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I have a pdf that doesn't yield trivial derivatives, so I cannot differentiate it and find the root to determine where its max exactly occurs. However, I have a general formula to express all its moments (mean, var, skew, kurtosis, etc).

The mean doesn't coincide with the mode because the function is not symmetrical. Is there a way to use the mean and the skewness (and/or higher moments) to characterize the mode?

Legendary
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Paul Cwave
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    The position of the maximum probability density (if such exists and is unique) is called the _mode_. – Nick Cox Dec 21 '17 at 13:11
  • Thanks Nick, apologies i'm not familiar with statistics terminology. I'll correct it. – Paul Cwave Dec 21 '17 at 13:12
  • The accepted answer does not address the recently-edited version of this question (not the answerer's fault, of course). – mkt Dec 21 '17 at 13:19
  • @mkt Stephan answered that moments cannot determine the max's location (in the comments) – Paul Cwave Dec 21 '17 at 13:22
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    In a reply at https://stats.stackexchange.com/questions/25010/identity-of-moment-generating-functions/25017#25017 I show, with plots of the PDFs, a family of distributions that all share the same (infinitely many) moments, yet clearly have different modes. – whuber Dec 21 '17 at 14:29

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Without more information, no. A maximum may not even exist (e.g., for the normal distribution), and even if it is known to be finite, your moments won't determine the distribution, nor the maximum.

Stephan Kolassa
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  • Thanks Stephan. Can you please explain what do you mean by saying 'a maximum may not even exist'? In my case I have a bell shape kind of curve, so there is a maximum and it is unique. Isn't that the same for the normal distribution? I'm not at all a statistician but I figured you guys would be used to this kind of questions... so here I am ;) – Paul Cwave Dec 21 '17 at 13:01
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    The normal distribution is unbounded. As are many other distributions. Ah, I see now what you mean, you are referring to the maximum of the PDF (I was thinking about the possible maximum of the random variable). OK, this maximum often exists. (Or it may not, as for many gamma distributions.) But even so, you cannot infer either the location nor the value of the maximum from the distribution's moments, per the answer I linked to. – Stephan Kolassa Dec 21 '17 at 13:06
  • (Incidentally, I wasn't the downvoter, if you wonder.) – Stephan Kolassa Dec 21 '17 at 13:06
  • Ok thanks Stephan, i was really wondering if moments could help me characterize the position of that max, because in my understanding the skewness could describe a deviation of the mean position from the max position... – Paul Cwave Dec 21 '17 at 13:10
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    Even that is only true in a very general sense. Skewness is the third moment, so it is really far more influenced by the tails of the distribution than by the mean and max of the PDF (which typically appear near the center, not in the tails). Related is [Westfall, *TAS*, 2014](http://www.tandfonline.com/doi/abs/10.1080/00031305.2014.917055). – Stephan Kolassa Dec 21 '17 at 13:14
  • Watch out on terminology here: As most often reported, skewness and kurtosis are dimensionless ratios based on moments, rather than moments themselves. – Nick Cox Dec 21 '17 at 13:19
  • @NickCox I used the following expression : $$\int_{0}^\infty x^n f(x) dx$$. Isn't that correct? – Paul Cwave Dec 21 '17 at 13:27
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    You've defined no symbol, but implicitly your variable is assumed zero or positive. Otherwise that looks like moments about zero to me. Variance is the second moment about the mean, etc. – Nick Cox Dec 21 '17 at 14:17
  • Ok @NickCox i see better what moments are, and the formula I've been working with only represents moments about 0. Thank you for pointing that out! – Paul Cwave Dec 21 '17 at 14:57