I am studying the conditions for a Counting process to be a Poisson process.

Here, I understand that the "little$-o$" notation is defined as follows:


Now, I am trying to apply this notation to calculate related limits.

Specifically, what happens when, instead of tending to $0$, $n$ tends to $infinity$, as seen below?

$\lim_{n\to\infty}$ $e^{-n o(1/n)}$


$\lim_{n\to\infty}$ $e^{-n / o(n)}$

I am new to the "little-$o$" notation in terms of how it's used here in the Poisson process. It looks like I am not able to simplify and calculate these limits further without knowing the actual functional form of the functions $o(1/n)$ and $o(n)$.

Any guidance or insights on how to approach this would be very helpful for me to understand how the function $o(n)$ works. Thank you so much!


Here's the context behind the above examples:

I have $n$ independent Poisson processes: $[N_k(t): t >= 0]$, where $k=1, 2, ..., n$.

Each of these Poisson processes has a rate of $1$.

And $E$ is the event that none of the events in any of the processes occur in the time interval $(0, a_n]$.

So, $E$ $=$ $[N_k(a_n) = 0$, for all $k = 1, 2, ..., n]$

Given this, I am trying to find the below limit:

$\lim_{n\to\infty}$ $P(E)$

First, I find $P(E)$ $=$ $P[N_k(a_n) = 0$, for all $k = 1, 2, ..., n]$

So $P(E)$ $=$ $P[(N_1(a_n) + N_2(a_n) + ... + N_n(a_n)) = 0]$

The sum of $n$ independent Poisson processes is also a Poisson process, with rate being equal to the sum of the rates of the $n$ processes.

So $[(N_1(a_n) + N_2(a_n) + ... + N_n(a_n))$: $a_n >= 0]$ is a Poisson process with rate $(n)$, because each of the $n$ independent Poisson processes has a rate of $1$.

This means, $[N_1(a_n) + N_2(a_n) + ... + N_n(a_n)]$ is a Poisson random variable with parameter $(na_n)$. Is this correct?

Therefore, I get:

$\lim_{n\to\infty}$ $P(E)$

$=$ $\lim_{n\to\infty}$ $P[(N_1(a_n) + N_2(a_n) + ... + N_n(a_n)) = 0]$

$=$ $\lim_{n\to\infty}$ $e^{-na_n}$

Now, I have $2$ cases for which I am trying to evaluate the above limit.

Case $(i)$: When $a_n$ $=$ $o(1/n)$

Here, $\lim_{n\to\infty}$ $P(E)$

$=$ $\lim_{n\to\infty}$ $e^{-na_n}$

$=$ $\lim_{n\to\infty}$ $e^{-n o(1/n)}$

$=$ $e^{-0}$

$=$ $1$

Case $(ii)$: When $1/a_n$ $=$ $o(n)$

Here, $\lim_{n\to\infty}$ $P(E)$

$=$ $\lim_{n\to\infty}$ $e^{-na_n}$

$=$ $\lim_{n\to\infty}$ $e^{-n / o(n)}$ --- Because $1/a_n$ $=$ $o(n)$ implies $a_n = 1/o(n)$

I am stuck at this point now.

I wonder whether I have been on the right track here. I'd be grateful for any advice or insights on how to proceed further. Thank you so much.

  • 2
    $\begingroup$ $f(n)=o(g(n))$ as $n\to\infty$ means $\lim_{n\to\infty}(f(n)/g(n))=0$. $\endgroup$ Commented Nov 17, 2023 at 8:30
  • $\begingroup$ @GerryMyerson Thank you so much for your reply! This helps me see that $\lim_{n\to\infty}$ $e^{-n o(1/n)}$ $= 1$. For $\lim_{n\to\infty}$ $e^{-n / o(n)}$, I guess the limit depends on how $o(n)$ compares to $n$. If $o(n)$ grows to infinity but at a slower rate than $n$, the denominator in the exponent $−n/o(n)$ will grow, causing the exponent to approach $-infinity$, and the limit of the whole expression to approach $0$. $\endgroup$ Commented Nov 17, 2023 at 10:29
  • $\begingroup$ I wonder whether we can evaluate this limit $\lim_{n\to\infty}$ $e^{-n / o(n)}$ without more specific information about the growth rate or behavior of $o(n)$ as $n$ approaches infinity. Should I try to work out different cases here, based on the specific behaviour of $o(n)$? $\endgroup$ Commented Nov 17, 2023 at 10:31
  • 2
    $\begingroup$ Have you ever actually seen anyone write $\lim_{n\to\infty}e^{-n/o(n)}$? $\endgroup$ Commented Nov 17, 2023 at 12:02
  • 1
    $\begingroup$ You can think $f(x)=o(g(x))$ $(x\to\text{something})$ by definition means $f(x)=\psi(x)g(x)$ and $\psi(x)\to 0$, $(x\to\text{something})$. So $e^{-n o(1/n)}=e^{-n\psi(n)\frac{1}{n}}=e^{-\psi(n)}$ where $\lim\limits_{n\to\infty}\psi(n)=0$. But $\lim\limits_{n\to\infty}e^{-n o(1/n)}$ is rather weird. $o$ seldom appears in a $\lim$ symbol. $\endgroup$
    – Asigan
    Commented Nov 17, 2023 at 12:12

1 Answer 1


$o(f(n))$ is the set of all functions g(n) where the limit of $g(n) / f(n)$ is 0. (You need to be careful if f(n) = 0).

Now your examples are very different. $n \cdot o(1/n)$ is n, multiplied by a function that becomes smaller and smaller compared to 1/n. So the product becomes smaller and smaller compared to 1.

n / o(n) on the other hand divides n by a number that becomes smaller and smaller compared to n. If you divide by smaller numbers then the result becomes bigger. So n / o(n) becomes larger and larger compared to 1. o(n) can even become zero or negative, so n / o(n) could be undefined or a large negative number.

Now you had $e^{-n \cdot o(1/n)}$. That’s e raised to a power that gets smaller and smaller compared to 1, so the limit is $e^0 = 1$. In your second case, if you could guarantee that your function that is $o(n)$ is always positive, then you would have e raised to larger and larger negative powers with a limit of 0. But sadly, you don’t know that so the values get larger and larger or smaller and smaller.

  • $\begingroup$ Thank you for this! I have explained the context behind my examples below. I'd be grateful for any advice or suggestions. Have a great day! $\endgroup$ Commented Nov 18, 2023 at 5:22

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