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What does $\dot\sim$ notation (dot over tilde) mean, in the context like $x \mathrel{\dot\sim} \mathcal N(0,1)$?

Turns out it is easier to find how to typeset it correctly: tex.SE explains that one should type \mathrel{\dot\sim} instead of simply \dot\sim to fix the spacing issue -- than to find what it actually means. It has only been used 4 times on CV until now; is it standard?

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    $\begingroup$ The fact that it has only been used 4 times on CV means that there's probably been a lot of technically inaccurate statements on CV. $\endgroup$
    – Cliff AB
    Commented Nov 22, 2015 at 23:50

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Unless there was some other clue as to the intended meaning, I'd interpret that as "is approximately distributed as".

It's fairly standard. Note that some of the other usual ways of indicating "approximation" by modifying a symbol don't really work with $\sim$.

Note that $\sim$ can be read as "is distributed as" and that adding the dot over a symbol at least sometimes indicates approximation -- compare $=$ with $\mathrel{\dot =}$.

So "$x \mathrel{\dot\sim} \mathcal N(0,1)$" could be read something like "$x$ is approximately distributed as standard normal". Personally, I don't mind the closer spacing in \dot\sim ($\dot\sim$) for that use.

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  • $\begingroup$ Thanks, @Glen_b. There are two equally good answers here posted almost simultaneously so I could not decide which one to accept. After hesitating for several days, I decided to accept yours because it was posted 2 minutes earlier than Cliff's. $\endgroup$
    – amoeba
    Commented Nov 27, 2015 at 12:20
  • $\begingroup$ @amoeba If you feel Cliff's is in some way better you should feel free to change your mind about that $\endgroup$
    – Glen_b
    Commented Nov 27, 2015 at 14:28
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"$\dot \sim$" means "approximately distributed as". It is often used as short hand for something like

$\sqrt n (\bar x - \mu)/\sigma \rightarrow_d N(0,1)$ as $n \rightarrow \infty$

i.e. convergence in distribution, but you are too lazy to write out the necessary $n \rightarrow \infty$ to make the statement actually mathematically rigorous.

(Of course, in the above statement, this is exactly distributed if the $x_i \sim_{iid} N(\mu, \sigma)$. But if $x_i$ are not normal, it would only converge in distribution to $N(0,1)$.)

During grad school, one my professors went on a technical, but justified, rampage about how this notation is often used in an abusive manner. For example, if you were to write

$ \hat p \mathrel{\dot\sim} N(p, \sqrt{p(1-p)/n})$

where $\hat p$ is the standard MLE for a binomial distribution, this seems to imply that $\hat p$ is approximately normal for any n, which is of course not true. We were not allowed to use $\dot \sim$ notation in his class, but rather wrote everything out in the proper "converges in distribution" notation.

None of my other professors cared.

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    $\begingroup$ @amoeba: The statement below about $\hat p$ is an example of being too lazy to write out the full mathematically rigorous statement; $\sqrt n (\hat p - p) \rightarrow_d N(0, \sqrt{p(1-p)})$ as $n \rightarrow \infty$ is rigorous, but the statement above with $\dot \sim$ is not (because it implies approximate for all n). Calling $\dot \sim$ the "lazy" version may not be quite right: the actual amount of writing saved is minimal. But it's way easier to say "approximately distributed" to a layperson than "converges in distribution". $\endgroup$
    – Cliff AB
    Commented Nov 23, 2015 at 17:24
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    $\begingroup$ Thanks again, @Cliff. There are two equally good answers here posted almost simultaneously so I could not decide which one to accept. After hesitating for several days, I decided to accept Glen's answer because it was posted 2 minutes earlier. But I like the story about one professor not allowing to use $\dot\sim$ sign. I guess he would also object to using $\approx$ sign, that one is also vague and does not have a clear meaning. $\endgroup$
    – amoeba
    Commented Nov 27, 2015 at 12:24

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