एसीएफ और पीएसीएफ फॉर्मूला

Question

मैं समय-श्रृंखला डेटा से एसीएफ और पीएसीएफ की साजिश रचने के लिए एक कोड बनाना चाहता हूं। ठीक इसी तरह से मिनीटैब (नीचे) से उत्पन्न प्लॉट।

मैंने सूत्र को खोजने की कोशिश की है, लेकिन मैं अभी भी इसे अच्छी तरह से नहीं समझता हूं। क्या आप मुझे फार्मूला बताने की कृपा करेंगे और कृपया इसका उपयोग कैसे करें? ऊपर ACF और PACF प्लॉट पर क्षैतिज लाल रेखा क्या है? सूत्र क्या है?

धन्यवाद,

Comments

@javlacalle क्या सूत्र आपको सही प्रदान करता है? यह काम करता है, तो नहीं होगा n Σ टी = 1 ( y टी - ˉ y ) < $$\rho(k) = \frac{\frac{1}{n-k}\sum_{t=k+1}^n (y_t - \bar{y})(y_{t-k} - \bar{y})}{ \sqrt{\frac{1}{n}\sum_{t=1}^n (y_t - \bar{y})}\sqrt{\frac{1}{n-k}\sum_{t=k+1}^n (y_{t-k} - \bar{y})}} \,,$$ है ना? क्या यह निम्नलिखित की तरह होना चाहिए? \ rho (k) = \ frac {\ frac {1} {nk} \ sum_ {t = k + 1} ^ n (y_t - \ bar {y}) (y_ {tk} - \ bar \ _ \ _} )} {\ sqrt {\ frac {1} {n} \ sum_ {t = 1} ^ n (y_t - \ bar {y}) ^ 2} \ sqrt {\ frac {1} {nk} \ sumt {t} = k + 1} ^ n (y_ {tk} - \ bar {y}) ^ 2}} \ $$\sum_{t=1}^n (y_t - \bar{y}) < 0 \qquad \text{and/or} \qquad \sum_{t=k+1}^n (y_{t-k} - \bar{y}) < 0$$

— _

@conighion आप सही हैं, धन्यवाद। मैंने इसे पहले नहीं देखा था। मैंने इसे ठीक कर लिया है।

— javlacalle

Answer 1

Autocorrelations

दो चर $y_1, y_2$ बीच संबंध को इस प्रकार परिभाषित किया गया है:

$$\rho = \frac{\hbox{E}\left[(y_1-\mu_1)(y_2-\mu_2)\right]}{\sigma_1 \sigma_2} = \frac{\hbox{Cov}(y_1, y_2)}{\sigma_1 \sigma_2} \,,$$

जहां ई उम्मीद ऑपरेटर है, $\mu_1$ और $\mu_2$ क्रमशः के लिए साधन हैं $y_1$ और $y_2$ और $\sigma_1, \sigma_2$ उनके मानक विचलन कर रहे हैं।

एक एकल चर के संदर्भ में, यानी ऑटो- कॉन्ट्रैक्शन, $y_1$ मूल श्रृंखला है और $y_2$ इसका एक लंबित संस्करण है। उपर्युक्त परिभाषा पर, आदेश का नमूना निरूपण$k=0,1,2,...$मनाया श्रृंखला$y_t$ , t = 1 , 2 , के साथ निम्नलिखित अभिव्यक्ति की गणना करके प्राप्त किया जा सकता है । । । , एन$t=1,2,...,n$ :

$$\rho(k) = \frac{\frac{1}{n-k}\sum_{t=k+1}^n (y_t - \bar{y})(y_{t-k} - \bar{y})}{ \sqrt{\frac{1}{n}\sum_{t=1}^n (y_t - \bar{y})^2}\sqrt{\frac{1}{n-k}\sum_{t=k+1}^n (y_{t-k} - \bar{y})^2}} \,,$$

जहाँ $\bar{y}$ डेटा का नमूना माध्य है।

आंशिक निरंकुशताएँ

आंशिक चर दोनों चर को प्रभावित करने वाले अन्य चर (नों) के प्रभाव को हटाने के बाद एक चर की रैखिक निर्भरता को मापते हैं। उदाहरण के लिए, आदेश का आंशिक स्वायत्तता $y_{t-2}$ पर प्रभाव (रैखिक निर्भरता) को मापता है$y_t$ के प्रभाव को दूर करने के बाद$y_{t-1}$ दोनों पर$y_t$ और$y_{t-2}$ ।

