111 lines
3.4 KiB
Plaintext
111 lines
3.4 KiB
Plaintext
---
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title: "Lab10: Time Series"
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author: "Vladislav Litvinov <vlad@sek1ro>"
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output:
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pdf_document:
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toc_float: TRUE
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---
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Plotting data set
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```{r}
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setwd('/home/sek1ro/git/public/lab/ds/25-1/r')
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jj = scan("jj.dat")
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jj_ts = ts(jj, start = c(1960, 1), frequency = 4)
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jj_ts
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plot(jj_ts, ylab = "EPS", xlab = "Year")
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```
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In order to perform an ARIMA model, the time series will need to be transformed to remove any trend. Plot the difference of xt and xt-1, for all t > 0. Has this difference adequately detrended the series? Does the variability of the EPS appear constant over time? Why does the constant variance matter?
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```{r}
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jj_diff = diff(jj_ts)
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plot(jj_diff, xlab = "Year", ylab = "EPS diff")
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```
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Plot the log10 of the quarterly EPS vs. time and plot the difference of log10(xt ) and
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log10(xt-1) for all t > 0. Has this adequately detrended the series? Has the variability of the differenced log10(EPS) become more constant?
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```{r}
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log_jj = log10(jj_ts)
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log_jj_diff = diff(log_jj)
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plot(log_jj, xlab = "Year", ylab = "log10(EPS)")
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plot(log_jj_diff, xlab = "Year", ylab = "log10(EPS) diff")
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```
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Treating the differenced log10 of the EPS series as a stationary series, plot the ACF and PACF of this series. What possible ARIMA models would you consider and why?
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ACF(k) = Corr(x[t], x[t-k]) - Autocorrelation Function, показывает, насколько временной ряд коррелирует сам с собой
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PACF - Partial Autocorrelation Function - оказывает чистую связь после удаления влияния всех промежуточных значений между t и t-k, это последний коэффициент в AR(k)-регрессии
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xt = f1xt-1 + f2xt-2 + .. + fkxt-k + eps + f1xt-1 + f2xt-2 + .. + fkxt-k
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PACF(k) = fk
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ARMA(p, q)
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p - AR-часть, предыдущие значения - PACF
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q - MA-часть, ошибки предыдущих предсказаний - ACF
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ARIMA(p, d, q)
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d - I-часть, number of differences
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```{r}
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acf(log_jj_diff, lag.max = 20)
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ar(log_jj_diff)
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pacf(log_jj_diff, lag.max = 20)
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```
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Run the proposed ARIMA models from part d and compare the results. Identify an appropriate model. Justify your choice.
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Смысл: баланс между
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Качеством подгонки (чем лучше модель описывает данные, тем ниже ошибка)
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Сложностью модели (чем больше параметров, тем выше риск переобучения)
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AIC=2k−2ln(L)
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L > , k <
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Why is the choice of natural log or log base 10 in Problem 4.8 somewhat irrelevant to the transformation and the analysis?
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Why is the value of the ACF for lag 0 equal to one?
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```{r}
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library(forecast)
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fit_model = function(order) {
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Arima(log_jj_diff, order = order)
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}
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models <- list(
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"1, 0, 1" = fit_model(c(1,0,1)),
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"1, 1, 1" = fit_model(c(1,1,1)),
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"1, 0, 5" = fit_model(c(1,0,5)),
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"1, 1, 5" = fit_model(c(1,1,5))
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)
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print(models["1, 0, 5"])
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aic_values <- sapply(models, AIC)
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print(aic_values)
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```
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Arima(1, 0, 5)
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```{r}
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n = 10000
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phi4 = c(-0.18)
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AR <- arima.sim(n=n, list(ar=phi4[1]))
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plot(AR, main="AR series")
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acf(AR, main="ACF AR")
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pacf(AR, main="PACF AR")
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theta4 <- c(-0.65, -0.22, -0.28, 1, -0.4)
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MA <- arima.sim(n=n, list(ma=theta4))
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plot(MA, main="MA series")
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acf(MA, main="ACF MA")
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pacf(MA, main="PACF MA")
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```
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```{r}
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fit <- auto.arima(jj_ts)
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summary(fit)
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forecasted_values <- forecast(fit, h = 20)
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plot(forecasted_values)
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``` |