Correlation Analysis

Correlation Analysis using Pearson’s Correlation Coefficient, Spearman’s Rank correlation, Measure of the linear relationship between quantitative variables, strength, and direction of relationship, Coefficient of correlation

The Spearman Rank Correlation Test (Numerical Example)

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Consider the following data for the illustration of the detection of heteroscedasticity using the Spearman Rank correlation test. The Data file is available to download.

YX2X3
11208.1
16188.4
11228.5
14218.5
13278.8
17269
14258.9
15279.4
12309.5
18289.5

The estimated multiple linear regression model is:

$$Y_i = -34.936 -0.75X_{2i} + 7.611X_{3i}$$

The Residuals with the data table are:

YX2X3Residuals
11208.1-0.63302
16188.40.575564
11228.5-2.16954
14218.50.076455
13278.81.317102
172693.040825
14258.90.047951
15279.4-1.2497
12309.5-2.74881
18289.51.743171

We need to find the rank of absolute values of $u_i$ and the expected heteroscedastic variable $X_2$.

$Y$$X_2$$X_3$ResidualsRank of |$u_i$|Rank of $X_2$$d$$d^2$
11208.1-0.633 4224
16188.40.576 3124
11228.5-2.170 84416
14218.50.076 23-11
13278.81.317 67.5-1.52.25
172693.041 106416
14258.90.048 15-416
15279.4-1.250 57.5-2.56.25
12309.5-2.749 910-11
18289.51.743 79-24
       Total =070.5

Calculating the Spearman Rank correlation

\begin{align}
r_s&=1-\frac{6\sum d^2}{n(n-1)}\\
&=1-\frac{6\times 70.5)}{10(100-1)}=0.5727
\end{align}

Let us perform the statistical significance of $r_s$ by t-test

\begin{align}
t&=\frac{r_s \sqrt{n}}{\sqrt{1-r_s^2}}\\
&=\frac{0.5727\sqrt{8}}{\sqrt{1-(0.573)^2}}=1.977
\end{align}

The value of $t$ from the table at a 5% level of significance at 8 degrees of freedom is 2.306.

Since $t_{cal} \ngtr t_{tab}$, there is no evidence of the systematic relationship between the explanatory variables, $X_2$ and the absolute value of the residuals ($|u_i|$) and hence there is no evidence of heteroscedasticity.

Since there is more than one regressor (the example is from the multiple regression model), therefore, Spearman’s Rank Correlation test should be repeated for each of the explanatory variables.

Spearman Rank Correlation

As an assignment perform the Spearman Rank Correlation between |$u_i$| and $X_3$  for the data above. Test the statistical significance of the coefficient in the above manner to explore evidence about heteroscedasticity.

Read about Pearson’s Correlation Coefficient

R Language Interview Questions

Covariance and Correlation (2015)

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Introduction to Covariance and Correlation

Covariance and correlation are very important terminologies in statistics. Covariance measures the degree to which two variables co-vary (i.e. vary/change together). If the greater values of one variable (say, $X_i$) correspond with the greater values of the other variable (say, $X_j$), i.e. if the variables tend to show similar behavior, then the covariance between two variables ($X_i$, $X_j$) will be positive.

Similarly, if the smaller values of one variable correspond with the smaller values of the other variable, then the covariance between two variables will be positive. In contrast, if the greater values of one variable (say, $X_i$) mainly correspond to the smaller values of the other variables (say, $X_j$), i.e. both of the variables tend to show opposite behavior, then the covariance will be negative.

In other words, positive covariance between two variables means they (both of the variables) vary/change together in the same direction relative to their expected values (averages). It means that if one variable moves above its average value, the other variable tends to be above its average value.

Similarly, if covariance is negative between the two variables, then one variable tends to be above its expected value, while the other variable tends to be below its expected value. If covariance is zero then it means that there is no linear dependency between the two variables.

Mathematical Representation of Covariance

Mathematically covariance between two random variables $X_i$ and $X_j$ can be represented as
\[COV(X_i, X_j)=E[(X_i-\mu_i)(X_j-\mu_j)]\]
where
$\mu_i=E(X_i)$ is the average of the first variable
$\mu_j=E(X_j)$ is the average of the second variable

\begin{aligned}
COV(X_i, X_j)&=E[(X_i-\mu_i)(X_j-\mu_j)]\\
&=E[X_i X_j – X_i E(X_j)-X_j E(X_i)+E(X_i)E(X_j)]\\
&=E(X_i X_j)-E(X_i)E(X_j) – E(X_j)E(X_i)+E(X_i)E(X_j)\\
&=E(X_i X_j)-E(X_i)E(X_j)
\end{aligned}

Covariance

Note that, the covariance of a random variable with itself is the variance of the random variable, i.e. $COV(X_i, X_i)=VAR(X)$. If $X_i$ and $X_j$ are independent, then $E(X_i X_j)=E(X_i)E(X_j)$ and $COV(X_i, X_j)=E(X_i X_j)-E(X_i) E(X_j)=0$.

Covariance and Correlation

Covariance and Correlation

Correlation and covariance are related measures but not equivalent statistical measures.

