Was bedeutet negatives R-Quadrat?

Angenommen, ich habe einige Daten, und dann passe ich die Daten einem Modell an (einer nichtlinearen Regression). Dann berechne ich das R-Quadrat ( $R^2$ ).

Wenn R-Quadrat negativ ist, was bedeutet das? Heißt das, mein Modell ist schlecht? Ich kenne die Reichweite von$R^2$ [-1,1] sein kann. Wenn$R^2$ 0 ist, was bedeutet das auch?

kann negativ sein, es bedeutet nur, dass:$R^2$

1. Das Modell passt sehr schlecht zu Ihren Daten
2. Sie haben keinen Intercept gesetzt

Für die Leute, die sagen, dass zwischen 0 und 1 liegt, ist dies nicht der Fall. Während ein negativer Wert für etwas mit dem Wort "squared" (Quadrat) als Verstoß gegen die mathematischen Regeln empfunden werden kann, kann dies in einem R 2 -Modell ohne Intercept geschehen . Um zu verstehen, warum, müssen wir uns ansehen, wie R 2 berechnet wird.$R^2$$R^2$$R^2$

Dies ist etwas lang - Wenn Sie die Antwort ohne Verständnis haben möchten, fahren Sie mit dem Ende fort. Ansonsten habe ich versucht, dies in einfachen Worten zu schreiben.

Lassen Sie uns zunächst definieren 3 Variablen: , T S S und E S S .$RSS$$TSS$$ESS$

Berechnung von RSS :

Für jede unabhängige Variable haben wir die abhängige Variable y . Wir zeichnen eine lineare Linie der besten Anpassung, die den Wert von y für jeden Wert von x vorhersagt . Nennen wir die Werte von y der Linie prognostiziert y . Der Fehler zwischen der Vorhersage Ihrer Linie und dem tatsächlichen y- Wert kann durch Subtraktion berechnet werden. Alle diese Differenzen werden quadriert und addiert, was die Restsumme der Quadrate R S S ergibt .$x$$y$$y$$x$$y$$\hat y$$y$$RSS$

Putting , das in eine Gleichung, $RSS = \sum (y - \hat y)^2$

Berechnung der TSS :

Wir können den Durchschnittswert von berechnen , der ˉ y heißt . Wenn wir ˉ y zeichnen , ist dies nur eine horizontale Linie durch die Daten, da sie konstant ist. Was wir damit machen können, ist ˉ y (der Durchschnittswert von y) zu subtrahieren$y$$\bar y$$\bar y$$\bar y$$y$ ) von jedem tatsächlichen Wert von . Das Ergebnis wird quadriert und addiert, was die Gesamtsumme der Quadrate T S S ergibt .$y$$TSS$

Einfügen in eine Gleichung $TSS = \sum (y - \bar y)^2$

Berechnung des ESS :

Die Unterschiede zwischen y (die Werte von y durch die Linie vorhergesagt) und der durchschnittliche Wert ˉ y werden quadriert und addiert. Dies ist die Erklärung der Summe der Quadrate, die gleich Σ ( y - ˉ y ) 2$\hat y$$y$$\bar y$$\sum (\hat y - \bar y)^2$

Denken Sie daran, , aber wir können eine hinzufügen + y - y hinein, weil sie sich aufhebt. Daher T S S = Σ ( y - y + y - ˉ y ) 2 . Ausbau dieser Klammern, erhalten wir T S S = Σ ( y - y ) 2 $TSS = \sum (y - \bar y)^2$$+ \hat y - \hat y$$TSS = \sum (y - \hat y + \hat y -\bar y)^2$$TSS = \sum (y - \hat y)^2 + 2* \sum (y - \hat y)(\hat y - \bar y) + \sum (\hat y - \bar y)^2$

$2* \sum (y - \hat y)(\hat y - \bar y) = 0$$TSS = \sum (y - \hat y)^2 + \sum (\hat y - \bar y)^2$, which you may notice just means that $TSS = RSS + ESS$$TSS$$1 - \frac {RSS}{TSS} = \frac {ESS}{TSS}$.

Here's the important part:

$R^2$ is defined as how much of the variance is explained by your model (how good your model is). In equation form, that's $R^2 = 1 - \frac {RSS}{TSS}$. Look familiar? When the line is plotted with an intercept, we can substitute this as $R^2 = \frac {ESS}{TSS}$. Since both the numerator and demoninator are sums of squares, $R^2$ must be positive.

BUT

When we don't specify an intercept, $2* \sum (y - \hat y)(\hat y - \bar y)$ does not necessarily equal $0$. This means that $TSS = RSS + ESS + 2* \sum (y - \hat y)(\hat y - \bar y)$.

Dividing all terms by $TSS$, we get $1 - \frac{RSS}{TSS} = \frac {ESS + 2* \sum (y - \hat y)(\hat y - \bar y)}{TSS}$.

