Taylor expansion to contain sample mean, sample variance, sample skewness, and sample kurtosis

Question

I have the following expression:

$$\frac{1}{p} \ln\left(1+\frac{p^1}{1!n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \frac{p^3}{3!n} \sum_{i=1}^n x_i^3 + \frac{p^4}{4!n} \sum_{i=1}^n x_i^4 + \cdots \right)$$

Now let

$$Y = \frac{p^1}{1!n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \frac{p^3}{3!n} \sum_{i=1}^n x_i^3 + \frac{p^4}{4!n} \sum_{i=1}^n x_i^4 + \cdots$$

Then we have

$$\frac{1}{p}\ln\left(1+Y\right)$$

Using the Taylor series expansion on log, we have

$$\frac{1}{p}\ln(1+Y) = \frac{1}{p}\sum_{n=1}^{\infty} (-1)^{n+1} \frac{Y^n}{n} = \frac{1}{p}\left[Y - \frac{Y^2}{2} + \frac{Y^3}{3} - \frac{Y^4}{4} + \frac{Y^5}{5} - \cdots\right] $$

My question is, how can I expand the above expression so that my final approximation contains the following variables:

Sample mean: $\displaystyle \overline{x} = \frac{1}{n} \sum_{i=1}^n x_i$

Sample variance: $\displaystyle s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2 $

Sample skewness: $\displaystyle g_1 = \frac{\sum_{i=1}^n (x_i - \overline{x})^3}{(n-1)s^3}$

Sample kurtosis: $\displaystyle g_2 = \frac{\sum_{i=1}^n (x_i - \overline{x})^4}{(n-1)s^4}$

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To illustrate exactly what I mean, I know how to obtain the approximation so that it includes the sample mean and sample variance. Start with the expression as derived above:

$$\frac{1}{p}\left[Y - \frac{Y^2}{2} + \frac{Y^3}{3} - \frac{Y^4}{4} + \frac{Y^5}{5} - \cdots\right]\ \ \ \cdots \ \ (1)$$

Ignore the terms from $\frac{Y^3}{3}$ onwards and substitute in the original expression for $Y$:

$$\frac{1}{p}\left[ \left(\frac{p}{n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \cdots \right) - \frac{1}{2}\left(\frac{p}{n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \cdots\right)^2 + \cdots \right]$$

$$=\frac{1}{p} \left[\frac{p}{n} \sum_{i=1}^n x_i + \frac{p^2}{2n} \sum_{i=1}^n x_i^2 - \frac{p^2}{2n^2}\left(\sum_{i=1}^n x_i \right)^2 + \cdots\right] $$

$$\approx \frac{1}{n} \sum_{i=1}^n x_i + \frac{p}{2n} \sum_{i=1}^n x_i^2 - \frac{p}{2} \left( \frac{1}{n} \sum_{i=1}^n x_i\right)^2 $$

$$ = \overline{x} + \frac{p}{2} \left[\frac{1}{n}\sum_{i=1}^n x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i \right)^2\right]$$

$$= \overline{x} + \frac{p}{2} \left[\frac{n-1}{n} s^2 \right] \ \ \ \cdots \ \ (2)$$

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As can be seen above, the final approximation contains the sample mean and sample variance. However, I am not sure how exactly I can manipulate the expansion to contain sample skewness and sample kurtosis. Any help would be appreciated.

Answer 1

This answer is more an illustration of something that doesn't quite work (and an explanation of why).

Starting here:

Sample mean: $\displaystyle \overline{x} = \frac{1}{n} \sum_{i=1}^n x_i$

Sample variance: $\displaystyle s^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \overline{x})^2 $

Sample skewness: $\displaystyle g_1 = \frac{\sum_{i=1}^n (x_i - \overline{x})^3}{(n-1)s^3}$

Sample kurtosis: $\displaystyle g_2 = \frac{\sum_{i=1}^n (x_i - \overline{x})^4}{(n-1)s^4}$

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Sample third moment: $\frac{1}{n-1}\sum_{i=1}^n (x_i - \overline{x})^3=s^3 g_1$

Sample fourth moment: $\frac{1}{n-1}\sum_{i=1}^n (x_i - \overline{x})^4=s^4 g_2$

So if we let $S_k=\sum_{i=1}^n (x_i - \overline{x})^k$, then we have:

$S_2 = (n-1) s^2\quad\quad\quad\quad(A)$

$S_3 = (n-1) s^3 g_1$

$S_4 = (n-1) s^4 g_2$

Now let $T_k=\sum_{i=1}^n x_i^k$. Note that $T_1 = n\bar x$. We can expand as follows:

$T_k=\sum_{i=1}^n x_i^k = \sum_{i=1}^n (x_i-\bar x +\bar x) ^k$

From the binomial expansion of $(a+b)^k$, we have the particular cases:

$(a+b)^2 \;=\; a^2 \,+\, 2 a b^2 \,+\, b^2\,$,
$(a+b)^3 \;=\; a^3 \,+\, 3 a^2b \,+\, 3 a b^2 \,+\, b^3\,$, and
$(a+b)^4 \;=\; a^4 \,+\, 4 a^3b \,+\, 6 a^2 b^2 \,+\, 4 a b^3 \,+\, b^4\,$.