Equation of Correlation (Normalized Covariance

The correlation between two variables (Let, $X_i$ and $X_j$) is their normalized covariance, defined as
\begin{aligned}
\rho_{i,j}&=\frac{E[(X_i-\mu_i)(X_j-\mu_j)]}{\sigma_i \sigma_j}\\
&=\frac{n \sum XY – \sum X \sum Y}{\sqrt{(n \sum X^2 -(\sum X)^2)(n \sum Y^2 – (\sum Y)^2)}}
\end{aligned}
where $\sigma_i$ is the standard deviation of $X_i$ and $\sigma_j$ is the standard deviation of $X_j$.

Note that correlation is dimensionless, i.e. a number that is free of the measurement unit and its values lie between -1 and +1 inclusive. In contrast, covariance has a unit of measure–the product of the units of two variables.

For further reading about Correlation follow these posts

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Correlation Coefficient Range (2012)

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We know that the ratio of the explained variation to the total variation is called the coefficient of determination which is the square of the Correlation Coefficient Range lies between $-1$ and $+1$. This ratio (coefficient of determination) is non-negative, therefore denoted by $r^2$, thus

\begin{align*}
r^2&=\frac{\text{Explained Variation}}{\text{Total Variation}}\\
&=\frac{\sum (\hat{Y}-\overline{Y})^2}{\sum (Y-\overline{Y})^2}
\end{align*}

It can be seen that if the total variation is all explained, the ratio $r^2$ (Coefficient of Determination) is one and if the total variation is all unexplained then the explained variation and the ratio $r^2$ are zero.

The square root of the coefficient of determination is called the correlation coefficient, given by

\begin{align*}
r&=\sqrt{ \frac{\text{Explained Variation}}{\text{Total Variation}} }\\
&=\pm \sqrt{\frac{\sum (\hat{Y}-\overline{Y})^2}{\sum (Y-\overline{Y})^2}}
\end{align*}

and

\[\sum (\hat{Y}-\overline{Y})^2=\sum(Y-\overline{Y})^2-\sum (Y-\hat{Y})^2\]

Therefore

\begin{align*}
r&=\sqrt{ \frac{\sum(Y-\overline{Y})^2-\sum (Y-\hat{Y})^2} {\sum(Y-\overline{Y})^2} }\\
&=\sqrt{1-\frac{\sum (Y-\hat{Y})^2}{\sum(Y-\overline{Y})^2}}\\
&=\sqrt{1-\frac{\text{Unexplained Variation}}{\text{Total Variation}}}=\sqrt{1-\frac{S_{y.x}^2}{s_y^2}}
\end{align*}

where $s_{y.x}^2=\frac{1}{n} \sum (Y-\hat{Y})^2$ and $s_y^2=\frac{1}{n} \sum (Y-\overline{Y})^2$

\begin{align*}
\Rightarrow r^2&=1-\frac{s_{y.x}^2}{s_y^2}\\
\Rightarrow s_{y.x}^2&=s_y^2(1-r^2)
\end{align*}

Since variances are non-negative

\[\frac{s_{y.x}^2}{s_y^2}=1-r^2 \geq 0\]

Solving for inequality we have

\begin{align*}
1-r^2 & \geq 0\\
\Rightarrow r^2 \leq 1\, \text{or}\, |r| &\leq 1\\
\Rightarrow & -1 \leq r\leq 1
\end{align*}

Therefore, the Correlation Coefficient Range lies between $-1$ and $+1$ inclusive.

Correlation Coefficient Range

Alternative Proof: Correlation Coefficient Range

Since $\rho(X,Y)=\rho(X^*,Y^*)$ where $X^*=\frac{X-\mu_X}{\sigma_X}$ and $Y^*=\frac{Y-Y^*}{\sigma_Y}$

and as covariance is bi-linear and $X^*, Y^*$ have zero mean and variance 1, therefore

\begin{align*}
\rho(X^*,Y^*)&=Cov(X^*,Y^*)=Cov\{\frac{X-\mu_X}{\sigma_X},\frac{Y-\mu_Y}{\sigma_Y}\}\\
&=\frac{Cov(X-\mu_X,Y-\mu_Y)}{\sigma_X\sigma_Y}\\
&=\frac{Cov(X,Y)}{\sigma_X \sigma_Y}=\rho(X,Y)
\end{align*}

We also know that the variance of any random variable is $\ge 0$, it could be zero i.e. $(Var(X)=0)$ if and only if $X$ is a constant (almost surely), therefore

\[V(X^* \pm Y^*)=V(X^*)+V(Y^*)\pm2Cov(X^*,Y^*)\]

As $Var(X^*)=1$ and $Var(Y^*)=1$, the above equation would be negative if $Cov(X^*,Y^*)$ is either greater than 1 or less than -1. Hence \[1\geq \rho(X,Y)=\rho(X^*,Y^*)\geq -1\].

If $\rho(X,Y )=Cov(X^*,Y^*)=1$ then $Var(X^*- Y ^*)=0$ making $X^* = Y^*$ almost surely. Similarly, if $\rho(X,Y )=Cov(X^*,Y^*)=-1$ then $X^* = – Y^*$ almost surely. In either case, $Y$ would be a linear function of $X$ almost surely.

For proof of Cauchy-Schwarz Inequality please follow the link

We can see that the Correlation Coefficient range lies between $-1$ and $+1$.

Coefficient of Correlation Range

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