lars/FYS-STK4155
1
0
Fork 0
Code Issues Pull Requests Actions Packages Projects Releases Wiki Activity

updating regression slides

Browse Source
This commit is contained in:
mhjensen
2019-09-05 04:31:27 +02:00
parent 9d8be2597a
commit af4a4d8a46
114 changed files with 15782 additions and 15398 deletions
Download Patch File Download Diff File Expand all files Collapse all files
doc/src/Projects/2019/Exercises/addition.do.tx
+32 -15
View File
@@ -1,25 +1,42 @@
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 $$\left \langle \hat X \hat v_i \right \rangle$$:
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
$\left \langle \hat X \hat v_i \right \rangle$:
!bt
\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*}
!et
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 $\hat v_i$.
$$\begin{align*} <br/>\left \langle \hat X \hat v_i \right \rangle &= \frac{1}{N}\sum_j ( \hat X \hat v_i )_j \\ <br/>&= \frac{1}{N}\sum_j \sum_k x_{jk}\hat v_i(k)\\ <br/>&= \frac{1}{N}\sum_k \hat v_i(k) \sum_j x_{jk} \\<br/>&= \sum_k \hat v_i(k)\left( \frac{1}{N}\sum_j x_{jk} \right) \\ <br/>&= \sum_k \hat v_i(k) \left \langle \hat x_k \right \rangle <br/>\end{align*}$$
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
$\left \langle \hat X \hat v_i \right \rangle = 0$.
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 $\left \langle \hat u_i \right \rangle = 0$.
But $$\left \langle \hat X \hat v_i \right \rangle = \left \langle \hat u_i d_i \right \rangle = d_i \left \langle \hat u_i \right \rangle $$.
Regarding $\hat V$ and using the similar approach as above by
computing $\left \langle \hat X^T \hat u_i \right \rangle = d_i \left
\langle \hat v_i \right \rangle$, we have
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
$$\begin{align*} <br/>\left \langle \hat X^T \hat u_i \right \rangle &= \frac{1}{N}\sum_j ( \hat X^T \hat u_i )_j \\ <br/>&= \frac{1}{N}\sum_j \sum_k x_{kj} \hat u_i(k)\\ <br/>&= \frac{1}{N}\sum_k \hat u_i(k) \sum_j x_{kj} \\<br/>&= \frac{1}{N}\sum_k \hat u_i(k) \sum_j x_{kj} \\ <br/>&= \sum_k \hat u_i(k) \left( \frac{1}{N} \sum_j x_{kj} \right)\\ <br/>\end{align*} $$
We do not know anything about the sample mean over the rows of $$\hat X$$
!bt
\[
\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*}
!et