# 金融代写|金融实证代写Financial Empirical 代考|FINM6402

## 金融代写|金融实证代写Financial Empirical 代考|Square Root Transform

The first example is the U.S. Census Bureau monthly series of total inventory of nonmetallic mineral products in the USA, between the years 1992 and 2011. The default square root transform in X-12-ARIMA is $0.25+2\left(\sqrt{\mathbf{X}_t}-1\right)$, which is a shifted and scaled version of the basic $\sqrt{\mathbf{X}_t}$ square root transform, and the two adjusted log likelihoods are identical. Using X-12-ARIMA’s automdl and transform specs, we compare the square root transform to both a logarithmic transform and to no transform at all. Typically, the model with the smallest AICC would be preferred over other contenders, but since the same SARIMA $\left(\begin{array}{lll}0 & 2 & 1\end{array}\right)\left(\begin{array}{lll}0 & 1 & 1\end{array}\right)$ model was found to fit all three transforms of the data, the best transform would equivalently be indicated by the highest log likelihood. Table 1 displays the log likelihoods (adjusted for the transformations) along with the corresponding AICC. We see that the square root transform yields the highest log likelihood in the parent space and also the lowest value for AICC; this leads us to prefer the use of a square root transform for this total inventory series.

We proceed by using X-12-ARIMA to obtain an additive decomposition of the series $\left{X_t\right}$, where $X_t$ is just the square root of $\mathbf{X}_t$. In checking the difference series $X_t-\left(N_t+S_t\right)$, we note that the differences appear to be centered around 0 , with a maximum magnitude no greater than $5 \times 10^{-13}$; numerical error from rounding and computer precision explains why this difference is not identically 0 . Similar results hold for the difference between $\mathbf{X}_t$ and $\mathbf{N}_t \oplus \mathbf{S}_t$, which is just the application of $\varphi^{-1}$ to $N_t+S_t$. However, there are substantial discrepancies between $\mathbf{X}_t$ and $\mathbf{N}_t+\mathbf{S}_t$, as expected. For $\mathbf{N}_t=\varphi^{-1}\left(N_t\right)$ and $\mathbf{S}_t=\varphi^{-1}\left(S_t\right)$. Fig.5 shows a plot of the untransformed séries $\mathbf{X}_t$ along with $\mathbf{N}_t+\mathbf{S}_t$ on the top panel, and on the bottom panel, we hig. 5 confirms that the additive decomposition in transformed space does not translate to an additive decomposition in parent space, and the bottom panel shows that the deviations from 0 in this case are quite pronounced. Furthermore, while the lower panel of Fig. 5 indicates that the differences are roughly unbiased (the series is centered around zero), it also displays a highly seasonal pattern evincing some heteroskedasticity. We explain this behavior below.

Noting that the seasonal $S_t$ can be negative, it follows that $\mathbf{S}_t$ can be negative as well; however, if the original data $\mathbf{X}_t$ is always positive, it follows that
$$\mathbf{S}_t \oplus \mathbf{N}_t=\mathbf{S}_t+\mathbf{N}_t+\operatorname{sign}\left(\mathbf{S}_t \mathbf{N}_t\right) \sqrt{\left|\mathbf{S}_t\right|\left|\mathbf{N}_t\right|} .$$
Typically $\mathbf{N}_t$ is positive as well, so that
$$\mathbf{S}_t \oplus \mathbf{N}_t-\left(\mathbf{S}_t+\mathbf{N}_t\right)=\operatorname{sign}\left(\mathbf{S}_t\right) \sqrt{\left|\mathbf{S}_t\right|} \sqrt{\mathbf{N}_t}$$

## 金融代写|金融实证代写Financial Empirical 代考|Logistic Transform

The second example is the monthly unemployment rate for 16-19-year-old individuals of Hispanic origin between the years 1991 and 2011 ; the data was obtained from the Bureau of Labor Statistics. For rate data, the logistic transform $\varphi(\mathbf{a})=$ $\log (\mathbf{a})-\log (1-\mathbf{a})$ is sometimes warranted, as it ensures fits and predictions that are guaranteed to fall between 0 and 1. As in the previous example, we use X-12-ARIMA’s automdl and transform specs to help us compare the logistic transform to both a logarithmic transform and to no transform at all. Again, thê procedure selects the same SARIMA $\left(\begin{array}{llll}0 & 1 & 1\end{array}\right)\left(\begin{array}{lll}0 & 1 & 1\end{array}\right)$ model for all three transforms, so whichever transform has the highest $\log$ likelihood in the parent space will also have the lowest $\mathrm{AICC}$. Table 2 displays the log likelihoods (adjusted for the transformations) along with the corresponding AICC, and we see that the logistic transform does indeed result in a better model compared to the other two transformations.

We proceed by performing a logistic transform on $\mathbf{X}_t$ and then running $\mathrm{X}-12-$ ARIMA on the transformed series to obtain an additive seasonal decomposition. Checking the series of differences $X_t-\left(N_t+S_t\right)$, we find that the magnitude of the differences is bounded by $6 \times 10^{-15}$. These deviations from 0 are entirely explained by numerical error produced from passing the data through X-12-ARIMA. Similar results hold for $\mathbf{X}_t-\left(\mathbf{N}_t \oplus \mathbf{S}_t\right)$. But there are notable discrepancies between $\mathbf{X}_t$ and $\left(\mathbf{N}_t+\mathbf{S}_t\right)$, as in the previous illustration, as shown in Fig. 6. The top panel shows that the additive nature of the decomposition in transformed space is not preserved when mapped back to the parent space, while the bottom panel shows that this discrepancy (in the parent space) is a time series centered around $-0.5$. Also, the lower panel of discrepancies $\mathbf{X}_t-\left(\mathbf{N}_t+\mathbf{S}_t\right)$ exhibits seasonal structure; we explain this phenomenon next.
For the logistic transform, the composition operator $\oplus$ is defined as
$$\mathbf{S}_t \oplus \mathbf{N}_t=\frac{\mathbf{S}_t \cdot \mathbf{N}_t}{1-\mathbf{S}_t-\mathbf{N}_t+2 \mathbf{S}_t \cdot \mathbf{N}_t},$$

## 金融代写|金融实证代写Financial Empirical 代考|Square Root Transform

$$\mathbf{S}_t \oplus \mathbf{N}_t=\mathbf{S}_t+\mathbf{N}_t+\operatorname{sign}\left(\mathbf{S}_t \mathbf{N}_t\right) \sqrt{\left|\mathbf{S}_t\right|\left|\mathbf{N}_t\right|}$$

$$\mathbf{S}_t \oplus \mathbf{N}_t-\left(\mathbf{S}_t+\mathbf{N}_t\right)=\operatorname{sign}\left(\mathbf{S}_t\right) \sqrt{\left|\mathbf{S}_t\right|} \sqrt{\mathbf{N}_t}$$

## 金融代写|金融实证代写Financial Empirical 代考|Logistic Transform

$$\mathbf{S}_t \oplus \mathbf{N}_t=\frac{\mathbf{S}_t \cdot \mathbf{N}_t}{1-\mathbf{S}_t-\mathbf{N}_t+2 \mathbf{S}_t \cdot \mathbf{N}_t}$$

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