Suppose I have a probability distribution $A$ with continuous support over $\mathbb{R}$. Suppose $A$ has a sequence of finite (central) moments $\mu_1, \mu_2,\ldots,\mu_n$. I understand that $\mu_1$ is the mean, and $\mu_2$, $\mu_3$ and $\mu_4$ define variance, skewness, and kurtosis of the distribution, respectively.

I am wondering about the meaning of $\mu_5, \mu_6, \mu_7,\ldots$ When they are finite, what do they represent about the distribution $A$?

I understand that the odd central moments of the symmetric distribution are zero, so I am assuming that odd moments are related to the skew. What do even higher moments represent? I am particularly curious about $\mu_6$.

Also, suppose all moments of $A$ are finite. What does that say about $A$? Does it mean that $A$ has a specific representation? I've heard somewhere that all finite moments of $A$ with support $\mathbb{R}$ means that the tails of $A$ decay exponentially. Is that true? If so, can someone point me to a proof?


There is actually a quite simple visual interpretation to all the higher central moments.

To ease the interpretation, and without loss of generality, assume that the moments refer to centered and standardized random variables.

Let $Z = (X-\mu)/\sigma$ and $V = Z^k$, and let $p_k(v)$ denote the probability density or mass function of $V$. For odd $k$, $p_k(v)$ has positive and negative support; only non-negative support for even $k$.

Here is the visual: The $k$th central (standardized) moment of $X$ is equal to the point of balance of $p_k(v)$.

Since the transformation greatly dilates values where $|Z| >1$ and contracts values where $|Z| < 1$, the point of balance is mostly determined by the extremes where $|Z| > 1$.

In the case of odd $k$, this point of balance is determined by the relative extremity of the left and right tails of the distribution of $X$ in the portions of the tails that are most amplified by the given power $k$. Higher $k$ amplifies more extreme portions of the tails.

In the case of even $k$, this point of balance is determined by the overall extremity the tails of the distribution of $X$, without regard to whether left or right, again in the portion of the tail that is most amplified by the given power $k$. Again, higher $k$ amplifies more extreme portions of the tail.

It is worth noting that this visual representation shows that the appearance of the center of the distribution of $X$ is all but irrelevant. In particular, "peakedness" or "flatness" interpretations of even moments are erroneous.

  • 1
    $\begingroup$ +1 for taking the time to answer a 7+ year old question. I understand the visual that you describe, but for other readers an actual diagram might be useful. :) $\endgroup$
    – M.B.M.
    Mar 3 '19 at 21:26

For your final statement the classic distribution to consider is the log-normal distribution. It is an example of a case where all moments are finite, but the moment generating function does not exist and the moments do not determine the distribution.

Take for example the log-normal distribution with parameters $\mu=0$ and $\sigma = 1/\sqrt2$ which has a distribution on positive values with density $$f(x)=\frac{1}{ x^{1+\log_e x}\sqrt{\pi} }$$

This has finite moments about $0$ of $E[X^n]= \exp(n^2/4)$ but the density of the tail does not decay exponentially.

  • $\begingroup$ The moment generating function does not exist in a neighborhood around zero, which is what is key. It does, of course, exist, for $t \leq 0$ since the lognormal takes on only nonnegative values. $\endgroup$
    – cardinal
    Nov 10 '11 at 0:32
  • $\begingroup$ I see. But if MGF exists around 0, then, by Chernoff bound, the tails decay exponentially (this is my interpretation of Lemma 11.9.1 in Cover and Thomas's "Elements of Information Theory" 2nd edition). And if MGF exists around 0, all moments exist and are finite. Are those two statements correct? $\endgroup$
    – M.B.M.
    Nov 10 '11 at 0:58

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