Question

Consider the following short time series data set: t1234 5 6789101112 13 14 15 16 17 1819 20 x13 25 20 2 4 12 8 7 19 71 20 5 12 18 12 20 85 Evaluate each of the following expressions (i.e., give a numerical answer): a. ▽x13 c. ▽2x13 d. Vx13 e. 1-B1)x1 f. (1-0.7B +0.5B2)1-B)x13

Answer 1

Solution :

We will consider the given short time series data set where "t" denotes the time period and "x_t" denotes the observation at the time period "t". We have to evaluate the given expressions!

We know the formulae for the given expressions. They are given below :

$\triangledown \ is\ known\ as\ the\ \mathbf{Backward\ Difference\ Operator.}$

$where,\ \ \ \triangledown X_t=X_t-X_{t-1}\ \ \ \ \ \&\ \ \ \ \ \triangledown^2X_t=\triangledown(\triangledown X_t)$

$\mathbf{B} \ is\ known\ as\ the\ \mathbf{Backward\ Shift\ Operator.}$

$where,\ \ \ B X_t=X_{t-1}\ \ \ \ \ \&\ in\ general,\ \ \ \ B^mX_t=X_{t-m}\ ,\ m\geq 0$

$\triangle \ is\ known\ as\ the\ \mathbf{Forward\ Difference\ Operator.}$

$where,\ \ \ \triangle X_t=X_{t+1}-X_{t}\ \ \ \ \ \&\ \ \ \ \ \triangle^2X_t=\triangle(\triangle X_t)$

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$\mathbf{(a)\ \ \ To\ find\ \ :\ \ \triangledown x_{13}}$

$\triangledown x_{13}=x_{13}-x_{13-1}=x_{13}-x_{12}=20-1=19$

$\mathbf{Thus,\ \ we\ have,\ \ \triangledown x_{13}=19.......................(Ans)}$

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$\mathbf{(b)\ \ \ To\ find\ \ :\ \ B^2 x_{13}}$

$B^2x_{13}=B(Bx_{13})=B(x_{12})=x_{11}=7$

$\mathbf{Thus,\ \ we\ have,\ \ B^2x_{13}=7.......................(Ans)}$

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$\mathbf{(c)\ \ \ To\ find\ \ :\ \ \triangle^2 x_{13}}$

$\triangle^2 x_{13}=\triangle(\triangle x_{13})=\triangle(x_{14}-x_{13})=\triangle x_{14}-\triangle x_{13}=(x_{15}-x_{14})-(x_{14}-x_{13})$

  $=x_{15}-x_{14}-x_{14}+x_{13}=x_{15}-2*x_{14}+x_{13}=12-2*5+20=12-10+20=22$

$\mathbf{Thus,\ \ we\ have,\ \ \triangle^2x_{13}=22.......................(Ans)}$

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$\mathbf{(d)\ \ \ To\ find\ \ :\ \ \triangledown^2 x_{13}}$

$\triangledown^2 x_{13}=\triangledown(\triangledown x_{13})=\triangledown(x_{13}-x_{12})=\triangledown x_{13}-\triangledown x_{12}=(x_{13}-x_{12})-(x_{12}-x_{11})$

  $=x_{13}-x_{12}-x_{12}+x_{11}=x_{13}-2*x_{12}+x_{11}=20-2*1+7=25$

$\mathbf{Thus,\ \ we\ have,\ \ \triangledown^2 x_{13}=25.......................(Ans)}$

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$\mathbf{(e)\ \ \ To\ find\ \ :\ \ (1-B^{12}) x_{13}}$

$(1-B^{12}) x_{13}=x_{13}-B^{12}x_{13}=x_{13}-x_{13-12}=x_{13}-x_1=20-13=7$

$\mathbf{Thus,\ \ we\ have,\ \ (1-B^{12}) x_{13}=7.......................(Ans)}$

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$\mathbf{(f)\ \ \ To\ find\ \ :\ \ (1-0.7B+0.5B^2)(1-B) x_{13}}$

$(1-0.7B+0.5B^2)(1-B) x_{13}=[(1-0.7B+0.5B^2)-B(1-0.7B+0.5B^2)]x_{13}$

  $=(1-0.7B+0.5B^2-B+0.7B^2-0.5B^3)x_{13}$

  $=x_{13}-0.7Bx_{13}+0.5B^2x_{13}-Bx_{13}+0.7B^2x_{13}-0.5B^3x_{13}$

  $=x_{13}-0.7x_{13-1}+0.5x_{13-2}-x_{13-1}+0.7x_{13-2}-0.5x_{13-3}$

  $=x_{13}-0.7x_{12}+0.5x_{11}-x_{12}+0.7x_{11}-0.5x_{10}$

  $=20-0.7*1+0.5*7-1+0.7*7-0.5*19$

  $=20-0.7+3.5-1+4.9-9.5=17.2$

$\mathbf{Thus,\ \ we\ have,\ \ (1-0.7B+0.5B^2)(1-B) x_{13}}=17.2.......................(Ans)}$

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