-
@@ -161,6 +178,9 @@ MathJax.Hub.Config({
-We are obviously not limited to the above polynomial. We could replace the various powers of \( x \) with elements of Fourier series, that is, instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions. -For every set of values \( y_i,x_i \) we can then generalize the equations to +Defining the vectors $$ -\begin{align*} -y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ -y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ -\dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ -\dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}^{n-1,n-1}+\epsilon_{n-1}.\\ -\end{align*} +\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T, $$ -We redefine in turn the matrix \( \hat{X} \) as +$$ +\hat{\beta} = [\beta_0,\beta_1, \beta_2,\dots, \beta_{n-1}]^T, +$$ + +$$ +\hat{\epsilon} = [\epsilon_0,\epsilon_1, \epsilon_2,\dots, \epsilon_{n-1}]^T, +$$ + +and the matrix $$ \hat{X}= \begin{bmatrix} -x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ -x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ -x_{20}& x_{21}^1 &x_{22}2& \dots & \dots &x_{2,n-1}\\ +1& x_{0}^1 &x_{0}^2& \dots & \dots &x_{0}^{n-1}\\ +1& x_{1}^1 &x_{1}^2& \dots & \dots &x_{1}^{n-1}\\ +1& x_{2}^1 &x_{2}^2& \dots & \dots &x_{2}^{n-1}\\ \dots& \dots &\dots& \dots & \dots &\dots\\ -x_{n-1,00}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ +1& x_{n-1}^1 &x_{n-1}^2& \dots & \dots &x_{n-1}^{n-1}\\ \end{bmatrix} $$ -and without loss of generality we rewrite again our equations as +we can rewrite our equations as $$ \hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. $$ - -The left-hand side of this equation forms know. Our error vector \( \hat{\epsilon} \) and the parameter vector \( \hat{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
-We have defined the matrix \( \hat{X} \) +We are obviously not limited to the above polynomial. We could replace the various powers of \( x \) with elements of Fourier series, that is, instead of \( x_i^j \) we could have \( \cos{(j x_i)} \) or \( \sin{(j x_i)} \), or time series or other orthogonal functions. +For every set of values \( y_i,x_i \) we can then generalize the equations to $$ \begin{align*} y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ -y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_2\\ \dots & \dots \\ -y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_i\\ \dots & \dots \\ -y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}^{n-1,n-1}+\epsilon_{n-1}.\\ \end{align*} $$ - -We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as -$$ -\hat{\tilde{y}}= \hat{X}\hat{\beta}, -$$ - -and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parametrized values \( \tilde{y}_i \), namely -$$ -Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right), -$$ - -or using the matrix \( \hat{X} \) as -$$ -Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right). -$$
-The function +We redefine in turn the matrix \( \hat{X} \) as $$ -Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right), +\hat{X}= +\begin{bmatrix} +x_{00}& x_{01} &x_{02}& \dots & \dots &x_{0,n-1}\\ +x_{10}& x_{11} &x_{12}& \dots & \dots &x_{1,n-1}\\ +x_{20}& x_{21}^1 &x_{22}2& \dots & \dots &x_{2,n-1}\\ +\dots& \dots &\dots& \dots & \dots &\dots\\ +x_{n-1,00}& x_{n-1,1} &x_{n-1,2}& \dots & \dots &x_{n-1,n-1}\\ +\end{bmatrix} $$ -can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value of for example a numerical experiment. When linking below with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value +and without loss of generality we rewrite again our equations as $$ -y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, +\hat{y} = \hat{X}\hat{\beta}+\hat{\epsilon}. $$ -where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. - -
-In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring -$$ -\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0, -$$ - -which results in -$$ -\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0, -$$ - -or in a matrix-vector form as -$$ -\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right). -$$ - -
+The left-hand side of this equation forms know. Our error vector \( \hat{\epsilon} \) and the parameter vector \( \hat{\beta} \) are our unknow quantities. How can we obtain the optimal set of \( \beta_i \) values?
