From 253808705083aef7dc4a43cf4b29e29e88d8129d Mon Sep 17 00:00:00 2001 From: mhjensen Date: Wed, 28 Aug 2019 22:11:14 +0200 Subject: [PATCH] added some material --- .../Projects/2019/Exercises/addition.do.tx | 25 +++++++++++++++++++ 1 file changed, 25 insertions(+) create mode 100644 doc/src/Projects/2019/Exercises/addition.do.tx diff --git a/doc/src/Projects/2019/Exercises/addition.do.tx b/doc/src/Projects/2019/Exercises/addition.do.tx new file mode 100644 index 000000000..4102a8d4b --- /dev/null +++ b/doc/src/Projects/2019/Exercises/addition.do.tx @@ -0,0 +1,25 @@ +I think a possible way to show why $$\left \langle \hat u_i \right \rangle = 0$$ given that the columns of $$\hat X$$ is centered is by considering $$\left \langle \hat X \hat v_i \right \rangle$$: + + + +$$\begin{align*}
\left \langle \hat X \hat v_i \right \rangle &= \frac{1}{N}\sum_j ( \hat X \hat v_i )_j \\
&= \frac{1}{N}\sum_j \sum_k x_{jk}\hat v_i(k)\\
&= \frac{1}{N}\sum_k \hat v_i(k) \sum_j x_{jk} \\
&= \sum_k \hat v_i(k)\left( \frac{1}{N}\sum_j x_{jk} \right) \\
&= \sum_k \hat v_i(k) \left \langle \hat x_k \right \rangle
\end{align*}$$ + +where $$x_{jk}$$ being the element of $$\hat X$$ at row $$j$$ and column $$k$$, $$( \hat X \hat v_i )_j $$ the $$j$$-th element of the vector $$\hat X \hat v_i $$, $$\hat x_k$$ being the $$k$$-th column vector of $$\hat X$$, and $$\hat v_i(k)$$ the $$k$$-th element of the vector $$\hat v_i$$. + + + +Since the columns of $$\hat X$$ are assumed to be centered, $$\left \langle \hat x_k \right \rangle = 0$$ for all $$k$$. This gives that $$\left \langle \hat X \hat v_i \right \rangle = 0$$. + + + +But $$\left \langle \hat X \hat v_i \right \rangle = \left \langle \hat u_i d_i \right \rangle = d_i \left \langle \hat u_i \right \rangle $$. + +Since $$ \left \langle \hat X \hat v_i \right \rangle = 0$$, then $$d_i \left \langle \hat u_i \right \rangle = 0$$ also. Assuming that $$d_i \neq 0$$ (otherwise the variance in the exercise would just be zero), gives that $$\left \langle \hat u_i \right \rangle = 0$$. + + + +Regarding $$\hat V$$ and using the similar approach as above by computing $$\left \langle \hat X^T \hat u_i \right \rangle = d_i \left \langle \hat v_i \right \rangle$$, it seems that + +$$\begin{align*}
\left \langle \hat X^T \hat u_i \right \rangle &= \frac{1}{N}\sum_j ( \hat X^T \hat u_i )_j \\
&= \frac{1}{N}\sum_j \sum_k x_{kj} \hat u_i(k)\\
&= \frac{1}{N}\sum_k \hat u_i(k) \sum_j x_{kj} \\
&= \frac{1}{N}\sum_k \hat u_i(k) \sum_j x_{kj} \\
&= \sum_k \hat u_i(k) \left( \frac{1}{N} \sum_j x_{kj} \right)\\
\end{align*} $$ + +We do not know anything about the sample mean over the rows of $$\hat X$$