# What Is the Definition of Linear Predictive Coefficients When the Optimal Value Aren't Unique?

Linear predictive coefficients of signal y are defined as the best $k$ coefficients $a_i, i = 1, \ldots, k$, that will approximate $y_n$ by $-\sum_{i=1}^k{a_iy_{n-i}}$. (Best approximation is that which minimizes the sum of squared errors.)

How are LPCs defined when this problem has multiple solutions, e.g. when y$$is constant zero? • your example doesn't work: in case of a constant, only the first a_1 will be nonzero, the rest would be zero. Find a better example! – Marcus Müller Jun 9 '18 at 22:32 • Can you clarify what you are asking for? A definition or methods to compute optimal coefficients? – Jonas Schwarz Jun 10 '18 at 19:17 • Sorry, you are right, fixed it. I am asking for a definition. – seed Jun 11 '18 at 9:14 • In a real case I believe that it is very unlikelly that you get a full sequence of zeros. Even if this happens, it is very easy to detect this situation. – Filipe Pinto Jun 26 at 11:29 • Filipe, agreed, I just did not immediately realize that full zero sequence is the only problematic one. – seed Jul 1 at 9:52 ## 2 Answers Constant zero sequence is the only finite sequence for which a_i are not uniquely defined. Indeed, a_i are a solution to the equation$$ \begin{bmatrix} R_0 && R_1 && \ldots && R_{k-1} \\ R_1 && R_0 && \ldots && R_{k-2} \\ \vdots && \vdots && \ddots && \vdots \\ R_{k-1} && R_{k-2} && \ldots && R_0 \end{bmatrix} \begin{bmatrix} a_1 \\ a_2 \\ \vdots \\ a_k \end{bmatrix} = - \begin{bmatrix} R_0 \\ R_1 \\ \vdots \\ R_{k-1} \end{bmatrix} $$To prove their uniqueness, it is sufficient to prove that the autocorrelation matrix R is nondegenerate. If y_0, y_1, \ldots, y_n is the initial sequence, y_0 \ne 0, and$$ Y = \begin{bmatrix} y_0 && \ldots && y_n && 0 && \ldots && 0 \\ 0 && y_0 && \ldots && y_n && \ldots && 0 \\ \vdots && \ddots && \ddots && \vdots && \ddots && \vdots \\ 0 && \ldots && 0 && y_0 && \ldots && y_n \end{bmatrix}, $$then$$ R = YY^T, $$so R is the Gram matrix of the rows of Y. A Gram matrix is nondegenerate iff the vectors are linearly independent, which they obviously are. To deal with the constant zeros problem, I should probably just trim silence at the beginning and end of the audio. The least square solution is strictly convex. So you only have one minimum, although in cases, several combinations of inputs yield the same minimum. Let us go back to a simpler situation: assume that you are given two numbers, a and b, what is their best estimate? "Best" requires some more information: • How best (distance)? • What do variables relates (model)? • Where do a and b dwell (probability)? If how best is a least-squares error loss function, a quadratic measure can be minimized. If the model is a polynomial function, its degree should be fixed). If a and b have known properties, they should be assumed. The case of constant zero values renders most loss functions useless and reduces the space of variation to nil. In other words, if a and b are equal, and you don't did fancy behavior, one answer is, for any \alpha,$$ \alpha a + (1-\alpha)b$$but you could as well:$$ (a^\alpha b ^ \beta)^{\frac{1}{\alpha+\beta}}

and many others, depending on the model you wish. So, the proper "best estimate" is either:

• totally determined by your deterministic situation, and you ought to check input values and determine a predefined output based on logical testing (not based on stochastic optimization)
• set by "other" rules, imposed on the model ( sparse model on, $\alpha$ or $\beta$ for instance), that can give you a more "singular" solution.

So, in your case, the simplest model could be a solution. So what's the simplest?