i am new to machine learning and i am using housing price dataset from kaggle.com to solve regression problem. i want to know the difference between Correlation Coefficient and Correlation Determination and why people use one over the other. for instance, i can see the relation between YearBuild and SalePrice like this
now, what is the use of Coefficient Determination, why is it used
if R= Coeffiecient Corellation
then Coefficient Determination = R x R
is the percentage view of the Corellation Coeffiecient?
is it the relation of an individual feature with the rest of the feature?
The coefficient R squared tells you how much of the variance the regression model explains. If it is equal to 0.01 for example, it means that you have explained one percent of the variance. This is useful to know for obvious reasons. Unlike the correlation coefficient, R squared is always positive so just tells you that there is (or is not) a linear relationship, but not what its form is.
Related
Trying to understand how the r-squared (and also explained variance) metrics can be negative (thus indicating non-existant forecasting power) when at the same time the correlation factor between prediction and truth (as well as slope in a linear-regression (regressing truth on prediction)) are positive
R Squared can be negative in a rare scenario.
R squared = 1 – (SSR/SST)
Here, SST stands for Sum of Squared Total which is nothing but how much does the predicted points get varies from the mean of the target variable. Mean is nothing but a regression line here.
SST = Sum (Square (Each data point- Mean of the target variable))
For example,
If we want to build a regression model to predict height of a student with weight as the independent variable then a possible prediction without much effort is to calculate the mean height of all current students and consider it as the prediction.
In the above diagram, red line is the regression line which is nothing but the mean of all heights. This mean calculated without much effort and can be considered as one of the worst method of prediction with poor accuracy. In the diagram itself we can see that the prediction is nowhere near to the original data points.
Now come to SSR,
SSR stands for Sum of Squared Residuals. This residual is calculated from the model which we build from our mathematical approach (Linear regression, Bayesian regression, Polynomial regression or any other approach). If we use a sophisticated approach rather than using a naive approach like mean then our accuracy will obviously increase.
SSR = Sum (Square (Each data point - Each corresponding data point in the regression line))
In the above diagram, let's consider that the blue line indicates a sophisticated model with large mathematical analysis. We can see that it has obviously higher accuracy than the red line.
Now come to the formula,
R Squared = 1- (SSR/SST)
Here,
SST will be large number because it a very poor model (red line).
SSR will be a small number because it is the best model we developed
after much mathematical analysis (blue line).
So, SSR/SST will be a very small number (It will become very small
whenever SSR decreases).
So, 1- (SSR/SST) will be large number.
So we can infer that whenever R Squared goes higher, it means the
model is too good.
This is a generic case but this cannot be applied in many cases where multiple independent variables are present. In the example, we had only one independent variable and one target variable but in real case, we will have 100's of independent variables for a single dependent variable. The actual problem is that, out of 100's of independent variables-
Some variables will have very high correlation with target variable.
Some variables will have very small correlation with target variable.
Also some independent variables will have no correlation at all.
So, RSquared is calculated on an assumption that the average line of the target which is perpendicular line of y axis is the worst fit a model can have at a maximum riskiest case. SST is the squared difference between this average line and original data points. Similarly, SSR is the squared difference between the predicted data points (by the model plane) and original data points.
SSR/SST gives a ratio how SSR is worst with respect to SST. If your model can somewhat build a plane which is a comparatively good than the worst, then in 99% cases SSR<SST. It eventually makes R squared as positive if you substitute it in the equation.
But what if SSR>SST ? This means that your regression plane is worse than the mean line (SST). In this case, R squared will be obviously negative. But it happens only at 1% of cases or smaller.
Answer was originally written in quora by me -
https://qr.ae/pNsLU8
https://qr.ae/pNsLUr
we choose SSE(sum of squared error) for deciding the best fit line instead of sum of residual or sum of absolute residual
The purpose is to allow linear algebra to directly solve for equation coefficients in regression. The other fitting targets you mention cannot be used in this way. Using derivative calculus, it was found that a fitting target of lowest sum of squared error allowed a direct, non-iterative solution to the problem of fitting experimental data to equations that are linear in their coefficients - such as standard polynomial equations.
James is right that the ability to formulate the estimates of regression coefficients as a form of linear algebra is one large advantage of the least squares estimate (minimizing SSE), but using the least squares estimate provides a few other useful properties.
With the least squares estimate you're minimizing the variance of the errors - which is often desired. This gives us the best linear unbiased estimator (BLUE) of the coefficients (given the Gauss–Markov assumptions are met). (Gauss-Markov assumptions and a proof showing why this formulation gives us the best linear unbiased estimates can be found here.)
With the least squares, you also end up with a unique solution (assuming you have more observations than estimated coefficients and no perfect multicollinearity).
As for using the sum of residual, this wouldn’t work well since this would be minimized by having all negative residuals.
But the sum of absolute residual is used in some linear models where you may want the estimates to be more robust to outliers and aren’t necessarily concerned with the variance of the residuals.
I am working on a simple AI program that classifies shapes using unsupervised learning method. Essentially I use the number of sides and angles between the sides and generate aggregates percentages to an ideal value of a shape. This helps me create some fuzzingness in the result.
The problem is how do I represent the degree of error or confidence in the classification? For example: a small rectangle that looks very much like a square would yield night membership values from the two categories but can I represent the degree of error?
Thanks
Your confidence is based on used model. For example, if you are simply applying some rules based on the number of angles (or sides), you have some multi dimensional representation of objects:
feature 0, feature 1, ..., feature m
Nice, statistical approach
You can define some kind of confidence intervals, baesd on your empirical results, eg. you can fit multi-dimensional gaussian distribution to your empirical observations of "rectangle objects", and once you get a new object you simply check the probability of such value in your gaussian distribution, and have your confidence (which would be quite well justified with assumption, that your "observation" errors have normal distribution).
