I am trying to show that the central moment of a symmetric distribution:
$${\bf f}_x{\bf (a+x)} = {\bf f}_x{\bf(a-x)}$$ is zero for odd numbers. So for instance the third central moment $${\bf E[(X-u)^3] = 0}.$$ I started by trying to show that $${\bf E[(X-u)^3] = E[X^3] -3uE[X^2] + 3u^2E[X] – u^3}.$$ I am not sure where to go from here, any suggestions? Is there a better way to go about proving this?
Solved – Central Moments of Symmetric Distributions
expected valuemathematical-statisticsmoments
Related Solutions
This question appears to use some statistical terminology in unconventional ways. Understanding this will help resolve the issues:
A "signal" appears to be a (measurable) function $x$ defined on a determined Real interval $[t_-, t_+)$. (This makes $x$ a random variable.) That interval is endowed with a uniform probability density. Taking $t$ as a coordinate for the interval, the probability density function therefore is $\frac{1}{t_+-t_-} dt$.
A "sample" of a signal is a sequence of values of $x$ obtained along an arithmetic progression of times $\ldots, t_0 - 2h, t_0-h, t_0, t_0+h, t_0+2h, \ldots$ = $(t_1,t_2,\ldots,t_n)$ (restricted, of course, to the domain $[t_-, t_+)$). These values may be written $x(t_i) = x_i$.
The "expectation" operator $E$ may refer either to (a) the expectation of the random variable $x$, therefore equal to $\frac{1}{t_+-t_-}\int_{t_-}^{t_+}x(t)dt$ or (b) the mean of a sample, therefore equal to $\frac{1}{n}\sum_{i=1}^n x_i$. This lets us translate formulas for expectations of signals to formulas for expectations of their samples merely by replacing integrals by averages.
"Statistical independence" of signals $x$ and $y$ means they are independent as random variables.
The importance of powers of $\sin$ and $\cos$ becomes evident when we notice that "most" signals can be written as convergent Fourier series -- that is, linear combinations of the functions $\sin(n t)$, $n=1, 2, \ldots$, and $\cos(m t)$, $m=0, 1, 2, \ldots$, and that these functions, in turn, can be written as (finite) linear combinations of non-negative integral powers of products, $\sin^p(t)\cos^q(t)$. Because the expectation operator is linear, we would naturally be interested in studying and computing expectations of such monomials.
Now, $\sin$ and $\cos$ are usually not "statistically independent" except when the length of the interval, $t_+ - t_-$, is a multiple of $2\pi$. The signals of interest in applications are those for which this length is huge. Thus, at least to a very good approximation, we can think of $\sin$ and $\cos$ as being independent. But what does this mean? How can we think about it?
To discuss the concepts, I am going to make an "ink = probability" metaphor by considering scatterplots of large independent samples of $(x,y)$. If $A$ is any (measurable) region within the scatterplot, then the proportion of ink covering $A$ closely approximates the probability of $A$ under the joint distribution of $(x,y)$.
$x$ is plotted on the horizontal axis and $y$ on the vertical. $A$ is the rectangle. The proportion of blue ink drawn on this rectangle equals the proportion of the 2000 dots used, which reflect the joint probability density of $(x,y)$.
The statistical definition of independence of two random variables $x$ and $y$ is that their joint distribution is the product of the marginal distributions. The marginal distribution of $x$ is obtained by taking thin vertical slices of the scatterplot: the chance that $x$ lies between $x'$ and $x'+dx$ equals the proportion of ink used for all points in this region, regardless of the value of $y$: that's a vertical strip.
Now, in any vertical strip, some of the ink appears more in some places than in others. Compare these two strips:
In the right-hand strip in this illustration, the blue ink tends to be higher (have larger $y$ values) than in the left-hand strip. This is lack of independence. Independence means that no matter where the strip is located, you see the same vertical distribution of ink, as here:
What is interesting about this last figure is how it was drawn. It is a scatterplot of a sample $(x(t), y(t))$ at 10,000 equally spaced times. Let's look at the first one percent of this sample:
There is a clear lack of independence here! During any small interval of time, two signals can be highly functionally dependent, but if over a long period of time that functional dependence "averages out," the signals are still considered independent. (In this case the signals were $x(t) = \cos((e/20)t)$ and $y(t) = \sin(t/20)$ sampled at times $t=1, 2, \ldots, 10000$.)
