# Predicting $x_t$ knowing something about $\Delta^2 x_t$

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

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

• What does $\Delta^2$ mean? Is $\epsilon_t$ a constant? Commented Apr 21, 2022 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. Commented Apr 21, 2022 at 12:53
• 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
Commented Apr 21, 2022 at 12:57
• 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. Commented Apr 21, 2022 at 13:05
• Does the answer key literally say "$x_{10} = 2x_8 - x_9$"? That's not true, it should say something like $\mathbb{E}(x_{10}|\mathcal{F}_9) = 2x_8 - x_9$. Commented Apr 21, 2022 at 13:34

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.

• Thank Whuber, always helpful, I did apparently forget the definition of the second difference Commented Apr 22, 2022 at 12:55