Hi,I want to find the sign of [$]\partial_z (\partial_z \log F(T1) - \partial_z \log F(T2) )[$]where [$]T1<T2[$] and[$]F(T) = \int_0^T e^{g(t) + z\, h(t)} dt[$]You can interpret [$]F()[$] as an asset price, [$]T[$] as a maturity and [$]z[$] as a latent factor but it is not really important.All we know is that [$]h(t)>0\forall t[$] and it is either monotone inreasing (m.i.) or decreasing (m.d.).Then I get[$]\frac{\int_0^{T1} e^{g(t) + z\, h(t)}h(t)^2 dt}{\int_0^{T1} e^{g(t) + z\, h(t)}dt} - \left(\frac{\int_0^{T1} e^{g(t) + z\, h(t)}h(t) dt}{\int_0^{T1} e^{g(t) + z\, h(t)} dt}\right)^2 - \frac{\int_0^{T2} e^{g(t) + z\, h(t)}h(t)^2 dt}{\int_0^{T2} e^{g(t) + z\, h(t)}dt} + \left(\frac{\int_0^{T2} e^{g(t) + z\, h(t)}h(t) dt}{\int_0^{T2} e^{g(t) + z\, h(t)} dt}\right)^2[$]Denote the four terms as[$]A1 - A2 - B1 + B2[$]then, [$]A1<B1[$] and [$]B2 > A2[$] if [$]h(t)[$] is m.i.or[$]A1>B1][$] and [$]B2<A2[$] if [$]h(t)[$] is m.d.Is the monotonicity of [$]h(t)[$] helpful to determine the sign of [$]A1-A2[$] and [$]B2-B1[$] and, then, is the condition [$]T2>T1][$] sufficient to determine the whole sign?Could it help to add the assumption that [$]h(t)>1\, \forall t[$] (and m.d.) ? Thanks a lot in advance

So my intuition is that we can interpret [$]A1 - A2[$] as a variance of [$]h(t)[$] over the support [$](0,T1)[$] and density given by [$]e^{g(t)+z\, h(t)} / \int_0^{T1} e^{g(t)+z\, h(t)} dt.[$]Similarly, we can interpret [$]B1 - B2[$] as a variance over the support [$](0,T2)[$].Moreover we can interpret [$]A1 - A2[$] as a variance of [$]h(t)[$] on [$](0,T2)[$] with zero weight on observation in [$](T1,T2)[$].Considering that [$]h(t)[$] is monotone decreasing, is this useful to infer about the sign of [$](A1-A2) - (B1-B2)[$]?Any hint is very appreciated!

I thought about this a bunch, but don't have an answer to the question. Since you haven't received a response I though I'd right down some thoughts that might help you. From the way I understand the problem g(t) is arbitrary and h(t) has some monotonicity constraints. I get the feeling that there are counterexamples and here is a transformation that might get you there:You can tame the integral down a bit by setting h(t) = log f(t) and choosing g(t) = log ((sign) f'(t)) where sign is chosen based on which direction of monotonicity you are working with.This gives you an integrand (+/-)f'(t)f(t)^z which can easily be integrated to f(t)^{z+1}/z+1Supposing that your claim is true, then as I understand it, the difference of the second derivatives wrt z would have the same sign for all time T2>T1. You could then conclude that d/dt d^2/dz^2 has the same sign for all T1. This way you can pull the time derivative inside and get rid of the log.

Hi mauddibthanks a lot for your reply. I'm going to think about it!