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In a time series, I want to predict $ b_{i+1} $ and I know $ ...,x_{i-1},x_i, x_{i+1} $ and $ ...b_{i-2},b_{i-1},b_i $.

I know by prior knowledge that they have a linear relationship:

$\alpha * x_i = b_i$

Normally I would just make a linear regression to understand the constant $ \alpha $ and find:

$ E[b_{i+1}] = \alpha * x_{i+1}$

But, instead of that what happens if I do this:

$\frac{\alpha * x_{i+1}}{\alpha *x_i } = \frac{b_{i+1}}{b_i}$

Or:

$\frac{x_{i+1}}{x_i} = \frac{b_{i+1}}{b_i}$

$E[b_{i+1}]=b_i*\frac{x_{i+1}}{x_i}$

Instead of estimating the constant $\alpha$, I am in this case estimating the growth rate $\frac{x_{i+1}}{x_i}$. Simulating this millions of time the standard deviation is the same in the two above models.

Example, normal model:

b = > 2,4,8,11,12,15,19,22,31,40

x = > 1,2,4,5 ,6 ,8 ,10,11,15,20

Example, changed model:

b = > 4, 8, 11, 12, 15, 19, 22, 31, 40

x = > 2*2/1, 4*4/2, 8*5/4, 11*6/5, 12*8/6, 15*10/8, 19*11/10, 22*15/11, 31*20/15

My question is: should we predict using past values or is better to ignore $b_i$ despite having it as information? Although $b_i$ is an observed value, not an expected, it is still a real life data. Shouldn't be better for a model to adapt to the real life data?

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  • $\begingroup$ Can't you create a meaningful series/variable using both x and b? $\endgroup$ – Guilherme Marthe May 24 '17 at 20:30
  • $\begingroup$ Also from your third equation, a loglog regression seems to make sense. $\endgroup$ – Guilherme Marthe May 24 '17 at 20:31
  • $\begingroup$ Where is the error term in the linear relationship? $\endgroup$ – Michael R. Chernick May 24 '17 at 20:32
  • $\begingroup$ I'm confused: if $b_i$ is an observed value, then it's by definition real life data. Are both $b_i$ and $x_i$ observed variables (i.e. they are measured separately), or is $x_i$ measured then used to calculate $b_i$ in conjunction with some prior knowledge about $\alpha$? $\endgroup$ – LmnICE May 24 '17 at 20:39
  • $\begingroup$ @MichaelChernick I am not a Statistician, so correct me if I am wrong. When we make the linear regression to understand the constant $\alpha$ or the growth rate, we will have a intercept and slope. In this case the intercept will "include" the error, thus making it expected value be zero. $\endgroup$ – Dinidiniz May 24 '17 at 20:53
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A regression model must include an error term in order to make sense

\begin{equation} b_i = \alpha \cdot x_i + \epsilon _i \end{equation}

Otherwise, you can't mathematically find a single $\alpha$ that works for all combinations of $(x_i, b_i)$. This is relevant because now it's clearer why your second approach might not work so well: by suggesting that

\begin{equation} E[b_{i+1}] = \alpha \cdot x_{i+1} = \dfrac{b_i}{x_i} \cdot x_{i+1} \end{equation}

you are ignoring the error term $\epsilon _i$:

\begin{align} b_i &= \alpha \cdot x_i + \epsilon _i \\ &\neq \alpha \cdot x_i \end{align}

The average of all error terms must be zero, yes, but any particular $\epsilon _i$ can be arbitrarily large (which would happen if the pair $(x_i, b_i)$ were an outlier)

In other words, the model parameter $\alpha$ contains information about all $N$ $(x_i, b_i)$ pairs. In estimating $E[b_{i+1}]$ using only the information from $(x_i, b_i)$, you are ignoring the information that the pairs $(x_j, b_j) _{j \neq i}$ contain.

