- #1
Usagi
- 45
- 0
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: $\displaystyle g_2 = \frac{\sum_{i=1}^n (x_i - \overline{x})^4}{(n-1)s^4}$----------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:
$$\frac{1}{p}\left[Y - \frac{Y^2}{2} + \frac{Y^3}{3} - \frac{Y^4}{4} + \frac{Y^5}{5} - \cdots\right]$$
Ignore the terms from $\frac{Y^3}{3}$ onwards and substitute in the original expression for $Y$:
$$\frac{1}{p}\left[ \left(\frac{p}{n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \cdots \right) - \frac{1}{2}\left(\frac{p}{n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \cdots\right)^2 + \cdots \right]$$
$$=\frac{1}{p} \left[\frac{p}{n} \sum_{i=1}^n x_i + \frac{p^2}{2n} \sum_{i=1}^n x_i^2 - \frac{p^2}{2n^2}\left(\sum_{i=1}^n x_i \right)^2 + \cdots\right] $$
$$\approx \frac{1}{n} \sum_{i=1}^n x_i + \frac{p}{2n} \sum_{i=1}^n x_i^2 - \frac{p}{2} \left( \frac{1}{n} \sum_{i=1}^n x_i\right)^2 $$
$$ = \overline{x} + \frac{p}{2} \left[\frac{1}{n}\sum_{i=1}^n x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i \right)^2\right]$$
$$= \overline{x} + \frac{p}{2} \left[\frac{n-1}{n} s^2 \right]$$----------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.
$$\frac{1}{p} \ln\left(1+\frac{p^1}{1!n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \frac{p^3}{3!n} \sum_{i=1}^n x_i^3 + \frac{p^4}{4!n} \sum_{i=1}^n x_i^4 + \cdots \right)$$
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: $\displaystyle g_2 = \frac{\sum_{i=1}^n (x_i - \overline{x})^4}{(n-1)s^4}$----------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:
$$\frac{1}{p}\left[Y - \frac{Y^2}{2} + \frac{Y^3}{3} - \frac{Y^4}{4} + \frac{Y^5}{5} - \cdots\right]$$
Ignore the terms from $\frac{Y^3}{3}$ onwards and substitute in the original expression for $Y$:
$$\frac{1}{p}\left[ \left(\frac{p}{n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \cdots \right) - \frac{1}{2}\left(\frac{p}{n}\sum_{i=1}^n x_i + \frac{p^2}{2!n} \sum_{i=1}^n x_i^2 + \cdots\right)^2 + \cdots \right]$$
$$=\frac{1}{p} \left[\frac{p}{n} \sum_{i=1}^n x_i + \frac{p^2}{2n} \sum_{i=1}^n x_i^2 - \frac{p^2}{2n^2}\left(\sum_{i=1}^n x_i \right)^2 + \cdots\right] $$
$$\approx \frac{1}{n} \sum_{i=1}^n x_i + \frac{p}{2n} \sum_{i=1}^n x_i^2 - \frac{p}{2} \left( \frac{1}{n} \sum_{i=1}^n x_i\right)^2 $$
$$ = \overline{x} + \frac{p}{2} \left[\frac{1}{n}\sum_{i=1}^n x_i^2 - \left(\frac{1}{n}\sum_{i=1}^n x_i \right)^2\right]$$
$$= \overline{x} + \frac{p}{2} \left[\frac{n-1}{n} s^2 \right]$$----------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.