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6. (12 points) Let

\left\{X_{t}\right\}

be a time series in which we are inte...

Sep 25, 2024

Solution by Steps

step 1

We start with the observed process defined as

Y_t = X_t + e_t

, where

X_t

is stationary with autocorrelation function

\rho_k

and

e_t

is the measurement noise

step 2

To show that

Y_t

is stationary, we need to find the autocorrelation function of

Y_t

. The autocorrelation function is given by:

\operatorname{Corr}(Y_t, Y_{t-k}) = \frac{\operatorname{Cov}(Y_t, Y_{t-k})}{\sqrt{\operatorname{Var}(Y_t) \operatorname{Var}(Y_{t-k})}}

Since

Y_t

and

Y_{t-k}

are both composed of

X_t

and

e_t

, we can express the covariance

step 3

The covariance can be expressed as:

\operatorname{Cov}(Y_t, Y_{t-k}) = \operatorname{Cov}(X_t + e_t, X_{t-k} + e_{t-k}) = \operatorname{Cov}(X_t, X_{t-k}) + \operatorname{Cov}(e_t, e_{t-k})

Since

X_t

and

e_t

are independent,

\operatorname{Cov}(e_t, e_{t-k}) = 0

. Thus, we have:

\operatorname{Cov}(Y_t, Y_{t-k}) = \operatorname{Cov}(X_t, X_{t-k}) = \rho_k \sigma_X^2

where

\sigma_X^2

is the variance of

X_t

step 4

The variances of

Y_t

and

Y_{t-k}

are given by:

\operatorname{Var}(Y_t) = \operatorname{Var}(X_t + e_t) = \operatorname{Var}(X_t) + \operatorname{Var}(e_t) = \sigma_X^2 + \sigma_e^2

Thus, the correlation becomes:

\operatorname{Corr}(Y_t, Y_{t-k}) = \frac{\rho_k \sigma_X^2}{\sqrt{(\sigma_X^2 + \sigma_e^2)(\sigma_X^2 + \sigma_e^2)}} = \frac{\rho_k}{1 + \frac{\sigma_e^2}{\sigma_X^2}}

This shows that

Y_t

is stationary with the given autocorrelation function

Answer

Y_t

is stationary with autocorrelation function

\operatorname{Corr}(Y_t, Y_{t-k}) = \frac{\rho_k}{1 + \frac{\sigma_e^2}{\sigma_X^2}}

Key Concept

The relationship between the autocorrelation functions of the signal and the observed process in the presence of noise.

Explanation

The observed process

Y_t

retains the stationary property of the signal

X_t

while incorporating the effects of measurement noise, as shown by the derived autocorrelation function.