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Given the following observation for ortime serie X

t=1,2,3,4,5,6,7,8,9,10
Xt=47,64,23,71,38,64,55,41,59,48
Find
r(2)​

1 Answer

3 votes

Answer:t he autocorrelation of the time series X at lag 2 is approximately -0.018.

Explanation:

To find the autocorrelation of the time series X at lag 2, we need to calculate the covariance between X(t) and X(t-2). This can be done using the following formula:

r(2) = cov(X(t), X(t-2)) / var(X(t))

where cov(X(t), X(t-2)) is the covariance between X(t) and X(t-2), and var(X(t)) is the variance of X(t).

To calculate the covariance, we'll use the following formula:

cov(X(t), X(t-2)) = sum((X(t) - mean(X(t))) * (X(t-2) - mean(X(t-2)))) / (n - 1)

where n is the number of observations in the time series. To calculate the variance, we'll use the following formula:

var(X(t)) = sum((X(t) - mean(X(t)))^2) / (n - 1)

First, let's calculate the mean of X(t):

mean(X(t)) = (47 + 64 + 23 + 71 + 38 + 64 + 55 + 41 + 59 + 48) / 10 = 51

Next, let's calculate the covariance between X(t) and X(t-2):

cov(X(t), X(t-2)) = sum((X(t) - 51) * (X(t-2) - 49)) / (10 - 1) = -10.3

Finally, let's calculate the variance of X(t):

var(X(t)) = sum((X(t) - 51)^2) / (10 - 1) = 563.7

Plugging in these values, we can find the autocorrelation at lag 2:

r(2) = cov(X(t), X(t-2)) / var(X(t)) = -10.3 / 563.7 = -0.018

So the autocorrelation of the time series X at lag 2 is approximately -0.018.

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