$T_2=\sum_{i=1}^n (x_i-\bar x +\bar x)^2 = S_2 +2 \bar x\sum_{i=1}^n (x_i-\bar x)+n\bar{x}^2=S_2+0 +n\bar{x}^2=S_2+n\bar{x}^2$

$T_3=\sum_{i=1}^n (x_i-\bar x +\bar x)^3=S_3+ 3\bar x \sum_{i=1}^n (x_i-\bar x )^2+3\bar{x}^2\sum_{i=1}^n (x_i-\bar x )+n\bar{x}^3\\ \quad =S_3+ 3\bar x S_2+0+n\bar{x}^3=S_3+3\bar x . S_2 + n\bar{x}^3$

$T_4=\sum_{i=1}^n (x_i-\bar x +\bar x)^4\\ \quad =S_4+4\bar x \sum_{i=1}^n (x_i-\bar x)^3+6\bar{x}^2 \sum_{i=1}^n (x_i-\bar x)^2+4\bar{x}^3 \sum_{i=1}^n (x_i-\bar x)+n\bar{x}^4\\ \quad =S_4+4\bar x S_3+6\bar{x}^2 S_2+0+n\bar{x}^4=S_4+4\bar x S_3+6\bar{x}^2 S_2+n\bar{x}^4$

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Given all that, what do you think about:

$Y = \frac{p^1}{1!n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \frac{p^3}{3!n} \sum_{i=1}^n x_i^3 + \frac{p^4}{4!n} \sum_{i=1}^n x_i^4 + \cdots\\ \quad =\frac{p^1}{1!n}T_1+ \frac{p^2}{2!n} T_2 + \frac{p^3}{3!n} T_3 + \frac{p^4}{4!n} T_4 + \cdots\\$

$\ln(1+Y) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{Y^n}{n} = \left[Y - \frac{Y^2}{2} + \frac{Y^3}{3} - \frac{Y^4}{4} + \frac{Y^5}{5} - \cdots\right]$
$\quad = \left[Y - \frac{1}{2}Y^2 + \frac{1}{3}Y^3 - \frac{1}{4}Y^4 + \frac{1}{5}Y^5 - \cdots\right]$

$\quad = \left[(\frac{p^1}{1!n}T_1+ \frac{p^2}{2!n} T_2 + \frac{p^3}{3!n} T_3 + \frac{p^4}{4!n} T_4 + \cdots)\\ - \frac{1}{2}(\frac{p^1}{1!n}T_1+ \frac{p^2}{2!n} T_2 + \frac{p^3}{3!n} T_3 + \frac{p^4}{4!n} T_4 + \cdots)^2 \\ + \frac{1}{3}(\frac{p^1}{1!n}T_1+ \frac{p^2}{2!n} T_2 + \frac{p^3}{3!n} T_3 + \frac{p^4}{4!n} T_4 + \cdots)^3 \\ - \frac{1}{4}(\frac{p^1}{1!n}T_1+ \frac{p^2}{2!n} T_2 + \frac{p^3}{3!n} T_3 + \frac{p^4}{4!n} T_4 + \cdots)^4 \\ + \frac{1}{5}(\frac{p^1}{1!n}T_1+ \frac{p^2}{2!n} T_2 + \frac{p^3}{3!n} T_3 + \frac{p^4}{4!n} T_4 + \cdots)^5 \\ - \cdots\right]$
$\quad = (\frac{p^1}{1!n}T_1\left[1 - \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}+ \frac{1}{5}\cdots\right]$
$+ \frac{p^2}{2!n} T_2 \left[1 - \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}+ \frac{1}{5}\cdots\right]$
$+ \frac{p^3}{3!n} T_3 \left[1 - \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}+ \frac{1}{5}\cdots\right]$
$+ \frac{p^4}{4!n} T_4 \left[1 - \frac{1}{2}+ \frac{1}{3}- \frac{1}{4}+ \frac{1}{5}\cdots\right]$
$ + \cdots)$

[There's a issue here I'll come back to.]

$\quad = \ln 2 (\frac{p^1}{1!n}T_1+ \frac{p^2}{2!n} T_2 + \frac{p^3}{3!n} T_3 + \frac{p^4}{4!n} T_4 + \cdots)$

$\quad = \ln 2 (\frac{p^1}{1!n}n\bar x+ \frac{p^2}{2!n} (S_2+n\bar{x}^2) + \frac{p^3}{3!n} (S_3+3\bar x . S_2 + n\bar{x}^3) + \frac{p^4}{4!n} (S_4+4\bar x S_3+6\bar{x}^2 S_2+n\bar{x}^4) + \cdots)$

etc (you then substitute the equations at $(A)\,$).

Now to the issue. That change in order of summation is really only legal if the series is absolutely convergent (the alternating harmonic series isn't). So I guess you can't actually do that.

The other thing is that we actually (in spite of me saying otherwise in comments) also need to keep track of the last few terms from the expansion of every higher moment as well (not that this would necessarily prevent us from proceeding), so perhaps this approach isn't fruitful even if there weren't the problem with the alternating harmonic series.

Another approach may be more useful. (When a good answer gets posted I may remove this one.)