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I have a result I am trying to understand: $\frac{\partial}{\partial \beta} \int_0^\beta b dF(b)^N = \beta N F(\beta)^{N-1} f(\beta)$

This is how I would like to think about the problem. The PDF of $F(b)^N$, $g(b) = N F(b)^{N-1} f(b)$. Using $\int x d F(x) = \int x f(x) dx \ \Rightarrow \frac{\partial}{\partial \beta} \int_0^\beta b dF(b)^N = \int_0^\beta b N F(b)^{N-1} f(b) db$. Since I am taking a derivative wrt to a specific $b = \beta$, I can treat the integral like a discrete summation and make this simplification: $\frac{\partial}{\partial \beta} \int_0^\beta b N F(b)^{N-1} f(b) db = \frac{\partial}{\partial \beta} \beta N F(\beta)^{N-1} f(\beta) = N F(\beta)^{N-1} f(\beta) + \beta N (N-1) F(\beta)^{N-2} f(\beta)^2 + N \beta F(\beta)^{N-1} f(\beta)^2 \neq \beta N F(\beta)^{N-1} f(\beta)$.

I'd like to know why my logic doesn't work.

I know I should use Leibniz's Rule, but also get stuck doing that.

$\frac{\partial}{\partial \beta} \int_0^{\beta} b d F(b)^N = \int_0^\beta \frac{\partial}{\partial \beta} bdF(b)^N$. Here I think I should treat $\partial = d$, but am unsure how to continue. Should I just treat $\beta = b$ and take derivatives wrt b?

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You had the calculation correct up to $$\frac{d}{d \beta} \int_0^\beta b N F(b)^{N-1} f(b) db$$ At this point you need to apply the fundamental theorem of calculus, not Leinbiz's rule. $$\frac{d}{d y} \int_0^y f(x)dx = f(y)$$ For your problem it yields $$\frac{d}{d \beta} \int_0^\beta b N F(b)^{N-1} f(b) db = \beta N F(\beta)^{N-1} f(\beta)$$

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  • $\begingroup$ Doh!! Thanks! Just curious, why treating the integral like a discrete sum does not work now. $\endgroup$ – hipHopMetropolisHastings Jul 17 '15 at 11:13
  • $\begingroup$ If you look at the final statement, that really is "treating the integral like it is a discrete summation". The summand around $b = \beta$ is approximately $\beta N F(\beta)^{N-1} f(\beta)$. That's basically what the fundamental theorem of calculus says. $\endgroup$ – muaddib Jul 17 '15 at 11:20

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