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For $f_0$ and $f_1$ two continuos probability density functions on $\mathbb{R}$, by Hölder, I know that $f_1^x f_0^{1-x}$ is integrable on $\mathbb{R}$, where $0 \leq x \leq 1$. Let $l=f_1/f_0$, then $g:=l^xf_0=f_1^x f_0^{1-x}$. Is it true that derivative of $g$, of any order, w.r.t $x$ is also integrable on $\mathbb{R}$? For example $g^{''}=\ln(l)^2 l^xf_0$?

I have a solution to this problem at for the case when $l$ is increasing, and for $g^{''}$, here. I think it should still be true when $l$ is not necessarily increasing, and even for the higher order derivatives, but I am not able to show it.

Thanks for your help.

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Is $f_2$ supposed to be $f_0$? – Gerry Myerson Apr 7 '14 at 3:21
@Gerry Myerson Yes a typo. – Seyhmus Güngören Apr 7 '14 at 7:56

Consider $f_1^xf_0^{1-x}$ for $x$ in the strip $0<Re(x)<1$ and use Cauchy's formula.

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Thanks for the answer. I had a quick look at complex integrals and from your answer I guessed that I had to take a contour integral of $f_1^xf_0^{1-x}$, for complex $x$, and one segment of the contour would be the real line from $0$ to $1$. If I am not incorrect, $f_1^xf_0^{1-x}$ has no residue, therefore the contour integral should give $0$. I calculated the integral over the real line and it was $(f_1-f_0)/\ln(f_1-f_0)$. I dont have any idea about the part of the arc and I also dont know how to conclude about integrability. Could you please further help? – Seyhmus Güngören Apr 7 '14 at 10:49
For any x in the strip, $f_1^xf_0^{1-x}$ is integrable. Now express the derivatives as contour integrals of the function using Cauchy's formula. – Michael Renardy Apr 7 '14 at 13:17

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