enthdegree
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 1h comment Is there a simpler function with this shape? Nice diagram. Did you make that in inkscape? Apr 25 comment Uniform Distribution Estimator (not MLE) Byron, No. Intuitively, the max drawn from the distribution is going to be very close to $\theta$, while the min is going to be very close, but bigger than $0$. Apr 24 answered evaluate the following limit else prove that limit does not exist Apr 23 accepted Intuition for integrating $1/z$ around the unit circle Apr 23 revised Intuition for integrating $1/z$ around the unit circle edited title Apr 23 asked Intuition for integrating $1/z$ around the unit circle Apr 18 comment Proving an Integral inequality from a given integral inequality @Cupid my complaint still holds even with my error Apr 18 comment Proving an Integral inequality from a given integral inequality I am not sure your question is equivalent to the one from the book... doing it for $xg(x)$ instead of $g(x)$ forces the integrand, $xg(x)e^{-f(x)}$, to be small near 0. This property is not imposed if the integrand were only $g(x)e^{-f(x)}$ Apr 18 comment Proving an Integral inequality from a given integral inequality I disagree with your "intuitive" assertion that $\int_0^1 e^{-f(x)} dx \geq \int_0^1 e^{-g(x)}dx\Rightarrow f(x)\leq g(x)$... Take $f(x)$ to be a bump near x=0.1, and $g(x)$ to be the same bump near x=0.9... then the integral inequality holds, but the functional inequality obviously does not. Apr 14 comment Does sum of the reciprocals of all the composite numbers converge? I like this answer, the sum of reciprocals of primes is a big jackhammer Apr 6 comment Show that $\lim_{n\to \infty}\int_{[0,\infty]}\left(1+\dfrac{x}{n}\right)^n\exp^{-ax}=\dfrac{1}{a-1}$ Yes, you can justify this using the squeeze theorem. MCT is a good way too Apr 6 comment Show that $\lim_{n\to \infty}\int_{[0,\infty]}\left(1+\dfrac{x}{n}\right)^n\exp^{-ax}=\dfrac{1}{a-1}$ Did you know that $\exp(c):=\lim_{n\rightarrow \infty} (1+c/n)^n$ and that further, this sequence is increasing? Mar 29 comment Statistics. Uniform distribution estimator variance What is $X_{(i)}$? Mar 28 comment Question about the existence of Riemann-Stieltjes integrals. What text are you learning this out of? In my construction, both those limits exist and equal 1 because as $|\Gamma|$ becomes small, any problems around 0 due to the left/right-discontinuousness go away. Mar 28 comment Question about the existence of Riemann-Stieltjes integrals. Further, ($\sup_\Gamma L_\Gamma = \inf_\Gamma L_\Gamma$) iff ($\lim_{|\Gamma|\to 0} L_\Gamma$ exists). The only possibility I see is if there is a distinction between your constructions $L_\Gamma$ and $R_\Gamma$. Mar 28 comment Question about the existence of Riemann-Stieltjes integrals. The difference between Riemann and Riemann-Stieltjes integrals is that RS integrals let you specify a function to integrate against (i.e. "to weigh different parts of the space differently"). The $\phi$ function is analogous to measures in Lebesgue integrals. Riemann integrals do not have a mechanism for $\phi,$ or rather it is baked into the definition of Riemann integration that $\phi(x)=x$. $${}$$ I don't think your middle section true. That is, I believe if that (the RS integral exists) iff ($\sup_\Gamma L_\Gamma = \inf_\Gamma L_\Gamma$). Can you provide a counterexample to this? Mar 28 revised Question about the existence of Riemann-Stieltjes integrals. deleted 1 character in body Mar 28 answered Question about the existence of Riemann-Stieltjes integrals. Mar 28 revised Verify the identity: $(1-x^2)\dfrac{\partial^2 \Phi}{\partial x^2}-2x\dfrac{\partial\Phi}{\partial x}+h\dfrac{\partial^2}{\partial h^2}(h\Phi)=0$ added 109 characters in body Mar 28 comment Verify the identity: $(1-x^2)\dfrac{\partial^2 \Phi}{\partial x^2}-2x\dfrac{\partial\Phi}{\partial x}+h\dfrac{\partial^2}{\partial h^2}(h\Phi)=0$ Using the product rule twice: \begin{array}{rcl}\frac{\partial^2}{\partial h^2} h \Phi &=& \frac{\partial}{\partial h} ( \frac{\partial}{\partial h} h \Phi) \\ &=& \frac{\partial}{\partial h} \left( \Phi + h \frac{\partial}{\partial h} \Phi \right) \\ &=& \frac{\partial}{\partial h} \Phi + \frac{\partial}{\partial h} \Phi + h\frac{\partial^2}{\partial h^2} \Phi \end{array}