Hi there stochastic calculus gurus,I have a question about the quantity below: [$]\int_{t_{0}}^{t} \sigma(t^{\prime})dW(t^{\prime}) - \sqrt{ \int_{t_{0}}^{t} \sigma^{2}(t^{\prime})d t^{\prime}/(t-t_{0}) } (W(t)-W(t_{0}) [$]Where [$]\sigma(t)[$] is a simple non-anticipating function of time This thing must have mean zero but looking at its higher moments with the correlation formula for nonanticipating functions it would seem to have fairly restrictive conditions on the gradients of sigma to get positive variances. i.e. [$] \sqrt{\int \sigma^{2}dt } >\int \sigma dt/ \sqrt{t}[$] Is this how it works?Thanks for any pointers.

I think I see your issue. Your need conditions ensuring that [$]X_t \not= 0[$] (almost surely) where [$]X_t[$] is your expression.Once you have [$]X_t \not= 0[$], you know [$]Var(X_t) = E[X_t^2] > 0[$].Let's take [$]t_0 = 0[$] and write [$]X_t = A_t - B_t[$] for your two terms.I think it's clear [$]A_t \stackrel{d}{=} B_t[$] with the common distribution being [$]N(0,\int_0^t \sigma^2(s) ds)[$].But having the same distribution doesn't necessarily imply [$]A_t = B_t[$], assuming [$]\sigma(t)[$] is not a constant. Good puzzle.Since [$]\sigma(t)[$] must not be a constant, I think I would explore the case where [$]\sigma(t)[$] consists of two piecewise constants. Does [$]X_t \not= 0[$] (a.s.) for that one?

Last edited by Alan on April 9th, 2015, 10:00 pm, edited 1 time in total.

I suppose what's funny is that the variance of your term would be less if you replaced the rms of sigma by its ordinary mean.

Last edited by savr on April 9th, 2015, 10:00 pm, edited 1 time in total.

I am not sure if I understand your question properly, but it seems to me that, under your assumptions, the condition you have written down is always true (since the root mean square is always larger than or equal to the arithmetic mean).

QuoteOriginally posted by: AlanI think I see your issue. Your need conditions ensuring that [$]X_t \not= 0[$] (almost surely) where [$]X_t[$] is your expression.Once you have [$]X_t \not= 0[$], you know [$]Var(X_t) = E[X_t^2] > 0[$].Let's take [$]t_0 = 0[$] and write [$]X_t = A_t - B_t[$] for your two terms.I think it's clear [$]A_t \stackrel{d}{=} B_t[$] with the common distribution being [$]N(0,\int_0^t \sigma^2(s) ds)[$].But having the same distribution doesn't necessarily imply [$]A_t = B_t[$], assuming [$]\sigma(t)[$] is not a constant. Good puzzle.Since [$]\sigma(t)[$] must not be a constant, I think I would explore the case where [$]\sigma(t)[$] consists of two piecewise constants. Does [$]X_t \not= 0[$] (a.s.) for that one?Thanks for the suggestions: it all seems a little better behaved than I initially thought.If it you take [$]\sigma \sim t^{\alpha}[$] and look at what happens when the volatility is a decreasing function it all seems to make sense. Even the seeming problem areas of [$]\alpha \sim -1/2[$] are not such a big deal..

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