Let's get back to the questions, which concern expectation as well as independence. We can also think of expectation geometrically: in a scatterplot of two signals (or their samples), $(x,y)$, $E[x]$ is the average horizontal location of the dots and $E[y]$ is the average vertical location. For instance, consider the signals $x(t)=\cos(t)$ and $y(t)=\sin(t)$ over the interval $[0, 2\pi)$, with this scatterplot:
The symmetrical form indicates the average of $x$ must be at $0$ and likewise for the average of $y$. Therefore $E[\cos(t)]$ = $E[\sin(t)]$ = $0$ (for this particular domain).
Consider now their squares. Here's the scatterplot:
Of course! $\cos^2(t) + \sin^2(t)=1$, so the squares must lie along the line $x+y=1$. The average of each coordinate is $1/2$. Naturally the averages cannot be zero: squares tend to be positive. So, if we wish to find the second central moment of (say) $x(t) = cos^2(t)$, often written $\mu'_2(x)$, then we first compute its expectation ($1/2$) and (by definition) integrate the second power of the residuals:
$$\mu'_2(x) = E[(x - E[x])^2] = \frac{1}{t_+ - t_-}\int_{t_-}^{t_+}\left(\cos^2(t) - 1/2\right)^2 dt.$$
In general, the $(p,q)$ central moment of a bivariate signal $(x,y)$, written $\mu'_{pq}(x,y)$, is obtained similarly: first compute the expectations of $x$ (written $\mu(x)$) and $y$ (written $\mu(y)$), and then find the expectation of the appropriate monomial in the residuals $x(t) - \mu(x)$ and $y(t) - \mu(y)$:
$$\mu'_{pq}(x,y) = \frac{1}{t_+-t_-}\int_{t_-}^{t_+}\left(x(t)-\mu(x))^p\right)\left(y(t)-\mu(y)\right)^q dt.$$
Continuing the example posed in the question, let $x(t) = \sin(t)$, $y(t) = \cos(t)$, and suppose the domain is a multiple of one common period of both signals; say, $[t_-, t_+)$ = $[0, 2\pi)$. As we have seen, $\mu(x)$ = $\mu(y)$ = 0. Therefore
$$\mu'_{23}(x,y) = \frac{1}{2\pi}\int_0^{2\pi}\left(\sin(t)-0\right)^2\left(\cos(t)-0\right)^3dt = 0.$$
More generally (in this case) $\mu'_{2m,2n}(x,y) = \frac{\pi ^{1/2} 2^{m+2 n-1}}{\Gamma \left(\frac{1}{2}-n\right) \Gamma \left(-m-n+\frac{1}{2}\right) \Gamma (2 m+2 n+1)}$ for non-negative integral $m,n$; all other central moments are zero. (This formula in terms of Gamma functions works for non-integral $m$ and $n$.)
A subtle and potentially confusing point concerns what we consider $x$ to be. If, instead of taking $x(t)=\sin(t)$, we consider the different signal $x(t)=\sin^2(t)$, then, as we have seen, $\mu(x)=1/2$ and therefore
$$\mu'_2(x) = \mu'_2(\sin^2(t)) = \frac{1}{2\pi}\int_0^{2\pi}\left(\sin^2(t)-1/2\right)^2 dt = 1/8.$$
Beware: because the expectation of $\sin^2$ is nonzero, its second central moment, $1/8$, does not necessarily equal the fourth central moment of $\sin$ (which equals $3/8$), even though both integrals involve fourth powers of $\sin$.
Finally, when working with samples, just replace the integrals by averages.
Use the recurrence relation
$$\mu_{r+1}=\lambda\left(\frac{d\mu_r}{d\lambda}+r\mu_{r-1}\right).$$
The initial conditions are
- 1st central moment = $0$
- 2nd central moment = $\lambda$
Using these in the equation you will find the 3rd central moment is $\lambda.$ (Bear in mind that all central moments are zero when $\lambda=0,$ implying the differential equation has a unique solution.)
