Linear regression and correlation
The ASTA team
The regression problem
We want to predict
- We will study the dataset
trees, which is on the course website (and actually also already available in R).
trees <- read.delim("https://asta.math.aau.dk/datasets?file=trees.txt")
- In this experiment we have measurements of 3 variables for 31 randomly chosen trees:
Girth numeric. Tree diameter in inches.Height numeric. Height in ft.Volume numeric. Volume of timber in cubic ft.- We want to predict the tree volume, if we measure the tree height and/or the tree girth (diameter).
- This type of problem is called regression.
- Relevant terminology:
- We measure a quantitative response \(y\), e.g.
Volume.
- In connection with the response value \(y\) we also measure one (later we will consider several) potential explanatory variable \(x\). Another name for the explanatory variable is predictor.
Initial graphics
- Any analysis starts with relevant graphics.
library(mosaic)
library(GGally)
ggscatmat(trees) # Scatter plot matrix from GGally package
- For each combination of the variables we plot the \((x,y)\) values.
- It looks like
Girth is a good predictor for Volume.
- If we only are interested in the association between two (and not three or more) variables we use the usual
gf_point function.
Simple linear regression
- We choose to use
x=Girth as predictor for y=Volume. When we only use one predictor we are doing simple regression.
- The simplest model to describe an association between response \(y\) and a predictor \(x\) is simple linear regression.
- I.e. ideally we see the picture \[y(x)=\alpha +\beta x\] where
- \(\alpha\) is called the
Intercept - the line’s intercept with the \(y\)-axis, corresponding to the response for \(x=0\).
- \(\beta\) is called
Slope - the line’s slope, corresponding to the change in response, when we increase the predictor by one unit.
gf_point(Volume ~ Girth, data = trees) %>% gf_lm()
## Warning: Using the `size` aesthetic with geom_line was deprecated in ggplot2 3.4.0.
## ℹ Please use the `linewidth` aesthetic instead.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was generated.
Model for linear regression
- Assume we have a sample with joint measurements \((x,y)\) of predictor and response.
- Ideally the model states that \[y(x)=\alpha +\beta x,\] but due to random variation there are deviations from the line.
- What we observe can then be described by \[y=\alpha + \beta x + \varepsilon, \] where \(\varepsilon\) is a random error, which causes deviations from the line.
- We will continue under the following fundamental assumption:
- The errors \(\varepsilon\) are normally distributed with mean zero and standard deviation \(\sigma\).
- We call \(\sigma\) the conditional standard deviation given \(x\), since it describes the variation in \(y\) around the regression line, when we know \(x\).
Least squares
- In summary, we have a model with 3 parameters:
- \((\alpha,\beta)\) which determine the line
- \(\sigma\) which is the standard deviation of the deviations from the line.
- How are these estimated, when we have a sample \((x_1,y_1),\ldots,(x_n,y_n)\) of pairs of \((x,y)\) values?
- To do this we focus on the errors \[\varepsilon_i=y_i-\alpha-\beta x_i\] which should be as close to 0 as possible in order to fit the data best possible.
- We will choose the line, which minimizes the sum of squares of the errors: \[\sum_{i=1}^n \varepsilon_i^2=\sum_{i=1}^n(y_i-\alpha-\beta x_i)^2.\]
- If we set the partial derivatives to zero we obtain two linear equations for the unknowns \((\alpha,\beta)\), where the solution \((a,b)\) is given by: \[b=\frac{\sum_{i=1}^n(x_i-\bar{x})(y_i-\bar{y})}{\sum_{i=1}^n(x_i-\bar{x})^2} \quad \mbox{ and } \quad a=\bar{y}-b\bar{x}\]
The prediction equation and residuals
- The equation given by the estimates \((\hat{\alpha},\hat{\beta})=(a,b)\), \[\hat{y}=a+bx\] is called the regression equation or the prediction equation, since it can be used to predict \(y\) for any value of \(x\).
- Note: The prediction equation is determined by the current sample. I.e. there is an uncertainty attached to it. A new sample would without any doubt give a different prediction equation.
- Our best estimate of the errors is \[e_i=y_i-\hat{y}=y_i-a-bx_i, \] i.e. the vertical deviations from the prediction line.
- These quantities are called residuals.
