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Chapter 9: Problem 6

Let \(\left\\{X_{n}\right\\}_{n=-\infty}^{+\infty}\) be a zero-mean covariance stationary process having covariance function \(R(v)=\gamma^{|0|}, v=0, \pm 1, \ldots\), where \(|\gamma|<1 .\) Find the minimum mean square error linear predictor of \(X_{n+1}\) given the entire past \(X_{n}, X_{n-1} \ldots\)

Short Answer

Expert verified
The MMSE linear predictor for \(X_{n+1}\) given the entire past \(X_{n}, X_{n-1}, \ldots\) is \(0\).

Step by step solution

01

Definition of Covariance

The given covariance function is \(R(v) = \gamma^0\) for any \(v\). Since any number to the power of 0 equals 1, the covariance function simplifies to \(R(v) = 1\).
02

Minimum Mean Square Error (MMSE) Linear Prediction

To find the MMSE linear predictor of \(X_{n+1}\), given past values, we need to formulate it in terms of previous observations, known as filter coefficients. For MMSE prediction, an optimal linear combination of past observations is chosen to achieve the smallest possible mean squared error. The process of finding an MMSE estimator in general involves differentiation of an error function, setting it to zero, and subsequently solving for predictor coefficients.
03

Formulate the MMSE Predictor

Now, to formally derive a predictor, one could follow these steps:1. Establish the error term \(e_n = X_{n+1} - a_nX_n\), where \(a_n\) is the predictor coefficient to be found. The prediction of \(X_{n+1}\) via linear regression is thus \(\hat{X}_{n+1} = a_nX_n\).2. Given the task of minimizing the mean squared error, we need to find \(a_n\) that minimizes \(E[e_n^2]\).3. Solve for the coefficient \(a_n\) by setting the derivative \(dE[e_n^2]/da_n = 0\) and solving for \(a_n\). As we have \(E[e_n^2] = a_n ^ 2 - 2a_n R(1) + R(0)\), and \(R(v) = 1\) for all \(v\), we end up with \(a_n = 0\).
04

Conclude the Solution

Therefore, the best linear predictor for \(X_{n+1}\) given the process is \(X_{n}, X_{n-1}, \ldots\) is \(0\). This means that the future values cannot be predicted based on past values, thus predicting 0 error, implying a completely random process.

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Most popular questions from this chapter

Let \(\left\\{X_{s}\right\\}\) be a finite-state irreducible Markov chain having the transition probabilities \(\| P_{i j} \mid N_{i, j=1^{*}}\). There then exists a stationary distribution \(\pi\), i.e., a vector \(\pi(1), \ldots, \pi(N)\) satisfying \(\pi(i) \geq 0, i=1, \ldots, N, \sum_{i=1}^{N} \pi(i)=1\), and $$ \pi(j)=\sum_{i=1}^{N} \pi(i) P_{i j}, \quad j=1, \ldots, N $$ Suppose \(\operatorname{Pr}\left\\{X_{0}=i\right\\}=\pi(i), i=1, \ldots, N\). Show that \(\left\\{X_{n}\right\\}\) is weakly mixing, hence ergodic.

Let \(\left\\{X_{n}\right\\}\) be a zero-mean covariance stationary process having covariance function \(R_{X}(v)\) and spectral density function \(f_{X}(\omega),-\pi \leq \omega \leq \pi\). Suppose \(\left\\{a_{n}\right\\}\) is a real sequence for which \(\sum_{b, 0}^{\infty}\left|a_{i} a_{j} R(i-j)\right|<\infty\), and define $$ Y_{n}=\sum_{k=0}^{\infty} a_{k} X_{n-k} $$ Show that the spectral density function \(f_{Y}(\omega)\) for \(\left\\{Y_{n}\right\\}\) is given by $$ \begin{aligned} f_{Y}(\omega) &=\frac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\left|\sum_{k=0}^{\infty} a_{k} e^{i k \omega}\right|^{2} f_{x}(\omega) \\ &=\frac{\sigma_{X}^{2}}{\sigma_{Y}^{2}}\left[\sum_{X=0}^{\infty} a_{j} a_{k} \cos (j-k) \omega\right] f_{X}(\omega), \quad-\pi \leq \omega \leq \pi . \end{aligned} $$

Suppose \(X_{0}\) has probability density function $$ f(x)= \begin{cases}2 x, & \text { for } 0 \leq x \leq 1, \\ 0, & \text { eleewhere, }\end{cases} $$ and that \(X_{n+1}\) is uniformly distributed on \(\left(1-X_{n}, 1\right]\), given \(X_{0}, \ldots, X_{n}\). Show that \(\left\\{X_{n}\right\\}\) is a stationary ergodic process.

Let \(\left\\{X_{k}\right\\}\) be a moving average process $$ X_{n}=\sum_{j=0}^{\infty} \alpha_{j} \xi_{n-j}, \quad \alpha_{0}=1, \quad \sum_{=0}^{\infty} \alpha_{j}^{2}<\infty $$ where \(\left\\{\xi_{n}\right\\}\) are zero-mean independent random variables having common variance \(\sigma^{2}\). Show that $$ U_{n}=\sum_{k=0}^{n} X_{k-1} \bar{\xi}_{k}, \quad n=0,1, \ldots $$ and $$ V_{n}=\sum_{k=0}^{n} X_{k} \xi_{h}-(n+1) \sigma^{2}, \quad n=0,1, \ldots $$ are martingales with respect to \(\left\\{\zeta_{n}\right\\}\).

Show that a predictor $$ \hat{X}_{n}=\alpha_{1} X_{n-1}+\cdots+\alpha_{p} X_{n-p} $$ is optimal among all linear predictors of \(X_{n}\) given \(X_{n-1}, \ldots, X_{n-p}\) if and only if $$ 0=\int_{-\pi}^{n} e^{i k \lambda}\left[1-\sum_{t=1}^{R} \alpha_{1} e^{-i t}\right] d F(\lambda), \quad k=1, \ldots, p $$ where \(F(\omega),-\pi \leq \omega \leq \pi\), is the spectral distribution function of the covariance stationary process \(\left\\{X_{n}\right\\} .\)
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