User:GFauxPas/Sandbox

Welcome to my sandbox, you are free to play here as long as you don't track sand onto the main wiki. --GFauxPas 09:28, 7 November 2011 (CST)

Continuous Random Variables
Boo, there's not enough foundation for continuous random variables to do stuff on normal distributions. That means more work for me -_-


 * $\displaystyle \operatorname{var} \left({X}\right) = \int_{x \in \Omega_X} \left({x - \mu}\right)^2 \operatorname{pdf}\left({x}\right) \, \mathrm dx$

Do we have stuff up about density functions? I couldn't find anything. --GFauxPas (talk) 15:21, 13 December 2012 (UTC)


 * The relevant articles are Definition:Probability Distribution and Definition:Probability Mass Function. Most of the abstract measure-theoretic foundation for analysis has been laid down, but indeed the application to random variables is still (largely) missing. --Lord_Farin (talk) 15:31, 13 December 2012 (UTC)

To do
I'll need this theorem in order to solidify some results with the definite integrals above.

Let $f$ be a differentiable real function. Then:


 * $\dfrac {f(x+h) - f(x-h)} {2h} \to f\,'(x)$ as $h \to 0$ --GFauxPas (talk) 22:28, 19 December 2012 (UTC)


 * Is the "2" in the denominator extraneous? --GFauxPas (talk) 22:31, 19 December 2012 (UTC)


 * Not unless $f'(x) = 0$. Try multiplying both sides with $2$. --Lord_Farin (talk) 23:30, 19 December 2012 (UTC)


 * Oh, okay, right. So to prove this I'm going to use that $f\,'_+ = f\,'_-$, from Limit iff Limits from Left and Right, right? And then what's the next step, can I have a hint?


 * It may be worthwhile to add definitions for "derivative from the left" and "derivative from the right". This also allows for neater structuring of Definition:Differentiable/Real Function/Interval.


 * Furthermore, it appears that the following equivalent formulation of differentiability is not covered yet (it's rather trivial that they're equivalent but needs proof nonetheless):


 * Let $x \in \R$ be a real number.


 * Let $f$ be a real function whose domain is a neighborhood of $x$.


 * Then $f$ is said to be differentiable at $x$ iff the following limit exists:


 * $f' \left({x}\right) := \displaystyle \lim_{h \to 0} \frac {f \left({x + h}\right) - f \left({x}\right)} {h}$


 * If this is the case, $f' \left({x}\right)$ is called the derivative of $f$ at $x$.


 * With this definition, we observe that your expression is (one half times) $\displaystyle \lim_{h \to 0} \frac {f \left({x + h}\right) - f \left({x}\right)} {h} + \frac {f \left({x - h}\right) - f \left({x}\right)} {-h} = 2 f' \left({x}\right)$, using that $-h \to 0$ iff $h \to 0$. --Lord_Farin (talk) 08:16, 20 December 2012 (UTC)


 * Equivalence of Definitions of the Derivative, Definition:Right-Hand Derivative, Definition:Left-Hand Derivative --GFauxPas (talk) 12:10, 20 December 2012 (UTC)

Is there a such thing as a one-sided (real valued) definite integral? --GFauxPas (talk) 13:54, 20 December 2012 (UTC)


 * The closest thing I can think of is Definition:Stieltjes Function of Measure on Real Numbers. --Lord_Farin (talk) 14:24, 20 December 2012 (UTC)

Continuity of Inverse Functions
Larson, appendix, A12

Theorem
Let $f: \mathbb I \to \mathbb J$ be a real function, where $\mathbb I$ is some real interval.

Let $f$ admit an inverse.

Then, if $f$ is continuous on its domain, then $f^{-1}$ is continuous on its domain.

Comment: yowzers. That's ... a bit of a mess. First off, the part two should be separate from the part one, and they can be linked together corollaryishly. Nextishwise, refactor: Is there anything else? --Dfeuer (talk) 03:35, 1 February 2013 (UTC) Ah yes, one more:
 * A strictly monotone bijection between tosets is an order isomorphism.
 * An order isomorphism between linearly ordered spaces is a homeomorphism.
 * The order topology on an interval in a linearly ordered space is the same as its subspace topology. --Dfeuer (talk) 03:37, 1 February 2013 (UTC)


 * Well I know it's messy, I'm just trying to make the proof work before I make it neat. Unfortunately I don't know enough to generalize results to everything you have said, as I have not learned order theory or topology. --GFauxPas (talk) 03:41, 1 February 2013 (UTC)


 * The order topology on a totally ordered set is typically one of the first ones introduced in a topology text. You could probably get through those basics in a few hours. The usual topology on the reals is the order topology, so the ideas are very similar. The basic idea is that a set is "open" iff it is a union of "basic" open sets, which take one of three forms: open rays to the left, open rays to the right, and (bounded) open intervals. The last bit, about the order topology on an interval being the subspace topology, means you don't have to make yourself crazy about the endpoints. You could probably express all these ideas directly in terms of real numbers &hellip; somehow. --Dfeuer (talk) 03:50, 1 February 2013 (UTC)


