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Let $X_t$ be a Squared Bessel Process ($BESQ$). Define $Y_t=∫_0^tX_sds$. Do we know whether $\lim_{t→∞}Y_t=∫_0^∞ X_sds$ is finite or infinite? Does it depend on $BESQ$ parameter?

Edit. It is obvious that $\int_0^{\infty}X_s ds = \infty$ when dimension parameter $\delta > 2$, since in this case $X_t$ is transient.

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Infinite, except possible in the case $dX_t = \sigma \sqrt {X_t} dW_t$. The recurrent case is handled as or more easily that the transient. Look at the time interval at a collection of return times, e.g. $T_n = \inf \lbrace t > T_{n-1} + 1 : X_t = 1 \rbrace$. The integral from $T_1$ to $T_n$ is a sum of n-1 i.i.d positive r.v.'s and so is going to $\infty$. I believe $dX_t = \sigma \sqrt {X_t} dW_t$ exists and absorbs at 0 with probability 1, which would make your integral finite, but either of those statements could be wrong.

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