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I'm trying to prove that the squared Brownian motion $(W_t^2)$ doesn't have independent increments, I tried using the covariance and it doesn't quite work, can anybody give me any pointer to how to prove this?

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up vote 1 down vote accepted

Fix some $t\gt s\gt0$ and consider the event $A=[W_t^2\gt W_s^2]$. Then $\mathrm P(A\mid W_s^2=0)=1$ and $\mathrm P(A\mid W_s^2=x)\to\frac12$ when $x\to+\infty$, hence the map $x\mapsto\mathrm P(A\mid W_s^2)$ is not constant.

In particular, $W_t^2-W_s^2$ and $W_s^2$ are not independent.

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Let $0<s<t$. Consider the difference: $$ W_t^2 - W_s^2 = \left( W_t-W_s \right) \left( W_t + W_s \right) = \left( W_t-W_s \right) \left( 2 W_s + (W_t - W_s) \right) $$ By the properties of the Brownian motion, $W_t - W_s$ is independent of $W_s$, but $W_t^2-W_s^2$ explicitly depends on $W_s$, hence $W_t^2-W_s^2$ is not independent on $W_s^2$.

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I'm sorry, I don't understand the hint :-$ – natorro May 18 '12 at 0:50

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