The available text books in most cases avoid greater details regarding some of the topics related to the moments of a probability distribution and I feel some of those topics are not really clear to me. I believe someone can explain me the following queries very clearly.

Query-1: I keep reading that characteristic functions are finite because it has a modulus less than or equal to 1 and also that the MGF doesn't always exist. That's why we need characteristic functions. Yes, I do understand that, $e^{itx}=\cos(tx)+i \sin(tx)$ and the modulus $|r|=\sqrt{\cos^2(tx)+\sin^2(tx)}=1$. But how does this determine characteristic functions always exist? What is the precise answer to why we need characteristic functions?

Query-2: I don't understand how the third and fourth central moments (and hence the Pearson's coefficients of skewness and kurtosis) determine the asymmetry and peakedness of a distribution.

  • 2
    $\begingroup$ @Glen_b makes some remarks on the relationship between kurtosis and heaviness of tails (another way of expressing "peakedness") at stats.stackexchange.com/a/86431/919. Note that no moment can tell you anything about the shape of a density function in a neighborhood of its mode(s) (as I argue in my answer in the same thread), so in this colloquial sense of "peakedness," kurtosis is irrelevant. Note, too, that skewness is not synonymous with asymmetry. $\endgroup$ – whuber Mar 3 '15 at 15:33
  1. Existence. If we have a random variable $x$ with the density function $f(x)$, then we know that $$\int_{-\infty}^\infty f(x) dx=1$$ rewrite this as follows $$1=\int_{-\infty}^\infty 1\cdot f(x) dx\ge |\int_{-\infty}^\infty(\cos tx + i \sin tx)\cdot f(x) dx|$$ simply because $$1\ge |\cos tx + i \sin tx|$$

Hence, characteristic function always exists $$E[e^{itx}]\equiv \int_{-\infty}^\infty e^{itx} f(x) dx$$

This is not a proof that a proper mathematician would accept, but you wanted an intution. Basically, $e^{iz}$ confines the outcome into $[-1,1]$, and we know already that the density function would produce a finite integral with even all 1s.

2. Symmetry is easy.

Skew is defined as $\int_{-\infty}^\infty x^3 f(x)$, so if your density is symmetrical, i.e. $f(x)=f(-x)$, the skew has got to be zero because: $$\int_{-\infty}^0 -x^3 f(-x) dx= -\int_0^\infty x^3 f(x)dx$$

Note, @whuber's comment: skew=0 is necessary for symmetry, it's not sufficient though. You can easily construct asymmetric distribution with zero skew. All you need is to make it that the left side of 0 adds up to the right side, and the shapes of these sides don't have to be the same.

I'm assuming the mean is zero here, i.e. symmetry is around the origin, but it's not important, you can always re-center.

Kurtosis is defined as $$\frac{\int_{-\infty}^\infty x^4 f(x)dx}{(\int_{-\infty}^\infty x^2 f(x)dx)^2}$$. If this was a discrete case then the numerator would have terms like $x_i^4 p_i$, where $p_i$ probability of a value $x_i$. On the other hand the denominator would have terms like $x_i^4 p_i^2$. So, the tail values with power 4 will enter with weight $p_i$ in numerator, but with weight $p_i^2$ in the denominator. The heavy tail will make kurtosis larger.

@whuber noticed an error in my previous explanation of kurtosis.

| cite | improve this answer | |
  • $\begingroup$ +1. Shouldn't all your integrals be over $\mathbb R$? Or do you mean that $\Omega=\mathbb R$ in this case? $\endgroup$ – ekvall Mar 3 '15 at 13:59
  • $\begingroup$ @Student001, by $\Omega$ I meant all space of $X$, so it could be denoted as $R$ too. $\endgroup$ – Aksakal Mar 3 '15 at 14:05
  • $\begingroup$ Ok, you mean that $X$ is $\Omega-$valued? In standard notation $\Omega$ would be the sample space but the density a function on $\mathbb R$ so the integral in that case doesn't make sense. Unless $\Omega = \mathbb R$ of course. $\endgroup$ – ekvall Mar 3 '15 at 14:10
  • $\begingroup$ @Student001, frankly, I doubt that OP would have confusion with my lazy usage of $\Omega$ and the standard $(\Omega,\mathcal{F},P)$, but I changed it to make you happy $\endgroup$ – Aksakal Mar 3 '15 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.