# What does $o_n(1)$ mean?

I'm trying to read the following article, and in the abstract they write: Let $$\xi$$ be a non-constant real-valued random variable with finite support, and let $$M_n(\xi)$$ denote a $$n\times n$$ random matrix with entries that are independent copies of $$\xi$$. For $$\xi$$ which is not uniform on its support, we show that: $$Pr[M_n(\xi) \text{ is singular}] = Pr[\text{zero row or column}]+ (1+o_n(1))Pr[\text{two equal (up to sign) rows or columns}]$$.

What does $$o_n(1)$$ mean here?

• It is hard to answer without any context… Mar 18 at 21:32
• My guess is that they mean $o(1)$ in the normal sense (Landau notation), but since they are working with several variables, they add $n$ as subscript to mean "as $n$ goes to infinity". Mar 18 at 23:25
• @JeanAbouSamra You should post it as an answer. Mar 19 at 8:57
• @JeanAbouSamra but here we only have 1 variable n Mar 19 at 10:42
• @L.breitman Actually, there is also the variable $p$. Mar 19 at 12:22

As I understand it, this $$o_n(1)$$ is meant as the standard “little o” notation (when $$v_n$$ takes non-zero values, $$u_n = o(v_n)$$ means $$\frac{u_n}{v_n} → 0$$), except that because several variables are involved ($$n$$, $$ξ$$, $$p$$), they added the $$n$$ subscript to clarify that the limit is as $$n$$ goes to infinity, for fixed values of the other variables. A perhaps more standard notation for what they write as $$x = o_n(y)$$ would be $$x \underset{n→+∞}{=} o(y)$$

• This is consistent with the definition of $O_\alpha, \Omega_\alpha, \Theta_\alpha$ in the "Notation" section of the paper, although strangely they omit defining $o_\alpha$. Mar 20 at 14:01
• Thanks for the answer! Mar 21 at 14:15

Given functions $$f(n)$$ and $$G(n)$$, from $$[0,\infty)\rightarrow [0,\infty)$$ we can define

Operands Meaning
$$f(n) = O_n(G(n))$$ $$\displaystyle \lim_{n\rightarrow \infty} \inf \frac{G(n)}{f(n)} > 0$$
$$f(n) = o_n(G(n))$$ $$\displaystyle \lim_{n\rightarrow \infty} \frac{G(n)}{f(n)} = 0$$
$$f(n) = \Theta_n(G(n))$$ $$f(n) = O(G(n)) \;\&\; G(n) = O(f(n))$$
$$f(n) \sim_{n} G(n)$$ $$\displaystyle \lim_{n\rightarrow \infty} \frac{G(n)}{f(n)} = 1$$
• The first one is incorrect. E.g. $1 + \sin(n) = O(1)$ but $\lim 1/(1 + \sin(n))$ does not exist. You can replace the limit by $\lim\inf$. Mar 20 at 12:27
• You are right. I corrected the answer. Mar 28 at 10:28