Testing Mean for Significance
Nao Mimoto - Dept. of Statistics : The University of Akron
1. Is the ture mean 0?
Suppose we have a random sample
set.seed(235) # set seed for RS generation (you don't need to do this)
mu = sample(c(0,2,1), 1) # mu is randomly drawn from 0, 1, or 2
X = mu + rnorm(100,0,5) # Y is mu + random noise
mean(X) # Calculate sample mean## [1] 2.016148 plot(X, type="o") # type="o" means both line and point
abline(h=0) # add horizontal line at 0
abline(h=mean(X), col="blue", lty=2) # add blue dashed line at sample mean
Question is if our ture mean \(\mu\) in this plot is \(0\) or not.
We know the sample mean \(\bar Y\) is not \(0\). But is this a random fluctuation? or is this an evidence that we have non-zero true mean?
2. Random Sample Case
We can say that above random sample comes from model like this: \[ X_t = \mu + \epsilon_t \hspace15mm \left\{ \begin{array}{ll} \mu : &\mbox{ Constant Trend (deterministic) }\\ \epsilon_t : &\mbox{ iid noise from } N(0, \sigma)\\ \end{array} \right. \]
\(\mu\) and \(\sigma\) are unknown.
This is equivalent to saying \(X_i\) is iid random sample from \(N(\mu,\sigma)\) distribution.
We will use \(\bar X\) to estimate \(\mu\).
CI for mean \(\mu\)
We know how ‘well’ \(\bar X\) estimates \(\mu\). \[ \bar X \sim N\Big(\mu, \frac{\sigma}{\sqrt n}\Big) \]
That gives us 95% CI for \(\mu\) \[ \bar X \pm 1.96 \frac{S}{\sqrt n} \] Since \(\sigma\) is unkown, it is replaced by the sample standard deviation \(S\).
If \(0\) is outside of the CI, then we reject \(H_0: \mu=0\) with 5% confidence level.
## [1] 2.016148## [1] 4.655382## [1] 0.9124548 CI.upper = mean(X) + ME
CI.lower = mean(X) - ME
c(CI.lower, mean(X), CI.upper) # show 3 numbers as vector## [1] 1.103693 2.016148 2.928603 plot(X, type="o") # plot Y
abline(h=0) # draw horizontal line at 0
abline(h=mean(X), col="blue", lty=2) # blue dash line at mean(Y)
abline(h=c(CI.upper, CI.lower), col="red", lty=2) # two red dash line at CI.upper and CI.lower
Since \(0\) is outside of the 95% CI for \(\mu\), with 95% confidence, we conclude that the true mean \(\mu\) is not \(0\).
3. Time Series Case
We want to do the same when the data shows autocorrelation.
Model
Suppose we observe \(X_t\), and model it as: \[ Y_t = \mu + X_t \hspace10mm \left\{ \begin{array}{ll} \mu : &\mbox{ Constant Trend (deterministic) }\\ X_t : &\mbox{ Stationary Time Series } \end{array} \right. \] We assume that \(E X_t=0\) and \(V(X_t) = \sigma^2\).
Note that if \(X_t\) has autocorrelation, then \(Y_t\) is also autocorrelated.
We will use \(\bar Y\) to estimate \(\mu\), as usual.
Expectation of Sample Mean
Does property of \(\bar Y\) changes? Expectation is \[ E (\bar Y) \hspace3mm = \hspace3mm E \Big(\frac{1}{n} \sum_{t=1}^n Y_t\Big) \hspace3mm = \hspace3mm \frac{1}{n} \sum_{t=1}^n E (Y_t) \]
For the mean of one \(Y\), we have \[ E(Y_t) \hspace3mm = \hspace3mm E(\mu + X_t) \hspace3mm = \hspace3mm \mu + E(X_t) \hspace3mm = \hspace3mm \mu + 0 \hspace3mm = \hspace3mm \mu \]
So we do have \(E(\bar Y) = \mu\) and that’s same as the RS case.
i.e. \(\bar Y\) is still unbiased estimator of \(\mu\).
Variance of Sample Mean
How about the variance of \(\bar Y\)?
