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I have a yearly time series which is the following:

1 2008 1.423832 2 2009 4.017000 3 2010 11.333000 4 2011 10.840000 5 2012 15.324000 6 2013 9.822000 7 2014 5.065000 8 2015 11.759000 9 2016 3.260000 10 2017 11.517000 11 2018 12.162000 12 2019 12.843123

I want to create a monthly time series from 2009 to 2018 such that the mean of each set of 12 months is the same as the years in this dataset, but months follow each other linearly, from July to June, without any abrupt gap between years.

I've tried using a system where for example January is imputated by 5.5 times the last year and 6.5 times this year, February is 4.5 last year and 7.5 this year, and so on.

But this makes, in some cases, June-July the minimum/maximum of the year and not the mean. The difference is very significant in some cases (on the left is the actual mean and )

[1,] 4.017 4.356189 [2,] 11.333 10.045469 [3,] 10.840 11.313115 [4,] 15.324 14.083493 [5,] 9.822 10.345170 [6,] 5.065 6.455427 [7,] 11.759 9.882330 [8,] 3.260 5.422764 [9,] 11.517 10.168153 [10,] 12.162 12.111386

I've multiplied the values in each year by the ratio between the actual mean and the mean I've got, so now means are already correct, but there's obviously a huge gap between each year:

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I've run out of ideas. Can you give me a hand?

Thanks

zest16
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  • Sorry if it is obvious to you but what do you mean when you say that you want "months to follow each other linearly"? And how do you exactly plan to do a monthly time series if you only have access to yearly information? – Milloupe May 07 '19 at 10:00
  • I want, for example, the difference from July to August to be the same as the one between August and September, and so on. There should be a slope change between June and July. Overall, the mean of the months in each year should be the same as the actual mean. I only have access to yearly information but I want to impute monthly data based on that information. – zest16 May 07 '19 at 11:30
  • I just realised that my main problem was that I didn't understand the financial term "impute". Now that this is corrected, I think the issue you talk about with June being the max of the year is necessary if the given year is larger than the two surrounding it. If you infer the values of each month depending on how far it is from the other years (which is what I understood of "imputing"), you end up putting values in June-July which depend nearly only on the current year, right? – Milloupe May 07 '19 at 14:36
  • Exactly. And conversely, if a given year is shorter than the two surrounding years, then the value between June-July is the minimum and not the mean of that year. Either way, the mean of the 12 values I get is never equal to the actual yearly mean. – zest16 May 07 '19 at 16:23

1 Answers1

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Please let me put some mathematical formalism in your question: Let's say we have three consecutive years of mean $A$, $B$, $C$. These are linked to monthly information $\{a_i\}_{i\in [1, 12]}, \{b_i\}_{i\in [1, 12]},\{c_i\}_{i\in [1, 12]}$ through:

$$ A = \frac{1}{12} \sum_{i=1}^{12} a_i $$ and so on, but you don't have access to these monthly values.

So you want to find some values $\{\tilde{a_j}\}_{j\in [1, 12]}, \{\tilde{b_j}\}_{j\in [1, 12]},\{\tilde{c_j}\}_{j\in [1, 12]}$ which verify:

$$\begin{cases} A = \frac{1}{12} \sum_{j=1}^{12} \tilde{a_j} \\ B = \frac{1}{12} \sum_{j=1}^{12} \tilde{b_j} \\ C = \frac{1}{12} \sum_{j=1}^{12} \tilde{c_j} \\ \end{cases}$$

If this is indeed what you wish and I understood it correctly, you then have chosen to define coefficients $\{\alpha_j\}_{j\in [1, 12]}, \{\beta_j\}_{j\in [1, 12]},\{\gamma_j\}_{j\in [1, 12]}$ such that:

$$ \tilde{b_j} = \alpha_jA + \beta_jB + \gamma_jC $$ (I'm taking the $\tilde{b_j}$'s as an example so that we have a year on each side)

You said you typically took ($\alpha_1 = 5.5$, $\beta_1 = 6.5$, $\gamma_1 = 0$) and ($\alpha_2 = 4.5$, $\beta_2 = 7.5$, $\gamma_2 = 0$), etc.

This sounds great, and if you compute the mean of these monthly estimated values, you get:

$$ \tilde{B} = \frac{1}{12} \sum_{j=1}^{12} \tilde{b_j} = \frac{1}{12} \sum_{j=1}^{12} (\alpha_jA + \beta_jB + \gamma_jC) $$ this directly gives:

$$ \tilde{B} = A (\frac{1}{12} \sum_{j=1}^{12} \alpha_j) + B (\frac{1}{12} \sum_{j=1}^{12} \beta_j) + C (\frac{1}{12} \sum_{j=1}^{12}\gamma_j) $$

And with this expression you get $\tilde{B} = B$ only if:

$$ B = A (\frac{1}{12} \sum_{j=1}^{12} \alpha_j) + B (\frac{1}{12} \sum_{j=1}^{12} \beta_j) + C (\frac{1}{12} \sum_{j=1}^{12}\gamma_j) \\ \Rightarrow 0 = A (\frac{1}{12} \sum_{j=1}^{12} \alpha_j) + B (\frac{1}{12} \sum_{j=1}^{12} \beta_j - 1) + C (\frac{1}{12} \sum_{j=1}^{12}\gamma_j) $$

This is one big equation, but it is what your coefficients must verify in order to be sure that you are inferring correct values for your monthly values.

And just to make sure, you can see that putting $\beta_j = 1, \forall j$ and the other coefficients to 0 (giving to each month the value of the yearly mean) does give you a valid set of parameters.

What I suggest now is that you play around with this equation and see what sets of parameters you can use!

Milloupe
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