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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}