Question

Seven successive observations on a stationary time-series are as follows:

12, 14, 13, 10, 15, 12, 15

(a) Calculate auto-covariances C0, C1, C2, C3 and C4.

(b) Calculate auto-correlation coefficients r1, r2, r3 and r4.

(c) Plot the correlogram.

20 Feb 2025
Answer :
Word Count : 588

(a) Auto-covariances C0,C1,C2,C3,C4C_0, C_1, C_2, C_3, C_4

To calculate the auto-covariances, we need the mean of the series first.

Given observations: 12, 14, 13, 10, 15, 12, 15

Step 1: Calculate the mean of the series:

Xˉ=12+14+13+10+15+12+157=917=13\bar{X} = \frac{12 + 14 + 13 + 10 + 15 + 12 + 15}{7} = \frac{91}{7} = 13

Now, we can calculate the auto-covariances. The general formula for CkC_k (auto-covariance) is:

Ck=1n−k∑t=1n−k(Xt−Xˉ)(Xt+k−Xˉ)C_k = \frac{1}{n-k} \sum_{t=1}^{n-k} (X_t - \bar{X})(X_{t+k} - \bar{X})

For C0C_0 (auto-covariance at lag 0):

C0=17∑t=17(Xt−Xˉ)2C_0 = \frac{1}{7} \sum_{t=1}^{7} (X_t - \bar{X})^2 C0=17[(12−13)2+(14−13)2+(13−13)2+(10−13)2+(15−13)2+(12−13)2+(15−13)2]C_0 = \frac{1}{7} \left[ (12 - 13)^2 + (14 - 13)^2 + (13 - 13)^2 + (10 - 13)^2 + (15 - 13)^2 + (12 - 13)^2 + (15 - 13)^2 \right] C0=17[1+1+0+9+4+1+4]=207≈2.857C_0 = \frac{1}{7} \left[ 1 + 1 + 0 + 9 + 4 + 1 + 4 \right] = \frac{20}{7} \approx 2.857

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