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I have this exercise question, what is your prediction of $x_{10}$ knowing that $\Delta^2 x_t = \epsilon_t $, knowing $x_9 = 1.56$ and $x_8 = 1.64$.

I take this to mean that

$$ x_t = x_{t-2} +\epsilon_t $$

where I assume the error term is iid with mean zero.

I would then answer that my best prediction is simply $x_8$ but the answer key gives a different answer of

$$ x_{10} = 2x_8 - x_9 = 1.72. $$

whuber
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Three Diag
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  • What does $\Delta^2$ mean? Is $\epsilon_t$ a constant? – Sextus Empiricus Apr 21 '22 at 12:52
  • I take $\Delta^2 x_t = x_t -x_{t-2}$ the second difference, $\epsilon_t$ is some iid noise with mean zero. – Three Diag Apr 21 '22 at 12:53
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    Usually $\Delta$ is the first difference and therefore $\Delta^2 = \Delta(\Delta)$ is computed as $$\Delta^2(x_t) = \Delta(x_t - x_{t-1}) = \Delta(x_t) - \Delta(x_{t-1}) = (x_t-x_{t-1})-(x_{t-1}-x_{t-2})=x_t-2x_{t-1} + x_{t-2}.$$ – whuber Apr 21 '22 at 12:57
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    Using whubers formulation you would get $x_{10} - 2 x_9 + x_8 = \epsilon_{10}$ if you also know $\epsilon$ (or the distribution of $\epsilon$) then you can make your estimate by rewriting the equation. – Sextus Empiricus Apr 21 '22 at 13:05
  • Does the answer key literally say "$x_{10} = 2x_8 - x_9$"? That's not true, it should say something like $\mathbb{E}(x_{10}|\mathcal{F}_9) = 2x_8 - x_9$. – Chris Haug Apr 21 '22 at 13:34

1 Answers1

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The first difference operator $\Delta$ is the difference between the identity $\mathbb I$ and the backshift operator $B$ given by

$$(Bx)_t = x_{t-1};$$

that is,

$$(\Delta x)_t = ((\mathbb I - B)x)_t = (\mathbb{I} x)_t - (Bx)_t = x_t - x_{t-1}.$$

Notice that iterations of $B$ shift the series even more:

$$(B^n x)_t = B(B^{n-1}x)_t = \cdots = x_{t-n}.$$

In particular, you have interpreted $\Delta^2$ as $B^2,$ which explains why your solution differs from the answer key.

The second difference operator $\Delta^2 = \Delta\circ\Delta$ results from two applications of $\Delta,$ whence

$$\left(\Delta^2x\right)_t = \left(\mathbb I - B)^2 x\right)_t = \left((\mathbb I - 2B + B^2)x\right)_t = x_t - 2x_{t-1} + x_{t-2}.$$

(Refer to Wikipedia for instance.)

We see that $\Delta^2$ differs from $B^2.$ Apply this result to $t=10$ to give

$$\epsilon_{10} = (\Delta^2 x)_{10} = x_{10} - 2x_9 + x_8 = x_{10} - 2(1.56) + 1.64.$$

Solving this equation yields a formula that differs from the solution in the question,

$$x_{10} = \epsilon_{10} + 2(1.56) - 1.64 = 1.48 + \epsilon_{10}.$$

I suspect you meant to stipulate that $x_9 = 1.64$ and $x_8=1.56$ rather than the other way around, so let's continue with that re-interpretation, which gives

$$x_{10} = 1.72 + \epsilon_{10}.$$

The best prediction is the conditional expectation

$$\hat x_{10} = E[\epsilon_{10} + 2(x_9) - x_8\mid x_8, x_9] = E[\epsilon_{10}] + 1.72 = 1.72$$

(because you have stated that the $\epsilon_t$ are independent and have expectations of zero).

For a fuller account of working with operators like $B$ and $\Delta$ see https://stats.stackexchange.com/a/568744/919.

whuber
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