- We have that
- The prediction line passes through the point \((\bar{x},\bar{y})\).
- The sum of the residuals is zero.
Estimation of conditional standard deviation
To estimate \(\sigma\) we need the Sum of Squared Errors \[
SSE=\sum_{i=1}^n e_i^2=\sum_{i=1}^n(y_i-\hat{y}_i)^2,
\] which is the squared distance between the model and data.
We then estimate \(\sigma\) by the quantity \[
s=\sqrt{\frac{SSE}{n-2}}
\]
Instead of \(n\) we divide \(SSE\) with the degrees of freedom \(df=n-2\). Theory shows, that this is reasonable.
The degrees of freedom \(df\) are determined as the sample size minus the number of parameters in the regression equation.
In the current setup we have 2 parameters: \((\alpha,\beta)\).
Example in R
model <- lm(Volume ~ Girth, data = trees)
summary(model)
##
## Call:
## lm(formula = Volume ~ Girth, data = trees)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.065 -3.107 0.152 3.495 9.587
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -36.9435 3.3651 -10.98 7.62e-12 ***
## Girth 5.0659 0.2474 20.48 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.252 on 29 degrees of freedom
## Multiple R-squared: 0.9353, Adjusted R-squared: 0.9331
## F-statistic: 419.4 on 1 and 29 DF, p-value: < 2.2e-16
- The estimated residuals vary from -8.065 to 9.578 with median 0.152.
- The estimate of
Intercept is \(a=-36.9435\)
- The estimate of slope of
Girth is \(b=5.0659\)
- The estimate of the conditional standard deviation (called residual standard error in R) is \(s=4.252\) with \(31-2=29\) degrees of freedom.
Test for independence
- We consider the regression model \[y=\alpha+\beta x+\varepsilon\] where we use a sample to obtain estimates \((a,b)\) of \((\alpha,\beta)\), the estimate \(s\) of \(\sigma\) and the degrees of freedom \(df=n-2\).
- We are going to test \[H_0:\, \beta=0 \quad \mbox{ against } \quad H_a:\, \beta\not=0\]
- The null hypothesis specifies, that \(y\) doesn’t depend linearly on \(x\).
- Observed values of \(b\) far away from zero are critical for the null-hypothesis?
- It can be shown that \(b\) has standard error \[se_b=\frac{s}{\sqrt{\sum_{i=1}^n(x_i-\bar{x})^2}}\] with \(df=n-2\) degrees of freedom.
- So, we want to use the test statistic \[t_\text{obs}=\frac{b}{se_b}\] which has to be evaluated in a t-distribution with \(df\) degrees of freedom.
Example
- Recall the summary of our example:
##
## Call:
## lm(formula = Volume ~ Girth, data = trees)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.065 -3.107 0.152 3.495 9.587
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -36.9435 3.3651 -10.98 7.62e-12 ***
## Girth 5.0659 0.2474 20.48 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.252 on 29 degrees of freedom
## Multiple R-squared: 0.9353, Adjusted R-squared: 0.9331
## F-statistic: 419.4 on 1 and 29 DF, p-value: < 2.2e-16
- As we noted previously \(b=5.0659\) and \(s=4.252\) with \(df=29\) degrees of freedom.
- In the second column(
Std. Error) of the Coefficients table we find \(se_b=0.2474\).
- The observed t-score (test statistic) is then \[t_\text{obs}=\frac{b}{se_b}=\frac{5.0659}{0.2474}=20.48\] which also can be found in the third column(
t value).
- The corresponding p-value is found in the usual way by using the t-distribution with 29 degrees of freedom.
- In the fourth column(
Pr(>|t|)) we see that the p-value is less than \(2\times 10^{-16}\). This is no surprise since the t-score was way above 3.
Confidence interval for slope
When we have both the standard error and the reference distribution, we can construct a confidence interval in the usual way: \[b\pm t_{crit} se_b, \] where the t-score is determined by the confidence level and we find this value using qdist in R.
In our example we have 29 degrees of freedom and with a confidence level of \(95\%\) we get \(t_{crit} =\) qdist("t", 0.975, df = 29)= 2.045.