Again use the 2nd and 3rd central moments to obtain the 4th and so on.
Best Answer
This answer aims to make a demonstration that is as elementary as possible, because such things frequently get to the essential idea. The only facts needed (beyond the simplest kind of algebraic manipulations) are linearity of integration (or, equivalently, of expectation), the change of variables formula for integrals, and the axiomatic result that a PDF integrates to unity.
Motivating this demonstration is the intuition that when $f_X$ is symmetric about $a$, then the contribution of any quantity $G(x)$ to the expectation $\mathbb{E}_X(G(X))$ will have the same weight as the quantity $G(2a-x)$, because $x$ and $2a-x$ are on opposite sides of $a$ and equally far from it. Provided, then, that $G(x) = -G(2a-x)$ for all $x$, everything cancels and the expectation must be zero. The relationship between $x$ and $2a-x$, then, is our point of departure.
Notice, by writing $y = x + a$, that the symmetry can just as well be expressed by the relationship
$$f_X(y) = f_X(2a-y)$$
for all $y$. For any measurable function $G$, the one-to-one change of variable from $x$ to $2a-x$ changes $dx$ to $-dx$, while reversing the direction of integration, implying
$$\mathbb{E}_X(G(X)) = \int G(x) f_X(x)dx = \int G(x) f_X(2a - x)dx = \int G(2a-x)f_X(x)dx.$$
Assuming this expectation exists (that is, the integral converges), the linearity of the integral implies
$$\int \left(G(x) - G(2a - x)\right)f_X(x)dx = 0.$$
Consider the odd moments about $a$, which are defined as the expectations of $G_{k,a}(X) = (X-a)^k$, $k = 1, 3, 5, \ldots$. In these cases
$$\eqalign{ G_{k,a}(x) - G_{k,a}(2a-x) &= (x-a)^k - (2a-x-a)^k \\&= (x-a)^k - (a-x)^k \\ &= (1^k - (-1)^k)(x-a)^k \\&= 2(x-a)^k,}$$
precisely because $k$ is odd. Applying the preceding result gives
$$0 = \int \left(G_{k,a}(x) - G_{k,a}(2a - x)\right)f_X(x)dx = 2\int (x-a)^k f_X(x)dx.$$
Because the right hand side is twice the $k$th moment about $a$, dividing by $2$ shows that this moment is zero whenever it exists.
Finally, the mean (assuming it exists) is
$$\mu_X = \mathbb{E}_X(X) = \int x f_X(x)dx = \int (2a-x)f_X(x)dx.$$
Once again exploiting linearity, and recalling that $\int f_X(x)dx=1$ because $f_X$ is a probability distribution, we can rearrange the last equality to read
$$2\mu_X = 2\int x f_X(x)dx = 2a\int f_X(x)dx = 2a\times 1 = 2a$$
with the unique solution $\mu_X = a$. Therefore all our previous calculations of moments about $a$ are really the central moments, QED.
Postword
The need to divide by $2$ in several places is related to the fact that there is a group of order $2$ acting on the measurable functions (namely, the group generated by the reflection in the line around $a$). More generally, the idea of a symmetry can be generalized to the action of any group. The theory of group representations implies that when the character of that action on a function is not trivial, it is orthogonal to the trivial character, and that means the expectation of the function must be zero. The orthogonality relations involve adding (or integrating) over the group, whence the size of the group constantly appears in denominators: its cardinality when it is finite or its volume when it is compact.
The beauty of this generalization becomes apparent in applications with manifest symmetry, such as in mechanical (or quantum mechanical) equations of motion of symmetrical systems exemplified by a benzene molecule (which has a 12 element symmetry group). (The QM application is most relevant here because it explicitly calculates expectations.) Values of physical interest--which typically involve multidimensional integrals of tensors--can be computed with no more work than was involved here, simply by knowing the characters associated with the integrands. For instance, the "colors" of various symmetric molecules--their spectra at various wavelengths--can be determined ab initio with this approach.