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Speech is a continuous signal, which means that consecutive samples of the signal are correlated (see figure on the right). In particular, if we know a previous sample xn-1, we can make a prediction of the current sample, \( \hat x_n = x_{n-1}, \) such that \( \hat x_n \approx x_n. \) By using more previous samples we have more information, which should help us make a better prediction. Specifically, we can define a predictor which uses M previous samples to predict the current sample xas

\[ \hat x_n = - \sum_{k=1}^M a_k x_{n-k}. \]

This is a linear predictor because it takes a linearly weighted sum of past components to predict the current one.

The error of the prediction, also known as the prediction residual is

\[ e_n = x_n - \hat x_n = x_n + \sum_{k=1}^M a_k x_{n-k} = \sum_{k=0}^M a_k x_{n-k}, \]

where a0=1. This explains why the definition \( \hat x_n \) included a minus sign; when we calculate the residual, the double negative disappears and we can collate everything into one summation.

A short segment of speech. Notice how consecutive samples are mostly near each other, which means that consecutive samples are correlated.

Using vector notation, we can make the expressions more compact

\[ e = Xa \]


\[ e = \begin{bmatrix}e_0\\e_1\\\vdots\\e_{N-1}\end{bmatrix},\qquad X = \begin{bmatrix}x_0 & x_{-1} & \dots & x_{M} \\x_1 & x_0 & \dots & x_{M-1} \\ \vdots & \vdots & & \vdots \\ x_{N-1} & x_{N-2} & \dots & x_{N-M} \end{bmatrix}, \qquad a = \begin{bmatrix}a_0\\a_1\\\vdots\\a_{M}\end{bmatrix}. \]

Here we calculated the residual for a length N frame of the signal.

Vector a holds the unknown coefficients of the predictor. To find the best possible predictor, we can minimize the minimum mean-square error (MMSE). The square error is the 2-norm of the residual, \( \|e\|^2=e^T e \) . The mean of that error is defined as the expectation

\[ E\left[\|e\|^2\right] = E\left[a^T X^T X a\right] = a^T E\left[X^T X\right] a = a^T R_x a, \]

where \( R_x = E\left[X^T X\right] \) .

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