-We can rewrite +We have defined the matrix \( \hat{X} \) $$ -\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right), +\begin{align*} +y_0&=\beta_0x_{00}+\beta_1x_{01}+\beta_2x_{02}+\dots+\beta_{n-1}x_{0n-1}+\epsilon_0\\ +y_1&=\beta_0x_{10}+\beta_1x_{11}+\beta_2x_{12}+\dots+\beta_{n-1}x_{1n-1}+\epsilon_1\\ +y_2&=\beta_0x_{20}+\beta_1x_{21}+\beta_2x_{22}+\dots+\beta_{n-1}x_{2n-1}+\epsilon_1\\ +\dots & \dots \\ +y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsilon_1\\ +\dots & \dots \\ +y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\ +\end{align*} $$ - -as -$$ -\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta}, -$$ - -and if the matrix \( \hat{X}^T\hat{X} \) is invertible we have the solution -$$ -\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}. -$$ - -The residuals \( \hat{\epsilon} \) are in turn given by -$$ -\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta}, -$$ - -and with -$$ -\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, -$$ - -we have -$$ -\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, -$$ - -meaning that the solution for \( \hat{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach. - -
-Here we derive the equations for the SVD. +We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as +$$ +\hat{\tilde{y}}= \hat{X}\hat{\beta}, +$$ + +and in order to find the optimal parameters \( \beta_i \) instead of solving the above linear algebra problem, we define a function which gives a measure of the spread between the values \( y_i \) (which represent hopefully the exact values) and the parametrized values \( \tilde{y}_i \), namely +$$ +Q(\hat{\beta})=\sum_{i=0}^{n-1}\left(y_i-\tilde{y}_i\right)^2=\left(\hat{y}-\hat{\tilde{y}}\right)^T\left(\hat{y}-\hat{\tilde{y}}\right), +$$ + +or using the matrix \( \hat{X} \) as +$$ +Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right). +$$
-
diff --git a/doc/pub/Regression/html/._Regression-bs009.html b/doc/pub/Regression/html/._Regression-bs009.html new file mode 100644 index 000000000..8ef81bf1f --- /dev/null +++ b/doc/pub/Regression/html/._Regression-bs009.html @@ -0,0 +1,219 @@ + + + + + + + +
+ + + + +
+The function +$$ +Q(\hat{\beta})=\left(\hat{y}-\hat{X}\hat{\beta}\right)^T\left(\hat{y}-\hat{X}\hat{\beta}\right), +$$ + +can be linked to the variance of the quantity \( y_i \) if we interpret the latter as the mean value of for example a numerical experiment. When linking below with the maximum likelihood approach below, we will indeed interpret \( y_i \) as a mean value +$$ +y_{i}=\langle y_i \rangle = \beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}+\epsilon_i, +$$ + +where \( \langle y_i \rangle \) is the mean value. Keep in mind also that till now we have treated \( y_i \) as the exact value. Normally, the response (dependent or outcome) variable \( y_i \) the outcome of a numerical experiment or another type of experiment and is thus only an approximation to the true value. It is then always accompanied by an error estimate, often limited to a statistical error estimate given by the standard deviation discussed earlier. In the discussion here we will treat \( y_i \) as our exact value for the response variable. + +
+In order to find the parameters \( \beta_i \) we will then minimize the spread of \( Q(\hat{\beta}) \) by requiring +$$ +\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = \frac{\partial }{\partial \beta_j}\left[ \sum_{i=0}^{n-1}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)^2\right]=0, +$$ + +which results in +$$ +\frac{\partial Q(\hat{\beta})}{\partial \beta_j} = -2\left[ \sum_{i=0}^{n-1}x_{ij}\left(y_i-\beta_0x_{i,0}+\beta_1x_{i,1}+\beta_2x_{i,2}+\dots+\beta_{n-1}x_{i,n-1}\right)\right]=0, +$$ + +or in a matrix-vector form as +$$ +\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right). +$$ + +
+
+
+ +
+ + +
+ + + + +