Since adding a constant does not change the variance, \[ \begin{aligned} V(\bar Y) \hspace3mm = \hspace3mm V(\bar Y - \mu ) \hspace3mm = \hspace3mm V \Big( \frac{1}{n} \sum_{i=1}^n (Y_i - \mu) \Big) \hspace3mm = \hspace3mm V \Big( \frac{1}{n} \sum_{i=1}^n X_i \Big) \hspace3mm = \hspace3mm V ( \bar X ) \end{aligned} \]
Now the variance of \(\bar X\) can be written as, \[ V (\bar X) \hspace3mm = \hspace3mm V \Big( \frac{1}{n} \sum_{i=1}^n X_i \Big) \hspace3mm = \hspace3mm \frac{1}{n^2} \, \mbox{Cov}\Big(\sum_{i=1}^n X_i, \, \sum_{j=1}^n X_j \Big) \hspace3mm = \hspace3mm \frac{1}{n^2} \Big[\mbox{ sum of Cov of all pairs }(X_i, X_j) \Big] \]
\(\Big[\mbox{ Sum of Cov of all pairs }(X_i, X_j) \Big]\) means, \[ \hspace3mm = \hspace3mm \mbox{ Sum of Cov of pairs } \left[ \begin{array}{c|cccc} & X_1 & X_2 & \cdots & X_n \\ \hline X_1 & \ddots & & \cdots & \\ X_2 & & \ddots & \cdots & \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ X_n & & & \cdots & \\ \end{array} \right] \hspace5mm = \hspace5mm \mbox{ Sum of } \left[ \begin{array}{cccc} \gamma(0) & \gamma(1) & \cdots & \gamma(n-1) \\ \gamma(-1) & \gamma(0) & \cdots & \gamma(n-2) \\ \vdots & \vdots &\ddots & \vdots \\ \gamma(n-1) & \gamma(n-2) & \cdots & \gamma(0) \\ \end{array} \right]\\ \] where \(\gamma(h)\) is ACVF at lag \(h\).
Conbining, we get \[ V(\bar Y) \hspace3mm = \hspace3mm V(\bar X) \hspace3mm = \hspace3mm \frac{1}{n^2} \sum_{h=-n}^n (n-|h|) \gamma(h) \hspace3mm = \hspace3mm \frac{1}{n^2} \sum_{h=-n}^n (n-|h|) \gamma(h) \hspace3mm = \hspace3mm \frac{1}{n} \sum_{h=-n}^n \Big(1 -\frac{|h|}{n}\Big) \gamma(h). \]
The variance of sample mean is different under the presence of autocorrelation.
Note that if \(X_t\) were iid as in the first section, then \(\gamma(0)=1\), and \(\gamma(h)=0\) for all \(h \ne 0\). Above formula reduces to \[
V (\bar X)
= \frac{1}{n} \Big(1 -\frac{|0|}{n}\Big) \gamma(0)
\hspace3mm = \hspace3mm \frac{\sigma^2}{n}
\]
CI for \(\mu\) When \(Y_t\) is TS
So we have \[ V (\bar Y) = \frac{\nu^2}{n} \hspace5mm \mbox{ where } \hspace5mm \nu^2 = \sum_{h=-n}^n \Big(1 -\frac{|h|}{n}\Big) \gamma(h). \] That means approximately, \[ \bar Y \sim N\Big(\mu , \frac{\nu}{\sqrt n} \Big). \] Then the confidence interval for \(\mu\) is \[ \bar Y \pm 1.96 \sqrt{\frac{ \nu^2 }{n}} \]
In practice, we don’t know the true value of \(\gamma(h)\), so we need to use the sample version \(\hat \gamma(h)\). (sample ACVF instead of theoretical ACVF) \[ \hat \nu^2 = \sum_{h=-\sqrt n}^{\sqrt n} \Big(1 -\frac{|h|}{n}\Big) \hat \gamma(h). \]
Also, we can’t really use \(\hat \gamma(h)\) with \(h\) close to \(n\). So sum goes from \(-\sqrt{n}\) to \(\sqrt{n}\) instead.
Finally, modify the sum to: \[
\hat \nu^2 = \sum_{h=-\sqrt n}^{\sqrt n} \Big(1 -\frac{|h|}{n}\Big) \hat \gamma(h)
\hspace3mm = \hspace3mm
\gamma(0) + 2 \sum_{h=1}^{\sqrt n} \Big(1 -\frac{|h|}{n}\Big) \hat \gamma(h).
\]
4. Ex: Testing Mean when There’s Autocorrelation
# Simulate the autocorrelated data
set.seed(1352)
mu = sample(c(0,1,2), 1)
Y = mu + arima.sim(n=70, list(ar = c(.7) ), sd=5 )
layout(matrix(1:2, 1, 2)) # plot side by side
plot(Y, type="o") # plot Y
abline(h=0) # draw at 0
abline(h=mean(Y), col="blue") # draw blue line at sample mean
acf(Y) # plot acf as 2nd plot
Because of the presence of significant ACF at lag 1, we need to use the new formula for CI.
