From 045d32c2985d045b103e552472e438d568777eaa Mon Sep 17 00:00:00 2001 From: paciic <70255895+Paciic@users.noreply.github.com> Date: Tue, 28 Oct 2025 18:40:34 +0000 Subject: [PATCH 1/4] Update index.tex --- .../6/sections/means_medians_modes_moments/index.tex | 6 ++++++ 1 file changed, 6 insertions(+) diff --git a/src/chapters/6/sections/means_medians_modes_moments/index.tex b/src/chapters/6/sections/means_medians_modes_moments/index.tex index 1f290f4f..a5ee953a 100644 --- a/src/chapters/6/sections/means_medians_modes_moments/index.tex +++ b/src/chapters/6/sections/means_medians_modes_moments/index.tex @@ -4,3 +4,9 @@ \subsection{problem 1} \input{problems/1} \subsection{problem 2} \input{problems/2} + + + + +\subsection{problem 11} +\input{problems/11} From 82ee053167ad853f6907edc421e23afa278a6adb Mon Sep 17 00:00:00 2001 From: paciic <70255895+Paciic@users.noreply.github.com> Date: Tue, 28 Oct 2025 18:41:07 +0000 Subject: [PATCH 2/4] Create 11.tex --- .../problems/11.tex | 99 +++++++++++++++++++ 1 file changed, 99 insertions(+) create mode 100644 src/chapters/6/sections/means_medians_modes_moments/problems/11.tex diff --git a/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex b/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex new file mode 100644 index 00000000..5590d693 --- /dev/null +++ b/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex @@ -0,0 +1,99 @@ +(a). Since, \[Z_j = \frac{X_j-\mu}{\sigma} \Longrightarrow X_j = \sigma Z_j +\mu\] +so, +\begin{equation} \label{gamma_as_e(z3)} +\begin{split} + \text{Skew}(X_j) &= \text{E} [(\frac{X_j - \mu}{\sigma})^3] \\ + &= \text{E} [(\frac{\sigma Z_j + \mu - \mu}{\sigma})^3] \\ + &= \text{E} [(\frac{\sigma Z_j }{\sigma})^3] \\ + &= \text{E} (Z_j^3). +\end{split} +\end{equation} +Since \(\text{E} (Z_j) = 0\), \(\text{Var}(Z_j) = 1\), \[\text{Skew} (Z_j) = \text{E} [(\frac{Z_j - \text{E} (Z_j)}{\sqrt{\text{Var}(Z_j)}})^3] = \text{E} (Z_j^3) = \text{Skew}(X_j).\] +\\ +Now, \(\bar{X}_n = \frac{1}{n} \sum_{j=1}^{n}X_j\), therefore by linearity of expectation, +\begin{flalign*} + \text{E} (\bar{X}_n) &= \text{E} (\frac{1}{n} \sum_{j=1}^{n}X_j) \\ + &= \frac{1}{n} \text{E} (\sum_{j=1}^{n}X_j) \\ + &= \frac{1}{n} \cdot n \mu \\ + &= \mu +\end{flalign*} +and, +\begin{flalign*} + \text{Var} (\bar{X}_n) &= \text{Var}(\frac{1}{n} \sum_{j=1}^{n}X_j) \\ + &= \frac{1}{n^2} \text{Var} (\sum_{j=1}^{n}X_j). +\end{flalign*} +Since each \(X_j\) is mutually independent, +\begin{flalign*} + \text{Var}(\bar{X}_n) &= \frac{1}{n^2} \cdot n\sigma^2\\ + &= \frac{\sigma^2}{n} +\end{flalign*} +thus, +\begin{flalign*} + \text{Skew}(\bar{X}_n) &= \text{E} [(\frac{\bar{X}_n - \text{E} (\bar{X}_n)}{\sqrt{\text{Var}(\bar{X}_n)}})^3] \\ + &= \text{E} [(\frac{\bar{X}_n - \mu}{\sqrt{\frac{\sigma^2}{n}}})^3] \\ + &= \text{E} [(\frac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma})^3] \\ +\end{flalign*} +Consider \(\bar{Z}_n\), +\begin{flalign*} + \bar{Z}_n &= \frac{1}{n} \sum_{j=1}^{n}Z_j \\ + &= \frac{1}{n} \sum_{j=1}^{n}\frac{X_j-\mu}{\sigma}\\ + &= \frac{(\sum_{j=1}^{n}X_j)-n\mu}{n \sigma} \\ + &= \frac{(n \cdot \frac{\sum_{j=1}^{n}X_j}{n})-n\mu}{n \sigma} \\ + &= \frac{n\bar{X}_n-n\mu}{n \sigma} \\ + &= \frac{\bar{X}_n-\mu}{\sigma} +\end{flalign*} +so, by linearity of expectation, \begin{equation} \label{expectation_z_bar} + \text{E} (\bar{Z}_n) = \frac{1}{n}\sum_{j=1}^{n}\text{E} (Z_j) = 