Hi

@Ali Ehsan Abbas *Re: Wouldn't a distribution with mean 0 and variance of 1 by default be a standard normal distribution. *Nope

a distribution with non-zero skew and/or non-zero excess kurtosis is non-normal. For example, a normal mixture distribution can mix two normals, each with µ=0 and variance of 1.0, such that the normal mixture distribution that mixes them equally has mean of zero, variance of 1.0, but heavy-tails (as do all variance mixture distributions). But that's just to illustrate that the definition of normal depends on different third (skew) and fourth (kurtosis) moments.

Re: elliptical, I like your thinking!

I'm not sure, I have my doubts only because I *think* elliptical includes fat-tailed distributions (

https://en.wikipedia.org/wiki/Elliptical_distribution) so I don't mean to be a stick in the mud, but couldn't you just "shift left" a heavy-tailed distribution ? ... The "problem" with 1.56 deviate in isolation is that you don't know where the mean is, so you don't know if that's in the tail or in the body, or the extreme tail etc! However,

**as usual, I could be wrong** Thanks,

append: I gave this a bit more thought, and I'm going to double-down on my previous response. I think the condition should be that the 1.56 is a

*standardized* deviate. Any non-normal deviate, say given by (D), can be standardized with (D - µ)/σ, but as mentioned, to standardize

**does not imply **the distribution normal or somehow "made normal." That is, a standardized 1.56 ought to have lighter-than-normal tails (aka, platykurtosis). To be real strict, since I can still imagine exceptions even in this case, I'd say the condition is actually: if the

*standardized* deviate is 1.56 and the distribution is

*unimodal*. Thanks!

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