Question

Your time series data are as follows. 0.12-0.671 -0.273-0.994 0.571 1.088 0.505 0 -0.389 1.316 1.328 2.06 5.292 4.396 4.177 2.94 5.501 4.896 4.588 3.436 For these data (length 20), plot the times series, plot the acf, and get the acf function numerically. Do the same for the differenced series of length 19 Part a) The serial correlations of lags 1,2,3 are: Part b) For the differenced series, the serial correlations of lags 1,2,3 are:

Answer 1

Solution: For a given a time series

$y_{t}$=(-0.12, -0.671,-0.273,-0.994,0.571,1.088,0.505,0,0.389,1.316,1.328,2.06,5.292,4.396,4.177,2.94,5.501,4.896,4.588,3.436)

Using R-programming, we plotting (1) time series plot, ACF plot (2) difference time series plot, difference ACF plot.

R-code:

y=c(-0.12, -0.671,-0.273,-0.994,0.571,1.088,0.505,0,-0.389,1.316,1.328,2.06,5.292,4.396,4.177,2.94,5.501,4.896,4.588,3.436)
y1=diff(x)
length(x)
par(mfrow=c(2,2))
acf1=acf(y,main="acf plot of original series")
plot(y,type="l",main="original series")
acfd=acf(y1,main="acf plot of difference series")
plot(y1,type="l",main="Difference of Time series")

(a) Mean of the series will be $\bar{y}= 1.98235$

By using given formula, we can get serial correlations

$\rho(k) = \frac{\frac{1}{n-k}\sum_{t=k+1}^n (y_t - \bar{y})(y_{t-k} - \bar{y})}{ \sqrt{\frac{1}{n}\sum_{t=1}^n (y_t - \bar{y})^2}\sqrt{\frac{1}{n-k}\sum_{t=k+1}^n (y_{t-k} - \bar{y})^2}}$

Serial correlation of lag 1,2, 3 as follows: 0.809, 0.646, 0.489

(b) For difference series mean will be $\bar{y}= 0.1871579$

We get serial correlation of lag 1,2, 3 for difference series as follows: -0.230 -0.038 -0.254