प्रत्येक आंशिक स्वसंबंध को प्रपत्र के पंजीकरण की एक श्रृंखला के रूप में प्राप्त किया जा सकता है:

$$\tilde{y}_t = \phi_{21} \tilde{y}_{t-1} + \phi_{22} \tilde{y}_{t-2} + e_t \,,$$

जहां $\tilde{y}_t$ मूल श्रृंखला शून्य से नमूना मतलब है, $y_t - \bar{y}$ । $\phi_{22}$ का अनुमान आदेश के आंशिक निरंकुशता का मूल्य देगा 2. $k$ अतिरिक्त लैग के साथ प्रतिगमन का विस्तार करना , अंतिम शब्द का अनुमान आदेश k का आंशिक निरंकुशता देगा।$k$ ।

नमूना आंशिक autocorrelations गणना करने के लिए एक वैकल्पिक तरीका प्रत्येक आदेश के लिए निम्नलिखित प्रणाली को हल करके है $k$ :

$$\begin{eqnarray} \left(\begin{array}{cccc} \rho(0) & \rho(1) & \cdots & \rho(k-1) \\ \rho(1) & \rho(0) & \cdots & \rho(k-2) \\ \vdots & \vdots & \vdots & \vdots \\ \rho(k-1) & \rho(k-2) & \cdots & \rho(0) \\ \end{array}\right) \left(\begin{array}{c} \phi_{k1} \\ \phi_{k2} \\ \vdots \\ \phi_{kk} \\ \end{array}\right) = \left(\begin{array}{c} \rho(1) \\ \rho(2) \\ \vdots \\ \rho(k) \\ \end{array}\right) \,, \end{eqnarray}$$

जहां $\rho(\cdot)$ नमूना autocorrelations हैं। नमूना autocorrelations और आंशिक autocorrelations के बीच यह मानचित्रण Durbin-Levinson पुनरावृत्ति के रूप में जाना जाता है । यह दृष्टिकोण चित्रण के लिए लागू करना अपेक्षाकृत आसान है। उदाहरण के लिए, R सॉफ़्टवेयर में, हम क्रम 5 का आंशिक ऑटोकरेक्शन प्राप्त कर सकते हैं:

# sample data
x <- diff(AirPassengers)
# autocorrelations
sacf <- acf(x, lag.max = 10, plot = FALSE)$acf[,,1]
# solve the system of equations
res1 <- solve(toeplitz(sacf[1:5]), sacf[2:6])
res1
# [1]  0.29992688 -0.18784728 -0.08468517 -0.22463189  0.01008379
# benchmark result
res2 <- pacf(x, lag.max = 5, plot = FALSE)$acf[,,1]
res2
# [1]  0.30285526 -0.21344644 -0.16044680 -0.22163003  0.01008379
all.equal(res1[5], res2[5])
# [1] TRUE

आत्मविश्वास बैंड

विश्वास बैंड नमूना autocorrelations के मूल्य के रूप में गणना की जा सकती $\pm \frac{z_{1-\alpha/2}}{\sqrt{n}}$$z_{1-\alpha/2}$$1-\alpha/2$

$\pm z_{1-\alpha/2}\sqrt{\frac{1}{n}\left(1 + 2\sum_{i=1}^k \rho(i)^2\right)}$.

Comments

(+1) Why the two different confidence bands?

— Scortchi - Reinstate Monica

@Scortchi Constant bands are used when testing for independence, while the increasing bands are sometimes used when identifying an ARIMA model.

— javlacalle

The two methods for calculating confidence bands are explained in a little more detail here.

— Scortchi - Reinstate Monica

Perfect explanation!

— Jan Rothkegel

@javlacalle, does the expression for $\rho(k)$ miss squares in the denominator?

— Christoph Hanck

Answer 2

"I want to create a code for plotting ACF and PACF from time-series data".

Although the OP is a bit vague, it may possibly be more targeted to a "recipe"-style coding formulation than a linear algebra model formulation.

