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Given, a random variable X whose mean , variance and fourth central moment are 0, 2 and 4 respectively. Now, how do I prove that

(1) third moment is 0

(2) distribute is symmetric about 0 and

(3) X is bounded.

With the above information the only thing I could find was that the distribution is platykurtic. Even if it is proved that the third moment is zero, how can this lead to symmetry. Can't symmetry be proved only by plotting the data?

Is there a mistake in the question?

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    $\begingroup$ I know of only one distribution which, up to location and scale, has those moments: a Bernoulli$(1/2)$. Note that its kurtosis is $1$ (its excess kurtosis is $-2$). You might therefore attempt to show that there are no other distributions with these moments. $\endgroup$ – whuber Dec 9 '16 at 17:19
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It is a fact that the function $$\phi_Y: n \to \mathbb{E}(|Y|^n)^{1/n},\ n \gt 0$$

(known as the $L_n$ norm of $Y$) is nondecreasing. The demonstration at https://stats.stackexchange.com/a/244221 uses Jensen's Inequality (as applied to a strictly convex function). That inequality is a strict inequality whenever $|Y|$ can take on more than one value with positive probability.

Letting $Y=X - \bar X$ be the centered version of $X$, from the given values of the variance and fourth (central) moment we deduce

$$\mathbb{E}(|Y|^4)^{1/4} = 4^{1/4} = \sqrt{2} = \operatorname{Var}(X)^{1/2} = \mathbb{E}(|Y|^2)^{1/2},$$

which shows $\phi_Y(4) = \phi_Y(2)$. Consequently, because $\phi$ has not decreased, $|Y|$ is almost surely constant, whence $X$ can take on at most two distinct values $\bar X \pm \sqrt{2}$ (almost surely). It is immediate that $X$ takes on each of those values with equal probability: that is, $X$ must be a shifted version of a Bernoulli$(1/2)$ variable that has been scaled by $\sqrt{8}$.

The demonstration of (1) (zero third moment), (2) (symmetry about $0$), and (3) (boundedness) is now trivial.


Notice that the same conclusions can be drawn whenever there are two moments $k \ne n$ for which $\phi_Y(k)=\phi_Y(n)$.

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  • $\begingroup$ In your answer I think you meant strict inequality when referring to Jensen's Inequality. There is also a condition that the function be convex or concave. $\endgroup$ – Michael Chernick Dec 9 '16 at 23:35
  • $\begingroup$ @Michael Thank you; indeed I meant "inequality." The function in question is $y\log(y)$ which is strictly convex, as demonstrated in the linked thread. $\endgroup$ – whuber Dec 9 '16 at 23:40
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Here's one approach. This answers $a$ , probably $b$, and hopefully $c$.

Summarizing what we know: $E[X]=0$, $E[X^2]=\mbox{Var}(X)=2$ and $E[X^4]=4$. Let $m_i:=E[X^i]$. Moments of any probability distribution must satisfy positive-definiteness, in the sense that any proper $n\times n$ sub matrix of the Hankel Moment Matrix be positive definite:

$$H:=\left(\begin{matrix} m_0 & m_1 & m_2 & \cdots \\ m_1 & m_2 & m_3 & \cdots \\ m_2 & m_3 & m_4 & \cdots \\ \vdots & \vdots & \vdots & \ddots \\ \end{matrix}\right)$$.

Picking $n=3$ gives us:

$$H_4=\left(\begin{matrix} m_0 & m_1 & m_2 \\ m_1 & m_2 & m_3 \\ m_2 & m_3 & m_4 \\ \end{matrix}\right)=\left(\begin{matrix} 1 & 0 & 2 \\ 0 & 2 & m_3 \\ 2 & m_3 & 4 \\ \end{matrix}\right),$$

and a quick hand calculation gives: $\mbox{det}(H_4)=-m_3^2$. Since $H_4$ must be positive definite, it follows that $m_3=0$.

To show that $X$ is symmetric about 0, it suffices to show that all odd moments are zero. I believe you can show this by induction on the Hankel sub-matrices.

To show that $X$ is bounded, the idea I had is the following equivalence:

$$P(|X|\leq R)=1 \Leftrightarrow E[|X|^k]\leq R^k, k=1,2,\cdots .$$

Maybe you can show this from the Hankel matrices?

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