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January 12th, 2018, 01:20 PM   #1
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Least Squares Problem

This problem projects $\displaystyle b = (b_1,b_2....,b_m)$ onto the line through $\displaystyle a = (1, 1, 1, ....1)$. We solve m equations ax = b in 1 unknown (by least squares).

(a) Solve $\displaystyle a_T~a~\hat{x} = a_T~b $ to show that \hat{x} is the mean (the average) of the b’s.
(b) Find $\displaystyle e = b - a \hat{x}$ and the variance $\displaystyle ||e||^2$ and the standard deviation $\displaystyle ||e||$.
(c) The horizontal line $\displaystyle \hat{b} = 3 $ is closest to b = (1, 2, 6). Check that p = (3, 3 3) is perpendicular to e and find the 3 by 3 projection matrix P.

Ans(a): Becauase a = (1,1,1,....1), therefore $\displaystyle a_T a = 1 + 1 + 1 +....+ 1 = 1~*~m = m$
And $\displaystyle a_T b = b_1 + b_2 + .... + b_m $
So $\displaystyle \hat{x} = \frac{b_1 + b_2 + b_3 + .... + b_m}{m} = b_{avg} $

Ans(b): Need help..

Ans(c): Need help..

Last edited by zollen; January 12th, 2018 at 01:42 PM.
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January 12th, 2018, 01:39 PM   #2
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b) Let $\displaystyle \epsilon_k=b_x-\hat{x} $. $\displaystyle \epsilon = (\epsilon_1,...\epsilon_m) $, $\displaystyle ||\epsilon ||^2=\frac{\epsilon_1^2+...\epsilon_m^2}{m} $.
Ill let you do (c).
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January 13th, 2018, 06:45 AM   #3
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Ans(b): $\displaystyle Variance = \frac{\sum_{k=1}^{m} (b_k - b_{avg})^2}{m} $

Ans(c): e = (1, 2, 6) - (3, 3, 3) = (-2, -1, 3)
(-2, -1, 3) dot_product (3, 3, 3) = 0

Since p = Pb, (3, 3, 3) = P (1, 2, 6)
Project Matrix (P) = \begin{bmatrix}
3&0&0\\
0&3/2&0\\
0&0&1/2\\
\end{bmatrix}
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January 14th, 2018, 06:23 AM   #4
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Ans(b): Given $\displaystyle \hat{x} = b_{avg} $
Therefore
$\displaystyle
e = b - a \hat{x} =
\begin{bmatrix}
b_1\\
b_2\\
..\\
b_m
\end{bmatrix}
~-~
\begin{bmatrix}
1\\
1\\
..\\
1
\end{bmatrix}
~b_{avg}
~=~
\begin{bmatrix}
e_1\\
e_2\\
..\\
e_m
\end{bmatrix}
$

$\displaystyle
||e||^2~=~\frac{{e_1}^2 + {e_2}^2 + ... + {e_m}^2}{m}
$
$\displaystyle
||e||~=~\sqrt{ \frac{{e_1}^2 + {e_2}^2 + ... + {e_m}^2}{m} }
$

Ans(c): $\displaystyle p=Pb~~and~~P= \frac{a * a^t}{a^t * a} $
$\displaystyle
a = \begin{bmatrix}
1\\
1\\
1
\end{bmatrix}
$
But...
$\displaystyle
a^t * a
~=~
\begin{bmatrix}
1&1&1\\
1&1&1\\
1&1&1
\end{bmatrix}
$ has no inverse....

Would anyone show me how to calculate the projection matrix?

Last edited by zollen; January 14th, 2018 at 06:25 AM.
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