After years' study in probability, I felt more and more happy with the construction of probability space, it's so simple seemingly but interesting that may be applied to many more areas rather than math.

Let’s review a little bit about the probability space in math. Given a space (\Omega, \mathbb{F}, \mathbb{P}), assume that there exists a r.v X s.t. \{\omega|X(\omega)>0\}\in\mathbb{F}, then we can measure the events \{X>0\} by using \mathbb{P}(X>0). What’s more, adding a time dimension to make it a stochastic process X_t(\omega), then we have properties like \mathbb{F}_t\subset\mathbb{F}_{t^+} and so on.

Secondly, let’s consider another fact, different people have different decisions when given tasks, why is that? That apparently depends on the knowledge we own, how do I relate this decision making base on probability space construction idea? This attracts me these days and I’ll post what I think later.

To be continued……


A comment on a simple proof of Cauchy inequality

In mathematics, the inequality

n({{x}_{1}}^{2}+\cdots +{{x}_{n}}^{2})\geq {{({{x}_{1}}+\cdots +{{x}_{n}})}^{2}}

is not hard to prove with quadratic root discriminant.

But the problem is that my computer science roommate want me to give an induction method. So I tried a new method to prove it, in which I used a small trick deserving taking a note.

Let me firstly restate the problem here: it’s known that

{{x}_{1}}+\cdots +{{x}_{n}}=1, prove that {{x}_{1}}^{2}+\cdots+{{x}_{n}}^{2}\geq \frac{1}{n}

The equality holds if and only if

{{x}_{1}}={{x}_{2}}=\cdots ={{x}_{n}}



the inequality obviously holds true.

Assume that when


the inequality holds true.

Then when


if  k+1=2m,

{{x}_{1}}^{2}+\cdots +{{x}_{n}}^{2}=\sum_{i=1}^{2m}{{{x}_{i}}^{2}}=\sum_{i=1}^{m}{({{x}_{i}}^{2}+{{x}_{i+m}}^{2})}\geq \sum_{i=1}^{m}{({{(\frac{{{x}_{i}}+{{x}_{i+m}}}{2})}^{2}}+{{(\frac{{{x}_{i}}+{{x}_{i+m}}}{2})}^{2}})}\geq \frac{1}{2}\sum_{i=1}^{m}{{{({{x}_{i}}+{{x}_{i+m}})}^{2}}}\geq \frac{1}{2}m{{(\frac{\sum_{i=1}^{2m}{{{x}_{i}}}}{m})}^{2}}=\frac{1}{2m}=\frac{1}{n}

if k+1=2m-1,

Therein lies the problem, we cannot break the formula into two parts equally if induction is used.

Instead of proving this inequality directly, we consider another problem which has a subtle link with the  issue we are dealing with. Say, when

{{x}_{1}}+\cdots +{{x}_{2m}}=1+\frac{1}{2m-1}

Prove that

\sum_{i=1}^{2m}{({{x}_{i}}^{2})}\geq 2m{{(\frac{1}{2m-1})}^{2}}.

Actually, by induction method, because


this statement holds true definitely. Further more, it gives

\sum_{i=1}^{2m-1}{({{x}_{i}}^{2})}\geq \frac{1}{(2m-1)} when \sum_{i=1}^{2m-1}{{{x}_{i}}}=1.

Otherwise, the new problem value would be even smaller.

The inspring point for this method is to add a free variable to cater the old problem, though technical when applying it. And I hope this trick will bring me more stuff.