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This results from applying Leibniz's Rule \begin{eqnarray*} \frac{d}{d\hat x} E_{ϱ_τ}(X-\hat x) &=& (τ -1)\frac{d}{d\hat x}\int_{-∞}^{\hat x}(x-\hat x)\,dF(x)+τ\frac{d}{d\hat x}\int_{\hat x}^{∞}(x-\hat x)\,dF(x) \\ &=& (τ -1)\left((x - \hat x)|_{\hat x} + \int_{-∞}^{\hat x} \frac{d}{d\hat x}(x-\hat x)\,dF(x)\right)+τ\left((x - \hat x)|_{\hat ...

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There are many interpolation and approximation methods available, depending on your requirements (in terms of smoothness, robustness to noise, computational cost...). Most of them are base on polynomials or piecewise polynomial functions. Interpolation makes sure that $f(x_i)=z_i$, while approximation realizes $f(x_i)\approx z_i$. ...

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It's because they don't only look for the derivative of the map $f:B\mapsto AB\in \mathbb R$, which is indeed just $A$; but rather for the gradient vector $\nabla f=\tfrac{df}{dB}$ such that $\forall H$, $$\langle\nabla f,H\rangle=df_B(H)=A(H)=AH=\langle A^T,H\rangle .$$

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Let $f:b\in\mathbb{R}^n\rightarrow 2y^TXb\in\mathbb{R}$ where $X\in M_{p,n},y\in\mathbb{R}^p$. $f$ is linear, then the derivative is $Df_b:h\in\mathbb{R}^n\rightarrow 2y^TXh\in \mathbb{R}$; note that the matrix associated to $Df_b$ is a row. We consider the scalar product over $\mathbb{R}^n$: $(u,v)=u^Tv$. Then the gradient $\nabla(f)$ is defined by: for ...

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Let our best fit line be described as $mx + b$ We would hope that we could solve for $m$ and $b$ without any contradictions using the system of equations: $\begin{cases} 2m+b = 1\\ 3m + b = 2\\ 3m + b = 3\\ 4m + b = 6\\ 5m + b = 5\end{cases}$ Which can be expressed as the following matrix equation: \begin{bmatrix} 2 & ... 1 I found where it was in my notes. We went over it very briefly, so it was a small section. Here's what you need to do. \frac{\partial{f}}{\partial{m}} = \sum_{i=1}^{5}2(y_i-mx_i-b)*(-x_i)\\ = -2(\sum x_iy_i-m\sum x_i^2-b \sum 1) = 0\\ \frac{\partial{f}}{\partial{b}} = -2(\sum y_i -mx_i-b)\\ -2(\sum y_i -m\sum x_i-b\sum i) = 0\\ \text{now we do some ... 1 In general \textbf Y and \textbf X are known because you have a sample. This sample have a dataset of m points: (x_{11},x_{12},\ldots,x_{1m},y_1), (x_{21},x_{22},\ldots,x_{2m},y_2), (x_{31},x_{32},\ldots,x_{3m},y_3), \ldots, (x_{n1},x_{n2},\ldots,x_{nm},y_m). The values of x_{ij} are represented by \textbf X and the values of y_i are represented ... 1 \begin{align} \text{E} \left ( \frac{\sum_{i=1}^{n} x_i Y_i}{\sum_{j=1}^{n} x_j^2} \right ) &= \left ( \sum_{j=1}^{n} x_j^2 \right )^{-1} \sum_{i=1}^{n} x_i \text{E}(Y_i) \\ &= \left ( \sum_{j=1}^{n} x_j^2 \right )^{-1} \sum_{i=1}^{n} x_i (\beta_0 + \beta_1 x_i) \\ &= \beta_0 n\bar{x} \left ( \sum_{j=1}^{n} x_j^2 \right )^{-1} + \beta_1 . ... 1 Denote x=(x_1,\dots,x_n). Assuming that \mathbb{E}[u_i|x]=0\mathbb{E}[\hat\beta_1|x]=\mathbb{E}\left[\frac{\sum x_i y_i}{\sum x_i^2}\mid x\right]=\mathbb{E}\left[\frac{\sum x_i (\beta_0 + \beta_1 x_i + u_i)}{\sum x_i^2}\mid x\right]=\frac{\beta_0\sum x_i+\beta_1\sum x_i^2+\sum x_i\mathbb{E}[u_i|x]}{\sum x_i^2}=\beta_1+\beta_0\frac{\sum x_i}{\sum ... 1 $$||A w+eb||^2=(Aw+eb)^T(Aw+eb)=mb^2+2be^TAw+w^TA^TAw\\b=-e^TAw/m$$ The effect is that you can removeb$by subtracting the average vector from each row of$A$. In matrix language, let$B=A-e(e^TA/m)$where$e$is a column vector of$m$ones. Now, using Lagrange Multipliers to make sure that$w\cdot w=1$:$\$F(w)=w^TB^TBw+\lambda(1-w^Tw)\\ ...

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