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In my head integration by parts has always been kind of vague and I just apply it without understanding whats really going on I guess. I.e. if I have the following:

$$\int_{u}^\infty (\ln x - \ln u) \mathrm dF(x) = \int_{u}^\infty \frac{1-F(x)}{x}\mathrm dx$$

I don't know how to handel it. For one I can't deal with the $\mathrm dF(x)$... I know it's like the "change in $F(x)$" and it's the "same" as $f(x)\mathrm dx$ so I write out the left had side like:

$$\int_{u}^\infty (\ln x - \ln u) \ f(x)\mathrm dx$$

and try to do integration by parts like:

$$u=(\ln x -\ln u) \qquad v= 1-F(x)$$

$$u'= \frac{1}{x} \qquad v'= f(x)$$

and then use the IBP formula to get the right hand side, but, for example, I don't really know "why" $v$ is $1-F(x)$ aside from maybe that the range of integration goes from $u$ to $\infty$, so like $ 1-F(x)$ would be $\operatorname{Pr}(X>u)$ sort of. It's all very loosely held together in my head and I usually get by in an exam since I'm fairly certain the answer is right but if I had to explain it, it kind of falls apart. Sort of the hazard of learning by mostly doing dozens of past papers I guess.

Can anyone fill the gaps. I'd appreciate multiple answers if possible (I tend to learn from what multiple sources say and have in common).

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  • $\begingroup$ For me it seems not totally true, the derivative of $1-F(x)$ is $-f(x)$ not $f(x)$ $\endgroup$
    – J. P. C.
    Commented Jun 11, 2017 at 9:08
  • $\begingroup$ Got something from an answer below? $\endgroup$
    – Did
    Commented Jul 13, 2017 at 11:03

1 Answer 1

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Integration by parts when there is no boundary term is often equivalent to some Fubini-type change of order of integration. To apply this general principle in the present case, one can start from the identity, valid for Lebesgue-almost every $x$, $$1-F(x)=P(X>x)=\int_x^\infty dP_X(t)$$ which implies that the RHS of the identity of interest is $$\int_u^\infty\frac{1-F(x)}x\,dx=\int_u^\infty\frac1x\left(\int_x^\infty dP_X(t)\right)dx=\int_u^\infty\left(\int_u^t\frac1xdx\right)dP_X(t)$$ that is, $$\int_u^\infty\frac{1-F(x)}x\,dx=\int_u^\infty\log\left(\frac{t}u\right)\,dP_X(t)$$ and the proof is complete.

The existence of the PDF of $X$ is not required and one only manipulates nonnegative terms in all this hence the identity holds for every random variable, with a PDF or not, whether the resulting integrals are finite or not (if they are not, one arrives at $+\infty=+\infty$, which is true). For an old mse related question, see here.

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  • $\begingroup$ In the 2nd last line of working, how did you go from the 2nd last step to the last step with the dPx(t)? Can you please explain how the bounds become u to t? $\endgroup$ Commented Jun 11, 2017 at 11:55
  • $\begingroup$ The first double integral is over the domain $$t>x\qquad x>u$$ and it coincides with the double integral over the domain $$u<x<t\qquad t>u$$ since, for every $u$, these two sets of conditions on $(x,t)$ are equivalent. $\endgroup$
    – Did
    Commented Jun 11, 2017 at 12:04

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