---

The ACF is rather straightforward: we have a time series, and basically make multiple "copies" (as in "copy and paste") of it, understanding that each copy is going to be offset by one entry from the prior copy, because the initial data contains $t$ data points, while the previous time series length (which excludes the last data point) is only $t-1$. We can make virtually as many copies as there are rows. Each copy is correlated to the original, keeping in mind that we need identical lengths, and to this end, we'll have to keep on clipping the tail end of the initial data series to make them comparable. For instance, to correlate the initial data to $ts_{t-3}$ we'll need to get rid of the last $3$ data points of the original time series (the first $3$ chronologically).

Example:

We'll concoct a times series with a cyclical sine pattern superimposed on a trend line, and noise, and plot the R generated ACF. I got this example from an online post by Christoph Scherber, and just added the noise to it:

x=seq(pi, 10 * pi, 0.1)
y = 0.1 * x + sin(x) + rnorm(x)
y = ts(y, start=1800)

Ordinarily we would have to test the data for stationarity (or just look at the plot above), but we know there is a trend in it, so let's skip this part, and go directly to the de-trending step:

model=lm(y ~ I(1801:2083))
st.y = y - predict(model)

Now we are ready to takle this time series by first generating the ACF with the acf() function in R, and then comparing the results to the makeshift loop I put together:

ACF = 0                  # Starting an empty vector to capture the auto-correlations.
ACF[1] = cor(st.y, st.y) # The first entry in the ACF is the correlation with itself (1).
for(i in 1:30){          # Took 30 points to parallel the output of `acf()`
  lag = st.y[-c(1:i)]    # Introducing lags in the stationary ts.
  clipped.y = st.y[1:length(lag)]    # Compensating by reducing length of ts.
  ACF[i + 1] = cor(clipped.y, lag)   # Storing each correlation.
}
acf(st.y)                            # Plotting the built-in function (left)
plot(ACF, type="h", main="ACF Manual calculation"); abline(h = 0) # and my results (right).

---

OK. That was successful. On to the PACF. Much more tricky to hack... The idea here is to again clone the initial ts a bunch of times, and then select multiple time points. However, instead of just correlating with the initial time series, we put together all the lags in-between, and perform a regression analysis, so that the variance explained by the previous time points can be excluded (controlled). For example, if we are focusing on the PACF ending at time $ts_{t-4}$, we keep $ts_t$, $ts_{t-1}$, $ts_{t-2}$ and $ts_{t-3}$, as well as $ts_{t-4}$, and we regress $ts_t \sim ts_{t-1} + ts_{t-2} + ts_{t-3}+ts_{t-4}$ through the origin and keeping only the coefficient for $ts_{t-4}$:

PACF = 0          # Starting up an empty storage vector.
for(j in 2:25){   # Picked up 25 lag points to parallel R `pacf()` output.
  cols = j        
  rows = length(st.y) - j + 1 # To end up with equal length vectors we clip.

  lag = matrix(0, rows, j)    # The storage matrix for different groups of lagged vectors.

for(i in 1:cols){
  lag[ ,i] = st.y[i : (i + rows - 1)]  #Clipping progressively to get lagged ts's.
}
  lag = as.data.frame(lag)
  fit = lm(lag$V1 ~ . - 1, data = lag) # Running an OLS for every group.
  PACF[j] = coef(fit)[j - 1]           # Getting the slope for the last lagged ts.
}

And finally plotting again side-by-side, R-generated and manual calculations:

That the idea is correct, beside probable computational issues, can be seen comparing PACF to pacf(st.y, plot = F).

---

code here.

Answer 3

Well, in the practise we found error (noise) which is represented by $e_t$ the confidence bands help you to figure out if a level can be considerate as only noise (because about the 95% times will be into the bands).

Comments

Welcome to CV, you might want to consider adding some more detailed information on how OP would go about do this specifically. Maybe also add some information on what each line represents?

— Repmat

Answer 4

Here is a python code to compute ACF:

def shift(x,b):
    if ( b <= 0 ):
        return x
    d = np.array(x);
    d1 = d
    d1[b:] = d[:-b]
    d1[0:b] = 0
    return d1

# One way of doing it using bare bones
# - you divide by first to normalize - because corr(x,x) = 1
x = np.arange(0,10)
xo = x - x.mean()

cors = [ np.correlate(xo,shift(xo,i))[0]  for i in range(len(x1)) ]
print (cors/cors[0] )

#-- Here is another way - you divide by first to normalize
cors = np.correlate(xo,xo,'full')[n-1:]
cors/cors[0]

Comments

Hmmm Code formatting was bad:

— Sada