## [1] 70 Ga = acf(Y, type="covariance", plot=FALSE) # calculate ACVF values without plotting
str(Ga) # See the structure inside Ga## List of 6
## $ acf : num [1:19, 1, 1] 35.06 19.04 9.88 1.38 -5.13 ...
## $ type : chr "covariance"
## $ n.used: int 70
## $ lag : num [1:19, 1, 1] 0 1 2 3 4 5 6 7 8 9 ...
## $ series: chr "Y"
## $ snames: NULL
## - attr(*, "class")= chr "acf" Ga.hat = Ga$acf # extract only "acf" part (it's actually ACVF)
# Calculate nu.sq using ACF
sum.limit = floor( sqrt(n) ) # make sqrt(n) into integer (for later use)
nu.sq = Ga.hat[1] + 2*sum( (1-(1:sum.limit)/n) * Ga.hat[2:(sum.limit+1)] )
#- Ga.hat[1] is ACVF at lag 0
mean(Y)## [1] 1.732918## [1] 1.197743## [1] 0.7127774 CI.u = mean(Y) + 1.96*sqrt(nu.sq/n)
CI.l = mean(Y) - 1.96*sqrt(nu.sq/n)
c(CI.l, mean(Y), CI.u) # show CI.l, mean(Y), CI.u as vector## [1] -0.6146584 1.7329183 4.0804951 plot(Y, type="o")
abline(h=0) # add black line at 0
abline(h=mean(Y), col="blue") # add blue horizontal line at mean(Y)
abline(h=c(CI.u, CI.l), col="red", lty=2) # add red dash lines at CI
Because 95% CI includes \(0\), we can not reject the null hypothesis that the true mean of \(Y\) is \(0\).
Note that if we used the RS version of the formula, we would have rejected.
Example: Color data
acf1 <- acf
library(TSA) # load package TSA
acf <- acf1
data(color)
layout(matrix(1:2, 1, 2)) # plot side by side
plot(color, type="o") # plot
abline(h=0) # draw at 0
abline(h=mean(color), col="blue") # draw blue line at sample mean
acf(color)
This time, let’s just get CI for the ture mean without testing to see if it is different from 0.
Note that I’m using the exact copy of the code used in the last section.
Y = color # So that I can reuse the above code without changing each Y
#--- CI for ture mean using ACVF ---
n = length(Y) # This is n
n # see what's inside## [1] 35 Ga <- acf(Y, type="covariance", plot=FALSE) # calculate ACVF values without plotting
str(Ga) # See the structure inside Ga## List of 6
## $ acf : num [1:16, 1, 1] 36.04 19.04 11.79 8.08 3.31 ...
## $ type : chr "covariance"
## $ n.used: int 35
## $ lag : num [1:16, 1, 1] 0 1 2 3 4 5 6 7 8 9 ...
## $ series: chr "Y"
## $ snames: NULL
## - attr(*, "class")= chr "acf" Ga.hat = Ga$acf # extract only "acf" part (it's actually ACVF)
sum.limit = floor( sqrt(n) ) # make sqrt(n) into integer (for later use)
nu.sq <- Ga.hat[1] + 2*sum( (1-(1:sum.limit)/n) * Ga.hat[2:(sum.limit+1)] )
#- Ga.hat[1] is ACVF at lag 0
mean(Y)## [1] 74.88571## [1] 1.799286## [1] 1.029621 CI.u <- mean(Y) + 1.96*sqrt(nu.sq/n)
CI.l <- mean(Y) - 1.96*sqrt(nu.sq/n)
c(CI.l, mean(Y), CI.u) # show CI.l, mean(Y), CI.u as vector## [1] 71.35911 74.88571 78.41232 plot(Y, type="o")
abline(h=0) # add black line at 0
abline(h=mean(Y), col="blue") # add blue horizontal line at mean(Y)
abline(h=c(CI.u, CI.l), col="red", lty=2) # add red dash lines at CI
5. Ex: Testing Mean for Random Sample
# Simulate the autocorrelated data
layout(matrix(1:2, 1, 2)) # plot side by side
plot(X, type="o") # plot Y
abline(h=0) # draw at 0
abline(h=mean(X), col="blue") # draw blue line at sample mean
acf(X)
Summary
- For random sample, the sample mean has approximately the distribution of: \[ N\left(\mu, \hspace{2mm} \frac{\sigma}{\sqrt{n}}\right). \] where \(\sigma\) can be estimated by the sample standard deviaiton.
- For autocorrelated data, the sample mean has approximately the distribution of: \[ N\left(\mu, \hspace{2mm} \frac{\nu}{\sqrt{n}}\right), \] where \(\nu\) can be estimated using sample ACVF via \[ \hat \nu^2 = \gamma(0) + 2 \sum_{h=1}^{\sqrt n} \Big(1 -\frac{|h|}{n}\Big) \hat \gamma(h). \]
- For autocorrelated data, variance of the sample mean is bigger than the random sample case. We must estimate the sample mean variance by summing up sample ACVF.