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<!-- navigation toc: --> <li><a href="._week45-bs001.html#___sec0" style="font-size: 80%;">Random forests</a></li>
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<h2 id="___sec7" class="anchor">Squared-Error Example and Iterative Fitting </h2>
<p>
To better understand what happens, let us develop the steps for the iterative fitting using the above squared error function.
<p>
For simplicity we assume also that our functions \( b(x;\gamma)=1+\gamma x \).
<p>
This means that for every iteration \( m \), we need to optimize
$$
(\beta_m,\gamma_m) = \mathrm{argmin}_{\beta,\lambda}\hspace{0.1cm} \sum_{i=0}^{n-1}(y_i-f_{m-1}(x_i)-\beta b(x;\gamma))^2=\sum_{i=0}^{n-1}(y_i-f_{m-1}(x_i)-\beta(1+\gamma x_i))^2.
$$
<p>
We start our iteration by simply setting \( f_0(x)=0 \).
Taking the derivatives with respect to \( \beta \) and \( \gamma \) we obtain
$$
\frac{\partial {\cal C}}{\partial \beta} = -2\sum_{i}(1+\gamma x_i)(y_i-\beta(1+\gamma x_i))=0,
$$
and
$$
\frac{\partial {\cal C}}{\partial \gamma} =-2\sum_{i}\beta x_i(y_i-\beta(1+\gamma x_i))=0.
$$
We can then rewrite these equations as (defining \( \boldsymbol{w}=\boldsymbol{e}+\gamma \boldsymbol{x}) \) with \( \boldsymbol{e} \) being the unit vector)
$$
\gamma \boldsymbol{w}^T(\boldsymbol{y}-\beta\gamma \boldsymbol{w})=0,
$$
which gives us \( \beta = \boldsymbol{w}^T\boldsymbol{y}/(\boldsymbol{w}^T\boldsymbol{w}) \). Similarly we have
$$
\beta\gamma \boldsymbol{x}^T(\boldsymbol{y}-\beta(1+\gamma \boldsymbol{x}))=0,
$$
<p>
which leads to \( \gamma =(\boldsymbol{x}^T\boldsymbol{y}-\beta\boldsymbol{x}^T\boldsymbol{e})/(\beta\boldsymbol{x}^T\boldsymbol{x}) \). Inserting
for \( \beta \) gives us an equation for \( \gamma \). This is a non-linear equation in the unknown \( \gamma \) and has to be solved numerically.
<p>
The solution to these two equations gives us in turn \( \beta_1 \) and \( \gamma_1 \) leading to the new expression for \( f_1(x) \) as
\( f_1(x) = \beta_1(1+\gamma_1x) \). Doing this \( M \) times results in our final estimate for the function \( f \).
<p>
<p>
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