If you are lazy (like most statisticians are):
## 2.5 % 97.5 %
## (Intercept) -43.825953 -30.060965
## Girth 4.559914 5.571799
- i.e. \((4.56, 5.57)\) is a \(95\%\) confidence interval for the slope of
Girth.
Correlation
- The estimated slope \(b\) in a linear regression doesn’t say anything about the strength of association between \(y\) and \(x\).
Girth was measured in inches, but if we rather measured it in kilometers the slope is much larger: An increase of 1km in Girth yield an enormous increase in Volume.- Let \(s_y\) and \(s_x\) denote the sample standard deviation of \(y\) and \(x\), respectively.
- The corresponding t-scores \[y_t=\frac{y}{s_y} \quad \mbox{ and } \quad x_t=\frac{x}{s_x}\] are independent of the chosen measurement scale.
- The corresponding prediction equation is then \[\hat{y}_t=\frac{a}{s_y}+\frac{s_x}{s_y}b x_t\]
- i.e. the standardized regression coefficient (slope) is \[r=\frac{s_x}{s_y}b\] which also is called the (sample) correlation between \(y\) and \(x\).
- It can be shown that:
- \(-1\leq r\leq 1\)
- The absolute value of \(r\) measures the (linear) strength of dependence between \(y\) and \(x\).
- When \(r=1\) all the points are on the prediction line, which has positive slope.
- When \(r=-1\) all the points are on the prediction line, which has negative slope.
- To calculate the sample correlation in R:
## Girth Height Volume
## Girth 1.0000000 0.5192801 0.9671194
## Height 0.5192801 1.0000000 0.5982497
## Volume 0.9671194 0.5982497 1.0000000
- There is a strong positive correlation between
Volume and Girth (r=0.967).
- Note, calling
cor on a data.frame (like trees) only works when all columns are numeric. Otherwise the relevant numeric columns should be extracted like this:
cor(trees[,c("Height", "Girth", "Volume")])
which produces the same output as above.
- Alternatively, one can calculate the correlation between two variables of interest like:
cor(trees$Height, trees$Volume)
## [1] 0.5982497
R-squared: Reduction in prediction error
R-squared: Reduction in prediction error
- We want to compare two different models used to predict the response \(y\).
- Model 1: We do not use the knowledge of \(x\), and use \(\bar{y}\) to predict any \(y\)-measurement. The corresponding prediction error is defined as \[TSS=\sum_{i=1}^n(y_i-\bar{y})^2\] and is called the Total Sum of Squares.
- Model 2: We use the prediction equation \(\hat{y}=a+bx\) to predict \(y_i\). The corresponding prediction error is then the Sum of Squared Errors \[SSE=\sum_{i=1}^n(y_i-\hat{y}_i)^2.\]
- We then define \[ r^2=\frac{TSS-SSE}{TSS} \] which can be interpreted as the relative reduction in the prediction error, when we include \(x\) as explanatory variable.
- This is also called the fraction of explained variation, coefficient of determination or simply r-squared.
- For example if \(r^2 = 0.65\), the interpretation is that \(x\) explains about \(65\%\) of the variation in \(y\), whereas the rest is due to other sources of random variation.
Graphical illustration of sums of squares
## Warning: Using `size` aesthetic for lines was deprecated in ggplot2 3.4.0.
## ℹ Please use `linewidth` instead.
## This warning is displayed once every 8 hours.
## Call `lifecycle::last_lifecycle_warnings()` to see where this warning was generated.
## `geom_smooth()` using formula = 'y ~ x'
- Note the data points are the same in both plots. Only the prediction rule changes.
- The error of using Rule 1 is the total sum of squares \(E_1 = TSS = \sum_{i=1}^n(y_i-\bar{y})^2\).
- The error of using Rule 2 is the residual sum of squares (sum of squared errors) \(E_2 = SSE = \sum_{i=1}^n(y_i-\hat{y}_i)^2\).
\(r^2\): Reduction in prediction error
- For the simple linear regression we have that \[r^2=\frac{TSS-SSE}{TSS}\] is equal to the square of the correlation between \(y\) and \(x\), so it makes sense to denote it \(r^2\).
- Towards the bottom of the output below we can read off the value \(r^2=0.9353=93.53\%\), which is a large fraction of explained variation.
