at 17:07
  • $\begingroup$ As I am aware of, no. $\endgroup$
    – Gordon
    Nov 21, 2019 at 17:35

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