Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello,

this message follows on from About Goldbach's conjecture and aims at modelling the quantity $N_{2}(n)$ by a random variable.

Let $n$ be a fixed natural number and let's define for every $\varepsilon\gt 0$ the quantity $Y_{k,\varepsilon}$ as the Bernoulli random variable equal to $1$ with probability $p=N_{1}(n)/P_{ord_{C}(n)}$ if $k\lt n$ and equal to $1$ with probability $\varepsilon/k^{2}$ if $k\ge n$. Assume that for any given $\varepsilon$ all of the quantities $Y_{k,\varepsilon}$ are independent. Let $S_{m,\varepsilon}$ be the random variable $\sum_{k=1}^{m}Y_{k,\varepsilon}$. I have been told that $S_{m,\varepsilon}$ converges almost surely towards $X=(S+T(\varepsilon))/np$, where $S$ is the sum of the $n-1$ first terms of the previous sum and $\displaystyle{T(\varepsilon)=\sum_{k=n}^{\infty}Y_{k,\varepsilon}}$, and that the mean value of this random variable $X$ is $1$ as $\varepsilon$ tends to $0$.

So my question is : does this entail that the limit, if it exists, as $n$ goes to infinity of the sequence $\left(\dfrac{N_{2}(n)P_{ord_{C}(n)}}{n.N_1(n)}\right)_{n\in\mathbb{N}}$ is equal to $1$?

Thank you in advance.

share|improve this question
    
How the behaviour of something not involving $N_2(n)$ at all could imply a result about $N_2(n)$? Even more basic: with your notations, $S_{m,\varepsilon}\to S+T(\varepsilon)$ BY DEFINITION. Hence dividing by $np$ is strange here and it is doubtful that any grand consequence could arise. –  Did Oct 8 '11 at 14:58
    
You are right, I made a mistake: it is $S_{m,\varepsilon}/np$ which converges towards $(S+T(\varepsilon))/np$. By the way, as I said, I aim at modelling $N_{2}(n)$ by a random variable, and this random variable is $S$. In other words, $S$ "mimics" $N_{2}(n)$ (read again the link given at the beginning of this thread), since this random variable $S$ is built in such a way that $S$ and $N_{2}(n)$ should be of the same "order of magnitude". Hence my question. –  Sylvain JULIEN Oct 8 '11 at 17:37
add comment

Know someone who can answer? Share a link to this question via email, Google+, Twitter, or Facebook.

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.