STAT429_sp24_HW03 - Copy

STAT 429 HW 03 Phillip Nguyen , pnguy6 If you choose to type the answers for this HW, you may find • Unless stated otherwise, { W t } t ≥ 0 is a white noise process with variance σ 2 w . – ( { W t } t ≥ 0 is a collection of random variables, independent of each other, with zero means and variance σ 2 w .) • Show your full work to receive full credit. Question 1 (a) Simulate a series of n = 1000 white noise observations and compute the sample ACF, ˆ ρ w ( h ) , to lag 30. Show the code and the ACF plot. What should be the (approximate) variance of ˆ ρ w ( h ) ( h ̸ = 0) ? Compare the sample ACF you obtain to the actual (theoretical) ACF of a white noise, ρ w ( h ) . w_a = rnorm ( 502 , 0 , 1 ) v_a = filter (w_a, sides= 2 , rep ( 1 , 3 ) / 3 ) print ( acf (v_a, 20 , na.action = na.pass)) 1

0 5 10 15 20 0.0 0.2 0.4 0.6 0.8 1.0 Lag ACF Series v_a ## ## Autocorrelations of series ’v_a’, by lag ## ## 0 1 2 3 4 5 6 7 8 9 10 ## 1.000 0.618 0.265 -0.034 0.055 0.040 -0.044 -0.121 -0.113 -0.062 -0.059 ## 11 12 13 14 15 16 17 18 19 20 ## -0.077 -0.114 -0.094 -0.098 -0.082 -0.079 -0.055 -0.037 -0.030 -0.010 he ACF of the sample size n = 500 has minimal positive and negative values whereas the theoretical ACF should be zero. This is due to the noise being added to the signal model with mean 0 and variance 1. (b) Repeat part (a) using only n = 50 . does changing n affect the results? w_b = rnorm ( 52 , 0 , 1 ) v_b = filter (w_b, sides= 2 , rep ( 1 , 3 ) / 3 ) print ( acf (v_b, 20 , na.action = na.pass)) 2

We need to check if X t and Y t are jointly stationary by verifying if their means, variances, and autocovariance functions are time-invariant. For series X t : Mean: E [ X t ] = E [ W t + 0 . 5 W t − 1 ] = E [ W t ] + 0 . 5 E [ W t − 1 ] = 0 (since the mean of a white noise process is zero) Variance: Var ( X t ) = Var ( W t + 0 . 5 W t − 1 ) = Var ( W t ) + 0 . 25 Var ( W t − 1 ) (since W t and W t − 1 are independent) = σ 2 w + 0 . 25 σ 2 w = 1 . 25 σ 2 w For series Y t : Mean: E [ Y t ] = E [ W t + U t ] = E [ W t ] + E [ U t ] = 0 (since the mean of a white noise process is zero) Variance: Var ( Y t ) = Var ( W t + U t ) = Var ( W t ) + Var ( U t ) (since W t and U t are independent) = σ 2 w + σ 2 u autocovariance functions for both series: For series X t : Cov ( X t , X t − k ) = Cov ( W t + 0 . 5 W t − 1 , W t − k + 0 . 5 W t − k − 1 ) = Cov ( W t , W t − k ) + 0 . 5 Cov ( W t − 1 , W t − k ) + 0 . 5 Cov ( W t , W t − k − 1 ) + 0 . 25 Cov ( W t − 1 , W t − k − 1 ) Since W t and W t − 1 are independent white noise processes, their covariance is zero unless k = 1 , in which case it’s σ 2 w . Therefore, Cov ( X t , X t − k ) = σ 2 w for k = 1 0 otherwise For series Y t : Cov ( Y t , Y t − k ) = Cov ( W t + U t , W t − k + U t − k ) = Cov ( W t , W t − k ) + Cov ( W t , U t − k ) + Cov ( U t , W t − k ) + Cov ( U t , U t − k ) Since W t and U t are independent white noise processes, their covariance is zero. Therefore, Cov ( Y t , Y t − k ) = σ 2 w for k = 0 0 otherwise Since the autocovariance function for series X t depends on the lag k , it is not time-invariant. However, the autocovariance function for series Y t is time-invariant. Thus, X t is not jointly stationary with Y t due to the time-varying autocovariance function. (b) Compute the CCF (cross-correlation function) of relating { X t } t ≥ 0 and { Y t } t ≥ 0 . 4

To compute the cross-correlation function (CCF) between the series { X t } and { Y t } , we need to find the correlation between each pair of corresponding observations at different lags. The CCF at lag k is given by: CCF ( k ) = Cov ( X t , Y t − k ) qqqqqqq Var ( X t ) · Var ( Y t − k ) Given that the series are jointly stationary, the covariance between X t and Y t − k will depend only on the lag k , and the variances of X t and Y t − k will remain constant. We already have expressions for the covariance and variances: For series X t : Mean: E [ X t ] = 0 Variance: Var ( X t ) = 1 . 25 σ 2 w For series Y t : Mean: E [ Y t ] = 0 Variance: Var ( Y t ) = σ 2 w + σ 2 u Now, let’s compute the cross-correlation function for different lags k : For k = 0 : CCF (0) = Cov ( X t , Y t ) qqqqqqq Var ( X t ) · Var ( Y t ) For k = 1 : CCF (1) = Cov ( X t , Y t − 1 ) qqqqqqq Var ( X t ) · Var ( Y t − 1 ) For k > 1 , Cov ( X t , Y t − k ) = 0 because the processes are independent. Thus, the CCF is given by: CCF (0) = σ 2 w qqqqqqq 1 . 25 σ 2 w · ( σ 2 w + σ 2 u ) CCF (1) = σ 2 w qqqqqqq 1 . 25 σ 2 w · σ 2 w CCF ( k ) = 0 , for k > 1 5

X <- W[ 2 : (T + 1 )] # Xt = Wt Y <- W[ 1 : T] + U # Yt = Wt-1 + Ut ccf_result <- ccf (X, Y, plot = FALSE ) plot (ccf_result, main = "Sample Cross-Correlation Function (CCF)" , xlab = "Lag" , ylab = "CCF" ) grid () -20 -10 0 10 20 0.0 0.2 0.4 0.6 Lag CCF Sample Cross-Correlation Function (CCF) Question 4 (a) Using R, create your own function that computes ACF (auto-correlation function). You may not use built-in functions like acf , acf1 , or acf2 . acf_my_function <- function (ts,maxh){ #function content goes here #ts: time series data, maxh: maximum h value #final line of the function should return the vector of size maxh, #containing acf values, c(rho(1), ..., rho(maxh)). n <- length (ts) acf_values <- numeric (maxh) for (h in 1 : maxh) { ts_shifted <- ts[h : n] ts_original <- ts[ 1 : (n - h + 1 )] 7

acf_values[h] <- cor (ts_original, ts_shifted) } return (acf_values) } (b) Compute the acf of soi series of astsa package using your own function in (4a). Check your answer using acf1 function of astsa package. library (astsa) ## Warning: package ’astsa’ was built under R version 4.3.2 # Load the SOI data data (soi) # Compute ACF using custom function acf_custom <- acf_my_function (soi, 30 ) # Compute ACF using acf1 function acf_astsa <- acf1 (soi, 30 ) 0.0 0.5 1.0 1.5 2.0 2.5 -0.4 -0.2 0.0 0.2 0.4 0.6 0.8 Series: soi LAG ÷ 12 ACF 8