0, +\end{equation}and, +\begin{flalign*} + \text{Var} (\bar{Z}_n) &= \text{Var}(\frac{1}{n} \sum_{j=1}^{n}Z_j) \\ + &= \frac{1}{n^2} \text{Var} (\sum_{j=1}^{n}Z_j) +\end{flalign*} +Since each \(Z_j\) is mutually independent and \(\text{Var}(Z_j) = 1\) for all j, +\begin{flalign*} + \text{Var}(\bar{Z}_n) &= \frac{1}{n^2} \cdot n \cdot 1^2\\ + &= \frac{1}{n} +\end{flalign*} +finally, +\begin{flalign*} + \text{Skew}(\bar{Z}_n) &= \text{E} [(\frac{\bar{Z}_n - \text{E} (\bar{Z}_n)}{\sqrt{\text{Var}(\bar{Z}_n)}})^3] \\ + &= \text{E} [(\frac{\frac{\bar{X}_n-\mu}{\sigma} - 0}{\sqrt{\frac{1}{n}}})^3] \\ + &= \text{E} [(\frac{\frac{\bar{X}_n-\mu}{\sigma}}{\frac{1}{\sqrt{n}}})^3] \\ + &= \text{E} [(\frac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma})^3] \\ + &= \text{Skew}(\bar{X}_n). +\end{flalign*} +And so \(\bar{Z}_n\) has the same skewness as \(\bar{X}_n\). \\ + +(b). From (a), +\begin{flalign*} + \text{Skew}(\bar{X}_n) &= \text{Skew}(\bar{Z}_n)\\ + &= \text{E} [(\frac{\bar{Z}_n - \text{E} (\bar{Z}_n)}{\sqrt{\text{Var}(\bar{Z}_n)}})^3] \\ + &= \frac{1}{(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot \text{E} [(\bar{Z}_n - \text{E} (\bar{Z}_n))^3] \\ + &= \frac{1}{(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot \text{E} ((\bar{Z}_n)^3) +\end{flalign*} +as \(\text{E}(\bar{Z}_n) = 0\) (by (\ref{expectation_z_bar})). \\ +Since \(\text{E} ((\bar{Z}_n)^3) = \text{E}[(\frac{1}{n}\sum_{j=1}^{n}\text{E} (Z_j))^3] = \frac{1}{n^3}\text{E}[(\sum_{j=1}^{n}\text{E} (Z_j))^3]\) (again by \ref{expectation_z_bar}), +\[ + \text{Skew}(\bar{X}_n) = \frac{1}{n^3(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot (\sum_{j=1}^{n}\text{E} (Z_j))^3 +\] +We now focus on \((\sum_{j=1}^{n}\text{E} (Z_j))^3\). +\[ + (\sum_{j=1}^{n} \text{E} (Z_j))^3= \sum_{i=1}^n \text{E}(Z_i)^3 + 3 \sum_{i=1}^{n} \sum_{\substack{j=1 \\ j \ne i}}^{n} \text{E}(Z_i)^2 \text{E}(Z_j) + 6 \sum_{i=1}^{n} \sum_{\substack{j=1 \\ j \ne i}}^{n} \sum_{\substack{k=1 \\ k \ne j, k \ne i}}^{n} \text{E}(Z_i) \text{E}(Z_j) \text{E}(Z_k) +\] +Since \(\text{E}(Z_j) = 0\) for all j, +\begin{flalign*} + (\sum_{j=1}^{n}\text{E} (Z_j))^3 &= \sum_{i=1}^n\text{E}(Z_i)^3 + 3\sum_{i=1}^{n} \sum_{\substack{j=1 \\ j \ne i}}^{n} \text{E}(Z_i)^2 \cdot 0 + 6\sum_{i=1}^{n}\sum_{\substack{j=1 \\ j \ne i}}^{n} \sum_{\substack{k=1 \\ k \ne j, k \ne i}}^{n} 0 \cdot 0 \cdot 0 \\ + &= \sum_{i=1}^n\text{E}(Z_i)^3. +\end{flalign*} +Hence, +\[ + \text{Skew}(\bar{X}_n) = \frac{1}{n^3(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot \sum_{j=1}^{n}(\text{E}(Z_j)^3). +\] +As each \(Z_j\) is independent, by (\ref{gamma_as_e(z3)}), \(\text{E}(Z_j)^3 = \text{E}(Z_j^3) = \text{Skew}(X_j) =\gamma\), and since \(\text{Var}(\bar{Z}_n)) = \frac{1}{n}\), +\begin{flalign*} + \text{Skew}(\bar{X}_n) &= \frac{1}{n^3(\frac{1}{n})^\frac{3}{2}} \cdot \sum_{j=1}^{n}\gamma\\ + &= \frac{1}{n^\frac{3}{2}} \cdot n\gamma\\ + &= \frac{\gamma}{\sqrt{n}}. +\end{flalign*} + +(c). Since \(\text{Skew}(\bar{X}_n) = \frac{\gamma}{\sqrt{n}}\), as n becomes large, skewness becomes small, hence the distribution of \(\bar{X}_n\) will be more symmetric. From 4e1023c0957ba5067ec495d082a5edc093b526aa Mon Sep 17 00:00:00 2001 From: paciic <70255895+Paciic@users.noreply.github.com> Date: Wed, 29 Oct 2025 00:07:07 +0000 Subject: [PATCH 3/4] Update 11.tex --- .../6/sections/means_medians_modes_moments/problems/11.tex | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex