# what do we actually study in machine learning?

I just started learning machine learning and learned a few basic algorithms and there is one stupid doubt in my mind and I am unable to find the answer of it. What do we actually study in machine learning algorithms that solves kinds of problems based on the data given to the algorithms?

Take an example of linear regression:

1. We give data.

2. We tell machine how to calculate error.

3. Using gradient descent also we are telling machine how to optimize and based on all above some optimized parameters come out.

Why there is actually a "learning" word in this algorithm? Although we are the telling how to perform steps and we have provided all the mathematics in this algorithm.

Let me try to explain you with an example,

Consider we have a dataset with features X1, X2 and Y is the target value. Now in Machine Learning our traget is to get an equation which should calculate the value as close as target value.

Consider a simple equation,

f(X) = w1X1 + w2X2 + b

and here our target is to identify best value of weights i.e. w1 and w2 such that f(X) ~Y.

Generally, initially f(X) gives you a very different value(as weights are not tuned), but gradually with each iteration(epoch), f(x) becomes better(weights getting tuned), that is called "parameter Learning".

• If you solve this using OLS: (X‘X)^-1 X‘y, it is a „one shot“ solution. The result will be the same as with gradient decent, but with an analytical solution. So your definition of learning does not hold here. It would however hold for Boosting or NN, so called „deep learning“. But since the question refers to linear regression, I find the answer not very striking. – Peter Jul 22 '19 at 18:31
• If we have different sets of X values, I dont think we have one shot solution. Thats the reason we have to trained ML model weights and optimize the cost function. I'm not sure where I'm incorrect. Appreciate if you share the example or good reference. – vipin bansal Jul 23 '19 at 4:22
• X is a matrix. This is the basic OLS (ordinary least squares) estimator. – Peter Jul 23 '19 at 5:59

I want to point out one aspect which is not adequately reflected in the previous answers. When you start with linear regression, you basically I) propose some model (a way how to represent the relation between $$y$$ and $$X$$), e.g. a linear function, and II) you specify a loss function (the sum of squared residuals). The aim is to find coefficients (or weights), which map $$X$$ to $$y$$ in a way so that there is no “better way” to predict $$y$$ (viz. minimise the sum of square residuals).

In the case of linear regression, you have two options: solve the problem analytically or by gradient decent. Optimal coefficients/weights (vector beta) are given by:

$$(X'X)^{-1} X'y=\beta.$$

So there is no need for gradient decent in this case. It is just simple matrix algebra.

Here is a minimal example in R, demonstrating how it works:

# Input some data
x0 <- c(1,1,1,1,1)
x1 <- c(1,2,3,4,5)
x2 <- c(8,4,3,1,8)
x <- as.matrix(cbind(x0,x1,x2))
y <- as.matrix(c(3,7,5,11,14))

# (X'X)^-1 X'y
beta1 = solve(t(x)%*%x) %*% t(x)%*%y

# R's regression command
beta2 = summary(lm(y ~ x[, 2:3]))

m <- nrow(y)
grad <- function(x, y, theta) {
gradient <- (1/m)* (t(x) %*% ((x %*% t(theta)) - y))
}

# Define gradient descent update algorithm
theta <- matrix(c(0, 0, 0), nrow=1) # Initialize the parameters

alpha = .05 # Learning rate
for (i in 1:maxit) {
theta <- theta - alpha  * grad(x, y, theta)
}
return(theta)
}

# Results
print(beta1)
print(beta2)

Results:

x0     x1        x2
[1,] -0.672238 2.6459 0.1530299
> print(beta1)
[,1]
x0 -0.6722955
x1  2.6459103
x2  0.1530343
> print(beta2)

Call:
lm(formula = y ~ x[, 2:3])

Residuals:
1       2       3       4       5
-0.1979  1.7683 -2.7245  0.9356  0.2185

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)  -0.6723     3.3565  -0.200   0.8598
x[, 2:3]x1    2.6459     0.7677   3.447   0.0748 .
x[, 2:3]x2    0.1530     0.3897   0.393   0.7325
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 2.399 on 2 degrees of freedom
Multiple R-squared:  0.8561,    Adjusted R-squared:  0.7122
F-statistic: 5.949 on 2 and 2 DF,  p-value: 0.1439

So as you can see, there is not much of a machine involved in solving the problem (apart of doing the math of course). So why machine learning? Increasing computational power has made it possible to solve very complex problems using gradient decent. The idea is that you depart from simple solutions and look for better, more complex ways to map $$X$$ to $$y$$.