##
## Call:
## lm(formula = Volume ~ Girth, data = trees)
##
## Residuals:
## Min 1Q Median 3Q Max
## -8.065 -3.107 0.152 3.495 9.587
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) -36.9435 3.3651 -10.98 7.62e-12 ***
## Girth 5.0659 0.2474 20.48 < 2e-16 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 4.252 on 29 degrees of freedom
## Multiple R-squared: 0.9353, Adjusted R-squared: 0.9331
## F-statistic: 419.4 on 1 and 29 DF, p-value: < 2.2e-16
Introduction to multiple regression model
Multiple regression model
- We look at data set containing measurements from the 67 counties of Florida.
- Our focus is on
- The response \(y\):
Crime which is the crime rate
- The predictor \(x_1\):
Education which is proportion of the population with high school exam
- The predictor \(x_2\):
Urbanisation which is proportion of the population living in urban areas
Example
FL <- read.delim("https://asta.math.aau.dk/datasets?file=fl-crime.txt")
head(FL, n = 3)
## Crime Education Urbanisation
## 1 104 82.7 73.2
## 2 20 64.1 21.5
## 3 64 74.7 85.0
library(mosaic)
splom(FL) # Scatter PLOt Matrix
Correlations
- There is significant (p\(\approx7\times 10^{-5}\)) positive correlation (r=0.47) between
Crime and Education
- Then there is also significant positive correlation (r=0.68) between
Crime and Urbanisation
## Crime Education Urbanisation
## Crime 1.0000000 0.4669119 0.6773678
## Education 0.4669119 1.0000000 0.7907190
## Urbanisation 0.6773678 0.7907190 1.0000000
cor.test(~ Crime + Education, data = FL)
##
## Pearson's product-moment correlation
##
## data: Crime and Education
## t = 4.2569, df = 65, p-value = 6.806e-05
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.2553414 0.6358104
## sample estimates:
## cor
## 0.4669119
Several predictors
- Both
Education (\(x_1\)) and Urbanisation (\(x_2\)) are pretty good predictors for Crime (\(y\)).
- We therefore want to consider the model \[y=\alpha + \beta_1x_1 + \beta_2x_2 + \epsilon\]
- The errors \(\epsilon\) are random noise with mean zero and standard deviation \(\sigma\).
- The graph for the mean response is in other words a 2-dimensional plane in a 3-dimensional space.
- We determine estimates \((a,b_1,b_2)\) for \((\alpha,\beta_1,\beta_2)\) via the least squares method, i.e deviations from the plane.
Example
model <- lm(Crime ~ Education + Urbanisation, data = FL)
summary(model)
##
## Call:
## lm(formula = Crime ~ Education + Urbanisation, data = FL)
##
## Residuals:
## Min 1Q Median 3Q Max
## -34.693 -15.742 -6.226 15.812 50.678
##
## Coefficients:
## Estimate Std. Error t value Pr(>|t|)
## (Intercept) 59.1181 28.3653 2.084 0.0411 *
## Education -0.5834 0.4725 -1.235 0.2214
## Urbanisation 0.6825 0.1232 5.539 6.11e-07 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Residual standard error: 20.82 on 64 degrees of freedom
## Multiple R-squared: 0.4714, Adjusted R-squared: 0.4549
## F-statistic: 28.54 on 2 and 64 DF, p-value: 1.379e-09
- From the output we find the prediction equation \[ \widehat y = 59 - 0.58x_1 + 0.68x_2 \]
- Not exactly what we expected based on the correlation.
- Now there appears to be a negative association between \(y\) and \(x_1\) (Simpsons Paradox)!
- We can also find the standard error (0.4725) and the corresponding t-score (-1.235) for the the slope of
Education
- This yields a p-value of 22%, i.e. the slope is not significantly different from zero.
Simpsons paradox
- The example illustrates Simpson’s paradox.
- When considered alone
Education is a good predictor for Crime (with positive correlation).
- When we add
Urbanisation, then Education has a negative effect on Crime (but not significant).
- A possible explanation is illustrated by the graph above.
Urbanisation has positive effect on both Education and Crime.- For a given level of
urbanisation there is a (non-significant) negative association between Education and Crime.
- Viewed alone
Education is a good predictor for Crime. If Education has a large value, then this indicates a large value of Urbanisation and thereby a large value of Crime.