As an example, consider your original example:

\begin{align} X_{LV} &= (1, 2, 4, 5, 6, 8, 10, 11, 15, 20, 30) \\ B_{LV} &= (2, 4, 8, 11, 12, 15, 19, 22, 31, 40) \end{align}

Doing the math on the first $N = 10$ elements of both vectors, $\alpha _{LV} = 2.01 \pm 0.08$. You would like to predict the next element in $B_{LV}$, so

\begin{align} E[b_{LV, 11}] &= \alpha _{LV} \cdot x_{LV, 11} \\ &= 60.06 \end{align}

Using your proposed alternative, we get

\begin{align} E'[b_{LV, 11}] &= \dfrac{b_{LV,10}}{x_{LV,10}} \cdot x_{LV, 11} \\ &= 60.0 \end{align}

which represents a $0.1\%$ difference. However, if we run the regression on these series instead

\begin{align} X_{HV} &= (1, 2, 4, 5, 6, 8, 10, 11, 15, 20, 30) \\ B_{HV} &= (2, 4, 8, 11, 12, 15, 19, 22, 21, 47) \end{align}

then we would get an $\alpha _{HV} = 2.0 \pm 0.4$. Note that the slope has a lot more variance in the high variance case. To predict $b_{HV, 11}$ we calculate

\begin{align} E[b_{HV, 11}] &= \alpha _{HV} \cdot x_{HV, 11} \\ &= 59.76 \end{align}

Your proposed alternative, however, yields

\begin{align} E'[b_{HV, 11}] &= \dfrac{b_{HV,10}}{x_{HV,10}} \cdot x_{HV, 11} \\ &= 70.5 \end{align}

which represents a much larger $18\%$ difference. This happens because the pair $(x_{HV,10}, b_{HV,10})$ is an outlier: Comparison between low variance and high variance regressions.

In other words, $\alpha _{HV}$ has information on all $(x_{HV, i}, b_{HV,i}) _{i \in [1\ldots 10]}$ pairs, which results in much better predictions. In contrast, the alternative amplifies any errors present in the $(x_{HV, 10}, b_{HV,10})$ pair.

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  • $\begingroup$ "The average of all error terms must be zero" >> does this follow from the fact that the distribution of the residuals must be normal? $\endgroup$ – redress May 24 '17 at 21:52
  • $\begingroup$ @redress No. This is what ordinary least squares regression is: find the parameters of a regression model (in this case, the $\alpha$) that minimizes the magnitude of the error terms (the $\epsilon _i$). Mathematically, that is equivalent to setting the mean of the error term distribution to zero, whatever that distribution might be. $\endgroup$ – LmnICE May 24 '17 at 23:26
  • $\begingroup$ Actually, the two models WORKS in simulations. What I understand of the second model: $E[b_{i+1}] = E[b_i] * E[\frac{x_{i+1}}{x_i}] = E[b_i] * (r + t * \frac{x_{i+1}}{x_i}) $ Where $r$ absorbs the error parameter. So what I see here the difference is that $E[b_i]$ isn't observed $b_i$. But that is for me even better, because your model is adjusting itself to reality. $\endgroup$ – Dinidiniz May 25 '17 at 4:26
  • $\begingroup$ Two points: 1) When you say the two models work, how are you measuring that? 2) If you're drawing both $x_i$ and $b_i$ from distributions with tight dispersion (low variance), then it makes sense that both approaches seem to work most of the time, because there would be few pairs with a high error term. If either $x_i$ or $b_i$ are drawn from a high variance distribution, then the alternative would not yield good results consistently. If the error term is absorbed by $r$, you wouldn't be able to know whether $r$ is representative of all your data or not unless you ran the regression. $\endgroup$ – LmnICE May 25 '17 at 11:23
  • $\begingroup$ When I am simulating, the first model is better to predict, showing less variance. When I try to put bigger errors in the simulation, the two models variance grows in the same exponential. But when I use $E[b_i]$ to calculate $b_{i+1}$ instead of $b_i$, the variance of the two models are the same. This is awkward, because I can image situations where the second model is better. For example: in the middle of a time series: $time < X$, all values will be correct; while $time >= X$, all values will be $b_i * 10$; the first model will have problems, the second won't. $\endgroup$ – Dinidiniz May 25 '17 at 17:05

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