+We can rewrite +$$ +\frac{\partial Q(\hat{\beta})}{\partial \hat{\beta}} = 0 = \hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right), +$$ + +as +$$ +\hat{X}^T\hat{y} = \hat{X}^T\hat{X}\hat{\beta}, +$$ + +and if the matrix \( \hat{X}^T\hat{X} \) is invertible we have the solution +$$ +\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}. +$$ + +
+
+
+ +
+ + +
+ + + + +
+The residuals \( \hat{\epsilon} \) are in turn given by +$$ +\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta}, +$$ + +and with +$$ +\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +$$ + +we have +$$ +\hat{X}^T\hat{\epsilon}=\hat{X}^T\left( \hat{y}-\hat{X}\hat{\beta}\right)= 0, +$$ + +meaning that the solution for \( \hat{\beta} \) is the one which minimizes the residuals. Later we will link this with the maximum likelihood approach. + +
+
+
+ +
+ + +
+ + + + +
+How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then +$$ +\hat{A} = \hat{U}\hat{\Sigma}\hat{V} +$$ +
+ +
+ +
+ + +-
@@ -161,6 +178,9 @@ MathJax.Hub.Config({
-
@@ -219,7 +220,15 @@ y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\e \end{align*} $$
Defining the vectors
@@ -283,7 +292,15 @@ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}
\end{align*}
$$
We redefine in turn the matrix \( \hat{X} \) as
@@ -329,7 +346,15 @@ y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,
\end{align*}
$$
We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as
@@ -403,7 +428,7 @@ $$
@@ -428,6 +453,16 @@ $$
$$
The residuals \( \hat{\epsilon} \) are in turn given by
-Here we derive the equations for the SVD.
+How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then
+
-
+
Defining the vectors
$$
\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,
@@ -220,7 +244,7 @@ $$
@@ -237,7 +261,16 @@ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsi
y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}^{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}
$$
+
+
We redefine in turn the matrix \( \hat{X} \) as
$$
\hat{X}=
@@ -262,7 +295,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo
@@ -278,7 +311,16 @@ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsi
y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}
$$
+
+
We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as
$$
\hat{\tilde{y}}= \hat{X}\hat{\beta},
@@ -299,7 +341,7 @@ $$
@@ -338,7 +380,7 @@ $$
@@ -357,6 +399,17 @@ $$
\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}.
$$
+
+
+
The residuals \( \hat{\epsilon} \) are in turn given by
$$
\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta},
@@ -381,11 +434,14 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
-Here we derive the equations for the SVD.
+How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then
+$$
+\hat{A} = \hat{U}\hat{\Sigma}\hat{V}
+$$
-
+
Defining the vectors
$$
\hat{y} = [y_0,y_1, y_2,\dots, y_{n-1}]^T,
@@ -225,7 +249,7 @@ $$
@@ -242,7 +266,16 @@ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsi
y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1}^{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}
$$
+
+
We redefine in turn the matrix \( \hat{X} \) as
$$
\hat{X}=
@@ -267,7 +300,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo
@@ -283,7 +316,16 @@ y_{i}&=\beta_0x_{i0}+\beta_1x_{i1}+\beta_2x_{i2}+\dots+\beta_{n-1}x_{in-1}+\epsi
y_{n-1}&=\beta_0x_{n-1,0}+\beta_1x_{n-1,2}+\beta_2x_{n-1,2}+\dots+\beta_1x_{n-1,n-1}+\epsilon_{n-1}.\\
\end{align*}
$$
+
+
We well use this matrix to define the approximation \( \hat{\tilde{y}} \) via the unknown quantity \( \hat{\beta} \) as
$$
\hat{\tilde{y}}= \hat{X}\hat{\beta},
@@ -304,7 +346,7 @@ $$
@@ -343,7 +385,7 @@ $$
@@ -362,6 +404,17 @@ $$
\hat{\beta} =\left(\hat{X}^T\hat{X}\right)^{-1}\hat{X}^T\hat{y}.
$$
+
+
+
The residuals \( \hat{\epsilon} \) are in turn given by
$$
\hat{\epsilon} = \hat{y}-\hat{\tilde{y}} = \hat{y}-\hat{X}\hat{\beta},
@@ -386,11 +439,14 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
-Here we derive the equations for the SVD.