Distance based, simple approach
Less statistical approach would be to directly take your model's decision factor and compress it to the [0,1] interaval. For example, if you simply measure distance from some perfect shape to your new object in some metric (which yields results in [0,inf)) you could map it using some sigmoid-like function, eg.
conf( object, perfect_shape ) = 1 - tanh( distance( object, perfect_shape ) )
Hyperbolic tangent will "squash" values to the [0,1] interval, and the only remaining thing to do would be to select some scaling factor (as it grows quite quickly)
Such approach would be less valid in the mathematical terms, but would be similar to the approach taken in neural networks.
Relative approach
And more probabilistic approach could be also defined using your distance metric. If you have distances to each of your "perfect shapes" you can calculate the probability of an object being classified as some class with assumption, that classification is being performed at random, with probiability proportional to the inverse of the distance to the perfect shape.
dist(object, perfect_shape1) = d_1
dist(object, perfect_shape2) = d_2
dist(object, perfect_shape3) = d_3
...
inv( d_i )
conf(object, class_i) = -------------------
sum_j inv( d_j )
where
inv( d_i ) = max( d_j ) - d_i
Conclusions
First two ideas can be also incorporated into the third one to make use of knowledge of all the classes. In your particular example, the third approach should result in confidence of around 0.5 for both rectangle and circle, while in the first example it would be something closer to 0.01 (depending on how many so small objects would you have in the "training" set), which shows the difference - first two approaches show your confidence in classifing as a particular shape itself, while the third one shows relative confidence (so it can be low iff it is high for some other class, while the first two can simply answer "no classification is confident")
Building slightly on what lejlot has put forward; my preference would be to use the Mahalanobis distance with some squashing function. The Mahalanobis distance M(V, p) allows you to measure the distance between a distribution V and a point p.
In your case, I would use "perfect" examples of each class to generate the distribution V and p is the classification you want the confidence of. You can then use something along the lines of the following to be your confidence interval.
1-tanh( M(V, p) )
I have derived and implemented an equation of an expected value.
To show that my code is free of errors i have employed the Monte-Carlo
computation a number of times to show that it converges into the same
value as the equation that i derived.
As I have the data now, how can i visualize this?
Is this even the correct test to do?
Can I give a measure how sure i am that the results are correct?
It's not clear what you mean by visualising the data, but here are some ideas.
If your Monte Carlo simulation is correct, then the Monte Carlo estimator for your quantity is just the mean of the samples. The variance of your estimator (how far away from the 'correct' value the average value will be) will scale inversely proportional to the number of samples you take: so long as you take enough, you'll get arbitrarily close to the correct answer. So, use a moderate (1000 should suffice if it's univariate) number of samples, and look at the average. If this doesn't agree with your theoretical expectation, then you have an error somewhere, in one of your estimates.
You can also use a histogram of your samples, again if they're one-dimensional. The distribution of samples in the histogram should match the theoretical distribution you're taking the expectation of.
If you know the variance in the same way as you know the expectation, you can also look at the sample variance (the mean squared difference between the sample and the expectation), and check that this matches as well.
EDIT: to put something more 'formal' in the answer!
if M(x) is your Monte Carlo estimator for E[X], then as n -> inf, abs(M(x) - E[X]) -> 0. The variance of M(x) is inversely proportional to n, but exactly what it is will depend on what M is an estimator for. You could construct a specific test for this based on the mean and variance of your samples to see that what you've done makes sense. Every 100 iterations, you could compute the mean of your samples, and take the difference between this and your theoretical E[X]. If this decreases, you're probably error free. If not, you have issues either in your theoretical estimate or your Monte Carlo estimator.
Why not just do a simple t-test? From your theoretical equation, you have the true mean mu_0 and your simulators mean,mu_1. Note that we can't calculate mu_1, we can only estimate it using the mean/average. So our hypotheses are:
H_0: mu_0 = mu_1 and H_1: mu_0 does not equal mu_1
The test statistic is the usual one-sample test statistic, i.e.
T = (mu_0 - x)/(s/sqrt(n))
where
mu_0 is the value from your equation
x is the average from your simulator
s is the standard deviation
n is the number of values used to calculate the mean.
In your case, n is going to be large, so this is equivalent to a Normal test. We reject H_0 when T is bigger/smaller than (-3, 3). This would be equivalent to a p-value < 0.01.
A couple of comments:
You can't "prove" that the means are equal.
You mentioned that you want to test a number of values. One possible solution is to implement a Bonferroni type correction. Basically, you reduce your p-value to: p-value/N where N is the number of tests you are running.
Make your sample size as large as possible. Since we don't have any idea about the variability in your Monte Carlo simulation it's impossible to say use n=....
The value of p-value < 0.01 when T is bigger/smaller than (-3, 3) just comes from the Normal distribution.
I want to use a linear regression model, but I want to use ordinary least squares, which I think it is a type of linear regression. The software I use is SPSS. It only has linear regression, partial least squares and 2-stages least squares. I have no idea which one is ordinary least squares (OLS).
Yes, although 'linear regression' refers to any approach to model the relationship between one or more variables, OLS is the method used to find the simple linear regression of a set of data.
Linear regression is a vast term that just says we are finding a relationship between the dependent and independent variable(s), no matter what technique we are using.
OLS is just one of the technique to do linear reg.
Lets say,
error(e) = (observed value - predicted value)
Observed values - blue dots in picture
predicted values - points on the line(vertically below to the observed values)
The vertical lines below represent 'e'. We square them -> add them and get total err. And we try to reduce this total error.
For OLS, as the name says (ordinary least squared method), here we reduce the sum of all e^2 i.e. we try to make the error least.