b/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex index 5590d693..c96b14c0 100644 --- a/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex +++ b/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex @@ -31,7 +31,7 @@ \begin{flalign*} \text{Skew}(\bar{X}_n) &= \text{E} [(\frac{\bar{X}_n - \text{E} (\bar{X}_n)}{\sqrt{\text{Var}(\bar{X}_n)}})^3] \\ &= \text{E} [(\frac{\bar{X}_n - \mu}{\sqrt{\frac{\sigma^2}{n}}})^3] \\ - &= \text{E} [(\frac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma})^3] \\ + &= \text{E} [(\frac{\sqrt{n}(\bar{X}_n - \mu)}{\sigma})^3] \end{flalign*} Consider \(\bar{Z}_n\), \begin{flalign*} From c0dcd2782f52b4b486d6330e31393835561bb6cb Mon Sep 17 00:00:00 2001 From: paciic <70255895+Paciic@users.noreply.github.com> Date: Wed, 29 Oct 2025 17:49:51 +0000 Subject: [PATCH 4/4] Update 11.tex --- .../means_medians_modes_moments/problems/11.tex | 10 +++------- 1 file changed, 3 insertions(+), 7 deletions(-) diff --git a/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex b/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex index c96b14c0..8ea16ef8 100644 --- a/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex +++ b/src/chapters/6/sections/means_medians_modes_moments/problems/11.tex @@ -9,7 +9,6 @@ \end{split} \end{equation} Since \(\text{E} (Z_j) = 0\), \(\text{Var}(Z_j) = 1\), \[\text{Skew} (Z_j) = \text{E} [(\frac{Z_j - \text{E} (Z_j)}{\sqrt{\text{Var}(Z_j)}})^3] = \text{E} (Z_j^3) = \text{Skew}(X_j).\] -\\ Now, \(\bar{X}_n = \frac{1}{n} \sum_{j=1}^{n}X_j\), therefore by linearity of expectation, \begin{flalign*} \text{E} (\bar{X}_n) &= \text{E} (\frac{1}{n} \sum_{j=1}^{n}X_j) \\ @@ -71,7 +70,7 @@ &= \frac{1}{(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot \text{E} [(\bar{Z}_n - \text{E} (\bar{Z}_n))^3] \\ &= \frac{1}{(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot \text{E} ((\bar{Z}_n)^3) \end{flalign*} -as \(\text{E}(\bar{Z}_n) = 0\) (by (\ref{expectation_z_bar})). \\ +as \(\text{E}(\bar{Z}_n) = 0\) (by (\ref{expectation_z_bar})). \hfill \\ Since \(\text{E} ((\bar{Z}_n)^3) = \text{E}[(\frac{1}{n}\sum_{j=1}^{n}\text{E} (Z_j))^3] = \frac{1}{n^3}\text{E}[(\sum_{j=1}^{n}\text{E} (Z_j))^3]\) (again by \ref{expectation_z_bar}), \[ \text{Skew}(\bar{X}_n) = \frac{1}{n^3(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot (\sum_{j=1}^{n}\text{E} (Z_j))^3 @@ -85,14 +84,11 @@ (\sum_{j=1}^{n}\text{E} (Z_j))^3 &= \sum_{i=1}^n\text{E}(Z_i)^3 + 3\sum_{i=1}^{n} \sum_{\substack{j=1 \\ j \ne i}}^{n} \text{E}(Z_i)^2 \cdot 0 + 6\sum_{i=1}^{n}\sum_{\substack{j=1 \\ j \ne i}}^{n} \sum_{\substack{k=1 \\ k \ne j, k \ne i}}^{n} 0 \cdot 0 \cdot 0 \\ &= \sum_{i=1}^n\text{E}(Z_i)^3. \end{flalign*} -Hence, -\[ - \text{Skew}(\bar{X}_n) = \frac{1}{n^3(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot \sum_{j=1}^{n}(\text{E}(Z_j)^3). -\] +Hence, \[\text{Skew}(\bar{X}_n) = \frac{1}{n^3(\text{Var}(\bar{Z}_n))^\frac{3}{2}} \cdot \sum_{j=1}^{n}(\text{E}(Z_j)^3).\] As each \(Z_j\) is independent, by (\ref{gamma_as_e(z3)}), \(\text{E}(Z_j)^3 = \text{E}(Z_j^3) = \text{Skew}(X_j) =\gamma\), and since \(\text{Var}(\bar{Z}_n)) = \frac{1}{n}\), \begin{flalign*} \text{Skew}(\bar{X}_n) &= \frac{1}{n^3(\frac{1}{n})^\frac{3}{2}} \cdot \sum_{j=1}^{n}\gamma\\ - &= \frac{1}{n^\frac{3}{2}} \cdot n\gamma\\ + &= \frac{1}{n^\frac{3}{2}} \cdot n\gamma \\ &= \frac{\gamma}{\sqrt{n}}. \end{flalign*}