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$1$ : It means "there exists something"

$0$ : It doesnt mean "there exists nothing", instead it means "there exist something but that is nothing"


Accept policy:

My accept policy is to accept the first completely correct answer. Acception is final and will never be changed. If no complete answer is available after a certain time, I accept partial answers.

Answering policy:

I try to answer at least two times more than the number of questions that I ask because my questions are usually two times more difficult than the questions that I am able to answer.

Voting policy:

I try to upvote all constructive answers to my questions and all nice questions/answers. I try to avoid downvoting and rather I use diplomatic ways to overcome the problems. All intentionally problematic questions or answers will be downvoted.

Respect policy:

I respect all persons who are willingly to help others. I show immediately in the comments that I dislike arrogant types.


Jul
16
comment Why $\int f_0 \mbox{d}(F_0+F_1)=\int f_0 \mbox{d}F_0+\int f_0 \mbox{d} F_1=1$ should be true?
@Ian I dont thing so. I am sure that both are probability measures. Here the first page projecteuclid.org/download/pdf_1/euclid.aoms/1177699803
Jul
16
comment Why $\int f_0 \mbox{d}(F_0+F_1)=\int f_0 \mbox{d}F_0+\int f_0 \mbox{d} F_1=1$ should be true?
@Ian it is also not guaranteed that $\int f_0 d F_0=1$. What I know is that $F_0$ and $F_1$ are some distinct probability measures. What does mutually singular mean? I have this from one paper. The author defines a dominating measure and take various integrals. I guess the choice of measure here doesnt change anything but my information here is unfortunately superficial..
Jul
16
asked Why $\int f_0 \mbox{d}(F_0+F_1)=\int f_0 \mbox{d}F_0+\int f_0 \mbox{d} F_1=1$ should be true?
Jul
15
comment Substitution in Matlab (Iteration Process)
if you are making iterations then you will first need some initial values A(-1) and A(-2) similarly for B. Matlab does accept only positive indices therefore you will need A(1), A(2) a-priori, similarly for B. If these are given and if $x_i$ are known then what remains is to write a single for loop in matlab to obtain the other values.
Jul
15
comment Substitution in Matlab (Iteration Process)
it seems strange to me. You have 2 different iterative formulas for the same thing. Why should they give the same results?
Jul
14
comment Operations on Random Variables
Use unitstep[x-1/(a+1)]-unitstep[x-1/a] in the place of $1_{...}$ then take the integrals only whenever the integral is non-zero. Change to unitstep definition for both $1$ functions.
Jul
14
answered Operations on Random Variables
Jul
10
revised Did I obtain this recursion in the Fourier domain correctly?
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Jul
9
revised Did I obtain this recursion in the Fourier domain correctly?
added 68 characters in body
Jul
9
revised Did I obtain this recursion in the Fourier domain correctly?
added 1 character in body
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jul
1
revised Does it hold? $\int_{0}^{\pi/2}\cos^{2k}xdx=\frac{(2k-1)!!}{2k!!}\frac{\pi}{2}$
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Jun
24
revised Did I obtain this recursion in the Fourier domain correctly?
edited title
Jun
24
comment Why is $fg$ integrable w.r.t. a probability measure if $f,g$ are Lebesgue integrable?
both are integrable so their integral is less than $\infty$. Now consider the Hölder inequality.
Jun
24
reviewed Approve suggested edit on Expected value of moment generating functions:
Jun
24
revised Did I obtain this recursion in the Fourier domain correctly?
edited title
Jun
23
revised Did I obtain this recursion in the Fourier domain correctly?
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Jun
23
revised Did I obtain this recursion in the Fourier domain correctly?
added 2 characters in body
Jun
22
asked Did I obtain this recursion in the Fourier domain correctly?