Finally, we substitute to get $R^2 = \frac {ESS + 2* \sum (y - \hat y)(\hat y - \bar y)}{TSS}$. This time, the numerator has a term in it which is not a sum of squares, so it can be negative. This would make $R^2$ negative. When would this happen? $2* \sum (y - \hat y)(\hat y - \bar y)$ would be negative when $y - \hat y$ is negative and $\hat y - \bar y$ is positive, or vice versa. This occurs when the horizontal line of $\bar y$ actually explains the data better than the line of best fit.

Here's an exaggerated example of when $R^2$ is negative (Source: University of Houston Clear Lake)

Put simply:

• When $R^2 < 0$, a horizontal line explains the data better than your model.

You also asked about $R^2 = 0$.

• When $R^2 = 0$, a horizontal line explains the data equally as well as your model.

I commend you for making it through that. If you found this helpful, you should also upvote fcop's answer here which I had to refer to, because it's been a while.

Neither answer so far is entirely correct, so I will try to give my understanding of R-Squared. I have given a more detailed explanation of this on my blog post here "What is R-Squared"

Sum Squared Error

The objective of ordinary least squared regression is to get a line which minimized the sum squared error. The default line with minimum sum squared error is a horizontal line through the mean. Basically, if you can't do better, you can just predict the mean value and that will give you the minimum sum squared error

R-Squared is a way of measuring how much better than the mean line you have done based on summed squared error. The equation for R-Squared is

Now SS Regression and SS Total are both sums of squared terms. Both of those are always positive. This means we are taking 1, and subtracting a positive value. So the maximum R-Squared value is positive 1, but the minimum is negative infinity. Yes, that is correct, the range of R-squared is between -infinity and 1, not -1 and 1 and not 0 and 1

What Is Sum Squared Error

Sum squared error is taking the error at every point, squaring it, and adding all the squares. For total error, it uses the horizontal line through the mean, because that gives the lowest sum squared error if you don't have any other information, i.e. can't do a regression.

As an equation it is this

Now with regression, our objective is to do better than the mean. For instance this regression line will give a lower sum squared error than using the horizontal line.

The equation for regression sum squared error is this

Ideally, you would have zero regression error, i.e. your regression line would perfectly match the data. In that case you would get an R-Squared value of 1

Negative R Squared

All the information above is pretty standard. Now what about negative R-Squared ?

Well it turns out that there is not reason that your regression equation must give lower sum squared error than the mean value. It is generally thought that if you can't make a better prediction than the mean value, you would just use the mean value, but there is nothing forcing that to be the cause. You could for instance predict the median for everything.

In actual practice, with ordinary least squared regression, the most common time to get a negative R-Squared value is when you force a point that the regression line must go through. This is typically done by setting the intercept, but you can force the regression line through any point.

When you do that the regression line goes through that point, and attempts to get the minimum sum squared error while still going through that point.

By default, the regression equations use average x and average y as the point that the regression line goes through. But if you force it through a point that is far away from where the regression line would normally be you can get sum squared error that is higher than using the horizontal line

In the image below, both regression lines were forced to have a y intercept of 0. This caused a negative R-squared for the data that is far offset from the origin.

For the top set of points, the red ones, the regression line is the best possible regression line that also passes through the origin. It just happens that that regression line is worse than using a horizontal line, and hence gives a negative R-Squared.

Undefined R-Squared

There is one special case no one mentioned, where you can get an undefined R-Squared. That is if your data is completely horizontal, then your total sum squared error is zero. As a result you would have a zero divided by zero in the R-squared equation, which is undefined.

As the previous commenter notes, r^2 is between [0,1], not [-1,+1], so it is impossible to be negative. You cannot square a value and get a negative number. Perhaps you are looking at r, the correlation? It can be between [-1,+1], where zero means there is no relationship between the variables, -1 means there is a perfect negative relationship (as one variable increases, the other decreases), and +1 is a perfect positive relationship (both variables go up or down concordantly).

If indeed you are looking at r^2, then, as the previous commenter describes, you are probably seeing the adjusted r^2, not the actual r^2. Consider what the statistic means: I teach behavioral science statistics, and the easiest way that I've learned to teach my students about the meaning of r^2 is " % variance explained." So if you have r^2=0.5, the model explains 50% of the variation of the dependent (outcome) variable. If you have a negative r^2, it would mean that the model explains a negative % of the outcome variable, which is not an intuitively reasonable suggestion. However, adjusted r^2 takes the sample size (n) and number of predictors (p) into consideration. A formula for calculating it is here. If you have a very low r^2, then it is reasonably easy to get negative values. Granted, a negative adjusted r^2 does not have any more intuitive meaning than regular r^2, but as the previous commenter says, it just means your model is very poor, if not just plain useless.