 * And what I'm calling an interval here is called a Convex Set. --Dfeuer (talk) 03:55, 1 February 2013 (UTC)


 * I'm not going to do that. Even if I weren't lazy, that would mean that someone at my current level would himself have to spend a few hours learning topology to understand the proof. In any event, the cases I put between the $\uparrow \downarrow$s need to be addressed, and I'd appreciate help with that. --GFauxPas (talk) 15:23, 1 February 2013 (UTC)

I'm going to attempt the topological approach, then see if I can adapt it smoothly to the special case of the reals. --Dfeuer (talk) 15:26, 1 February 2013 (UTC)

Proof
Let $f$ be invertible on $\mathbb I$.

Part 1:


 * $f$ is a bijection from Bijection iff Inverse is Bijection.


 * $f$ is then strictly monotone from Continuous Bijection of Interval is Strictly Monotone.

Part 2:

Because $f$ is a bijection, $\mathbb J = f\left({\mathbb I}\right)$.

From Image of Interval by Continuous Function is Interval, $\mathbb J$ is an interval.


 * $\downarrow$ PROVE THIS $\downarrow$

Let $a$ be an interior point of $\mathbb J$. Then $f^{-1} \left({a}\right)$ is an interior point of $\mathbb I$.


 * $\uparrow$ PROVE THIS $\uparrow$

Let $\epsilon > 0$. From Interval Defined by Betweenness there exists some $0 < \epsilon_1 < \epsilon$ such that:


 * $\mathbb I\,' = \left ({f^{-1}\left({a}\right) - \epsilon_1 \,.\,.\, f^{-1}\left({a}\right) + \epsilon_1} \right)$

is a subinterval of $\mathbb I$.

From Image of Subset is Subset of Image, $f\left({\mathbb I\,'}\right) \subseteq \mathbb J$.

From Interval Defined by Betweenness there exists some $\delta > 0$ such that:


 * $\left ({a - \delta \,.\,.\, a + \delta} \right)$

is a subinterval of $f\left({\mathbb I\,'}\right)$.

Because $ \left({\R, \vert \cdot \vert}\right)$ is a metric space, the above interval can be considered as an open ball:


 * $x \in \left ({a - \delta \,.\,.\, a + \delta} \right) \iff \left \vert {x - a} \right \vert < \delta$.

Similarly:


 * $\left \vert {x - a} \right \vert < \delta \implies \left \vert {f^{-1}\left({y}\right) - f^{-1}\left({a}\right)} \right \vert < \epsilon_1 < \epsilon$.

This proves that $f^{-1}$ is continuous at $a$, from the definition of continuity.

As $a$ was an arbitrary interior point, all that remains to be proven is to address the endpoints of $\mathbb J$, if $\mathbb J$ is half-open or closed.

--GFauxPas (talk) 03:24, 1 February 2013 (UTC)


 * Let me tend to the two indicated steps only. Second point simply follows by Image of Subset is Subset of Image. First point is more intricate. An attempt:
 * We have that $f^{-1}$ is strictly monotone by Inverse of Strictly Monotone Function.
 * Let $a$ be an interior point of $\mathbb J$; we find $b, c \in \mathbb J$ such that $b < a < c$.
 * By monotonicity of $f^{-1}$, $f^{-1} \left({b}\right) < f^{-1} \left({a}\right) < f^{-1} \left({c}\right)$.
 * By definition of $\mathbb J$, $f^{-1} (b), f^{-1} (c) \in \mathbb I$.
 * Thus $f^{-1} (a)$ is not an end point of $\mathbb I$.
 * Hence by Interior Point of Interval, $f^{-1} (a)$ is an interior point of $\mathbb I$.
 * Here, Interior Point of Interval is a reference to a result that a non-end point of an interval is an interior point. I hope that suffices. --Lord_Farin (talk) 13:55, 4 February 2013 (UTC)


 * That's very helpful, thanks! What about continuity at endpoints, do I go to the definition of one-sided limit? --GFauxPas (talk) 14:05, 4 February 2013 (UTC)


 * Yes, at least that's what arises from going into topology and back. --Lord_Farin (talk) 16:46, 4 February 2013 (UTC)


 * Wouldn't Interior Point of Interval follow directly from the definition of an interval? Is it as trivial as it seems, or am I missing something subtle? --GFauxPas (talk) 01:17, 5 February 2013 (UTC)


 * It's very intuitive and trivial, but since "interior point" is topological language, there is merit in having such a page. --Lord_Farin (talk) 14:11, 5 February 2013 (UTC)