+How can we use the singular value decomposition to find the parameters \( \beta_j \)? More details will come. We first note that a general \( m\times n \) matrix \( \hat{A} \) can be written in terms of a diagonal matrix \( \hat{\Sigma} \) of dimensionality \( n\times n \) and two orthognal matrices \( \hat{U} \) and \( \hat{V} \), where the first has dimensionality \( m \times n \) and the last dimensionality \( n\times n \). We have then
+$$
+\hat{A} = \hat{U}\hat{\Sigma}\hat{V}
+$$
+
+
+
+Rewriting the fitting procedure as a linear algebra problem, follows
+
$$
@@ -264,7 +273,7 @@ $$
Generalizing the fitting procedure as a linear algebra problem
+Generalizing the fitting procedure as a linear algebra problem
+Generalizing the fitting procedure as a linear algebra problem
+
$$
@@ -311,7 +328,7 @@ The left-hand side of this equation forms know. Our error vector \( \hat{\epsilo
Optimizing our parameters
+Optimizing our parameters
+Optimizing our parameters, more details
+
$$
@@ -355,7 +380,7 @@ $$
Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
+
+Interpretations and optimizing our parameters
+
$$
@@ -457,11 +492,16 @@ meaning that the solution for \( \hat{\beta} \) is the one which minimizes the r
The singular value decompostion
+The singular value decompostion
+$$
+\hat{A} = \hat{U}\hat{\Sigma}\hat{V}
+$$
+
Oct 17, 2017
Oct 18, 2017
@@ -168,6 +181,8 @@ where \( \epsilon_i \) is the error in our approximation.
Rewriting the fitting procedure as a linear algebra problem
@@ -184,7 +199,16 @@ y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\
y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\
\end{align*}
$$
+
+
+Rewriting the fitting procedure as a linear algebra problem, follows
+
-Generalizing the fitting procedure as a linear algebra problem
+Generalizing the fitting procedure as a linear algebra problem
+
+Generalizing the fitting procedure as a linear algebra problem
+
-Optimizing our parameters
+Optimizing our parameters
+
+Optimizing our parameters, more details
+
-Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
-Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
+
+Interpretations and optimizing our parameters
+
-The singular value decompostion
+The singular value decompostion
Oct 17, 2017
Oct 18, 2017
@@ -173,6 +186,8 @@ where \( \epsilon_i \) is the error in our approximation.
Rewriting the fitting procedure as a linear algebra problem
@@ -189,7 +204,16 @@ y_2&=\beta_0+\beta_1x_2^1+\beta_2x_2^2+\dots+\beta_{n-1}x_2^{n-1}+\epsilon_2\\
y_{n-1}&=\beta_0+\beta_1x_{n-1}^1+\beta_2x_{n-1}^2+\dots+\beta_1x_{n-1}^{n-1}+\epsilon_{n-1}.\\
\end{align*}
$$
+
+
+Rewriting the fitting procedure as a linear algebra problem, follows
+
-Generalizing the fitting procedure as a linear algebra problem
+Generalizing the fitting procedure as a linear algebra problem
+
+Generalizing the fitting procedure as a linear algebra problem
+
-Optimizing our parameters
+Optimizing our parameters
+
+Optimizing our parameters, more details
+
-Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
-Interpretations and optimizing our parameters
+Interpretations and optimizing our parameters
+
+Interpretations and optimizing our parameters
+
-The singular value decompostion
+The singular value decompostion