Understanding the Proof: Why Does It Work?

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Homework Statement



Screenshot (225).png


2. Questions

I want to know how this proof works, but there are several things I don't understand.

(1) How does the ##k\epsilon## appears? ##k## is the lipschitz constant, but what about ##\epsilon##? Is the author taking ##||x-y||<\epsilon##?

(2) What does the subscript ##1_\{||X_n-Y_n||>\epsilon\}## means and why is it there? I think the author divided the those values who were at a distance smaller than ##\epsilon## and those who weren't, but I'm not completely sure about this.

(3) Why is it ## E\{|f(X_n)-f(Y)n|1_{||X_n-Y_n||>\epsilon}\}\leq 2\alpha P\{||X_n-Y_n||\}##? This might be related to my second question, I am not completely sure about the meaning of the subscript. I see ##f## is bounded by ##\alpha##, but why is it ##2\alpha##?
 

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GabrielN00 said:

Homework Statement



View attachment 213380

2. Questions

I want to know how this proof works, but there are several things I don't understand.

(1) How does the ##k\epsilon## appears? ##k## is the lipschitz constant, but what about ##\epsilon##? Is the author taking ##||x-y||<\epsilon##?
I'm not sure, but I believe so.
GabrielN00 said:
(2) What does the subscript ##1_\{||X_n-Y_n||>\epsilon\}## means and why is it there? I think the author divided the those values who were at a distance smaller than ##\epsilon## and those who weren't, but I'm not completely sure about this.
It's the characteristic function for that set. In this case, it evaluates to 1 on the set where ##||X_n - Y_n|| > \epsilon##. The function evaluates to 0 on the complement of this set. See https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory).
GabrielN00 said:
(3) Why is it ## E\{|f(X_n)-f(Y)n|1_{||X_n-Y_n||>\epsilon}\}\leq 2\alpha P\{||X_n-Y_n||\}##? This might be related to my second question, I am not completely sure about the meaning of the subscript. I see ##f## is bounded by ##\alpha##, but why is it ##2\alpha##?
My guess: We have ##|f(x)| < \alpha## due to the Lipschitz continuity, so ##|f(x) - f(y)| < 2\alpha##.

BTW, don't surround things with $ signs -- they don't do anything on this site. Use a pair of $$ markers for standalone TeX, and a pair of ## markers for inline TeX.
 
Mark44 said:
I'm not sure, but I believe so.
It's the characteristic function for that set. In this case, it evaluates to 1 on the set where ##||X_n - Y_n|| > \epsilon##. The function evaluates to 0 on the complement of this set. See https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory).

My guess: We have ##|f(x)| < \alpha## due to the Lipschitz continuity, so ##|f(x) - f(y)| < 2\alpha##.

BTW, don't surround things with $ signs -- they don't do anything on this site. Use a pair of $$ markers for standalone TeX, and a pair of ## markers for inline TeX.

Thank you. In regard to the comment about Lipschitz continuity, isn't ##k## de Lipschitz constant? And what kind of inequality justified the change from expectation to probability in (3)?
 
GabrielN00 said:
Thank you. In regard to the comment about Lipschitz continuity, isn't ##k## de Lipschitz constant? And what kind of inequality justified the change from expectation to probability in (3)?
I don't know, but take a look here -- https://en.wikipedia.org/wiki/Slutsky's_theorem -- for an outline of the proof
 
GabrielN00 said:

Homework Statement



View attachment 213380

2. Questions

I want to know how this proof works, but there are several things I don't understand.

(1) How does the ##k\epsilon## appears? ##k## is the lipschitz constant, but what about ##\epsilon##? Is the author taking ##||x-y||<\epsilon##?

(2) What does the subscript ##1_\{||X_n-Y_n||>\epsilon\}## means and why is it there? I think the author divided the those values who were at a distance smaller than ##\epsilon## and those who weren't, but I'm not completely sure about this.

(3) Why is it ## E\{|f(X_n)-f(Y)n|1_{||X_n-Y_n||>\epsilon}\}\leq 2\alpha P\{||X_n-Y_n||\}##? This might be related to my second question, I am not completely sure about the meaning of the subscript. I see ##f## is bounded by ##\alpha##, but why is it ##2\alpha##?

The author splits up ##E|f(X_n)-f(Y_n)|## into two parts: (i) the part where ##|X_n - Y_n| < \epsilon## (so that the expectation on that part is ##< \epsilon k## because we have ##|f(X_n) - f(Y_n) | \leq k |X_n - Y_n| < k \epsilon##); and (ii) the part where ##|X_n - Y_n | > \epsilon##, over which we have ##|f(X_n) - f(Y_n)| \leq |f(X_n)| + |f(Y_n)| \leq 2 \alpha##, so that the expectation on that part is ##\leq 2 \alpha## times the probability of that part.
 
For avoidance of doubt and confusion: do not call that a "characteristic function".

For probability purposes: things like ##\mathbf 1_{\{X \leq x\}}## or ##\mathbb I_{\{X \leq x\}}## are referred to as Indicator Functions.

In most branches of math we could call this a characteristic function -- not so in probability (and stats). In probability a 'characteristic function' is a reserved term and only used to refer to a kind of transform (basically a Fourier transform). If you look at the link provided by @Mark44 you can verify this for yourself.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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