You can think about changing the base procedure, e.g. use decision trees instead of a linear function. Since one tree may be only a weak learner, you can combine (or ensemble) trees to a random forest. Or you go even further and use boosting, in which you repeatedly use information contained in the residual to improve your fit. Usually very small trees are used to do this, but you can boost on linear regression as well.

Here is a little example for L2 boosting:

library(glmnet)

# Simple boosting example from:
# P. Bühlmann, T. Hothorn (2007), "Boosting Algorithms: Regularization, Prediction and Model Fitting", Statistical Science 22(4), p. 477-505.
# L2-Boosting from Section 3.3 (p. 483)

#######################
# Data
x = matrix(c(130,165,150,150,140,198,307,350,318,304,302,429),nrow=6,ncol=2,byrow = TRUE)
y = c(18,15,18,16,17,15)

# Parameter
# This works like a learning rate
nu = 0.1
# Early stopping (SSR value to stop)
es = 7.58765583743327
# Max. iterations (of boosting)
maxiter = 10000

#######################
# OLS benchmark model
olsdata = data.frame(cbind(y,x))
ols = lm(y~.,data=olsdata)
olspreds = predict(ols,newdata=olsdata)
olsssr = sum((y-olspreds)^2)
print(paste0("OLS SSE: ",olsssr))

#######################
# Boosting

# Initialize f0
f0 = mean(y)
f0ssr = sum((y-f0)^2)

# Lists to store results per boosting iteration
ssrlist = list()
bstep = list()

# Boosting (p. 483, Sec. 3.3, L2-Boosting)
for (n in seq(1:maxiter)){
# I) Get residual
if(n==1){f=f0} # initialize in first step
u=y-f          # get residual from estimation
# II) Fit residual / predict, I use ridge (alpha=0)
reg = glmnet(x,u,alpha=0, lambda=2)
# III) Update f
f=f+nu*g
# Print feedback
cat(paste0("Step: ", n," SSR: ", sum(u^2), "\n" ))
# Save SSR/iter to list
ssrlist[[n]] <- sum(u^2)
bstep[[n]] <- n
# Early stopping rule
if(sum(u^2)<es){break}
}

#######################
# End statement
cat("\n ~~~ RESULTS ~~~ \n")
cat(paste0("OLS SSR:           ", olsssr, "\n"))
cat(paste0("Initialised SSR:   ", f0ssr, "\n"))
cat(paste0("Last SSR boosting: ", ssrlist[length(ssrlist)], " [after ",bstep[length(bstep)], " iterations] \n"))

The problem here is that you do not only need to solve one regression, but 10000 (in the example here). In real world applications, the amount of math that has to be done is immense (counts for boosting and neural nets as well).

So what is machine learning? You want to let “the machine learn things” (recognise patterns in data, recognise images, make prediction based on that etc) and you will likely be unable to do this learning task on your own using pen and paper. Improved computing capacity over time has enabled us to solve very complex problems. And because of this cheap resource, new tools and methods develop. You can do things today on your laptop which literally seemed impossible in the late 1990s.

Of course you need to tell “the machine” how to learn. This is a lot of statistics but also computer science. Neural nets for example have mostly been advanced by computer scientists. Ultimately, the machine is merely a tool, but a very useful one.

Machine learning is a very vast field and everybody will have a different answer to your question. Here is mine: In machine learning, the word "learning" refers to the process of iteratively improving your model by gradient descent (e.g. neural networks), or another algorithm (EM algorithm, K-Mean clustering).

For example, let's say you want to recognize images of dog/cat. During your gradient descent, you will tune your model coefficients to decrease your loss which should lead to a better image classifications. In other words, you learn to differentiate dogs from cat.

I guess your question is: within background preference 'the difference between Machine Learning and Statistical modeling', what's the meaning of 'learning'.

Seems like it is still necessary that everything should be clearly defined by us. But the word 'learning' here is not about subjective initiative.

System preference is very important for the answer of your question as like the algorithm or strategy of a specify task.

In this context, learning means that the algorithm has potential to accelerate its performance continually based on the same dataset and past experiences (without any change of algorithm, and forget about the capacity upper limit relating to parameters), which is the difference between Machine Learning and traditional Statistical modeling algorithm.