We get started with our exercise by collecting data. The data is generated by a function unknown to you. This precondition makes this example more realistic, as in real applications we do not know the exact specifications of the underlying data generation process either. At the end of this section we will uncover the secret of the data generation process.

This is the data, our observations, in tabular form. We have 25 data points, each data point is a \((x,y)\) pair.

`poly_data`

```
## x y
## 1 0.28757752 1.16443996
## 2 0.78830514 -0.88780200
## 3 0.40897692 0.17413059
## 4 0.88301740 -0.21743732
## 5 0.94046728 -0.07645417
## 6 0.04555650 0.26285704
## 7 0.52810549 -0.45020937
## 8 0.89241904 -0.88236568
## 9 0.55143501 -0.39313237
## 10 0.45661474 0.15201462
## 11 0.95683335 -0.64790568
## 12 0.45333416 0.25912870
## 13 0.67757064 -0.52350905
## 14 0.57263340 -0.49157948
## 15 0.10292468 0.19461123
## 16 0.89982497 -0.87515511
## 17 0.24608773 1.23942552
## 18 0.04205953 0.14918285
## 19 0.32792072 0.42349275
## 20 0.95450365 -0.49184749
## 21 0.88953932 -0.68494536
## 22 0.69280341 -0.62575853
## 23 0.64050681 -0.82552774
## 24 0.99426978 0.07943064
## 25 0.65570580 -1.95912808
```

Here is the data in form of a scatter plot:

`plot(poly_data$x, poly_data$y)`

The statistical software R provides powerful functionality to fit a
polynomial to data. One of these functions is the `lm()`

function, which we already know from simple linear regression. However,
in order to fit a \(k^{th}\)-order
polynomial we need to add additional arguments to the function call. In
addition, there are two different options of coding a polynomial
regression.

For a 3^{rd}-order polynomial the first option is to type
`lm(response ~ poly(predictor, 3))`

and the second option is
`lm(response ~ predictor + I(predictor^2) + I(predictor^3)))`

.
We need the `I(...)`

notation to prevent any special
interpretation of operators in the model formula.

The difference between those two options is, that in the first case
the `poly()`

function produces orthogonal polynomials, which have the advantage
that the inspection of the regression coefficient allows to assess which
polynomial significantly improved the regression. However, in this case
the numerical values for \(\beta_i\)
may no longer be plugged into the polynomial regression equation from
above. In contrast, if we apply the second option, the resulting
regression coefficients may be plugged into the regression equation and
yield the correct numerical result. However, as the predictors are
correlated (problem of multicollinearity) it is often difficult
to identify which polynomial significantly improved the regression.

Note:The prediction itself will not be affected at all by the choice of notation.

To alleviate the confusion we will calculate an example in R. For
this we construct two 2^{nd}-order polynomials for
`poly_data`

:

```
m1 <- lm(y ~ poly(x, 2), data = poly_data)
m2 <- lm(y ~ x + I(x^2), data = poly_data)
```

We review the output of both models:

`m1`

```
##
## Call:
## lm(formula = y ~ poly(x, 2), data = poly_data)
##
## Coefficients:
## (Intercept) poly(x, 2)1 poly(x, 2)2
## -0.2374 -1.9608 0.5614
```

`m2`

```
##
## Call:
## lm(formula = y ~ x + I(x^2), data = poly_data)
##
## Coefficients:
## (Intercept) x I(x^2)
## 0.8325 -2.8237 1.3861
```

As noted above, the regression coefficients differ. Let us predict
\(\hat y\) for \(x=0.5\) by applying the
`predict()`

function:

`predict(m1, newdata = list("x" = 0.5))`

```
## 1
## -0.2328107
```

`predict(m2, newdata = list("x" = 0.5))`

```
## 1
## -0.2328107
```

As expected, both models predict \(\hat y = -0.2328107\) for \(x=0.5\).

So now that we know the notation in R we start to build 6 different models, with \(k = 1,2,3,5,9,14\). For each model we calculate the RMSE. Finally, we plot the data together with the regression line of each particular model. For convenience we construct a loop to reduce the amount of coding.

```
# plotting setup
par(mfrow = c(3, 2), mar = c(2, 2, 1, 1)) # set up 6 subplots
# setup
RMSE <- data.frame("kth_order" = NA, "RMSE" = NA) # empty data frame to store RMSE
vals <- list("x" <- seq(min(poly_data$x), max(poly_data$y), by = 0.01)) # set up vector used for prediction
# run loop
k <- c(1, 2, 3, 5, 9, 14) # k-th order
for (i in 1:length(k)) {
# build models
model <- lm(y ~ poly(x, k[i]), data = poly_data)
# calculate RMSE and store it for further usage
RMSE[i, 1] <- k[i] # store k-th order
RMSE[i, 2] <- sqrt(sum((fitted(model) - poly_data$y)^2) / length(poly_data$y)) # calculate RMSE
# predict
predictions <- predict(model, newdata = vals)
# plot
plot(poly_data$x, poly_data$y, pch = 16, col = "blue",
ylim = c(min(poly_data$y) * 1.3, max(poly_data$y) * 1.3))
lines(vals[[1]], predictions, lwd = 2, col = "red")
text(x = 0.8, y = 0.95, paste0("k = ", k[i], ", RMSE = ", round(RMSE[i, 2], 3))) # annotate the plot
}
```

Awesome, pretty plots! The figure shows, that if we increase \(k\), the order of the polynomial, the curve becomes more flexible and it fits the data better and better. The better the data is fitted, the lower becomes the error (RMSE).

In order to visualize that behavior we plot the RMSE against \(k\):

```
plot(RMSE[, 1], RMSE[, 2],
xlab = "k-th order",
ylab = "RMSE",
type = "b",
col = "blue",
pch = 16
)
```

Hence, once again the question arises: which is the best polynomial
to fit the data? Do we believe that the 14^{th}-order polynomial
fits the underlying data generation process best? Even though we obtain
an excellent fit to the observation data by increasing the order of the
polynomial, it remains questionable if the high order polynomial
generalizes well. Imagine we conduct a new measurement campaign and
receive new data based on the same data generation process. Do you
believe the wildly oscillating curve of a high order polynomial will
still fit the data well? No, probably not!

This problem is known as **overfitting**. Recall, the goal is to
learn the model parameters from the data. Thus, we are interested in
achieving a good generalization of the model and not necessarily
perfectly fitted observation data.

So, how do we solve the problem? How do we determine the
*best* \(k^{th}\)-order
polynomial for our data set? Well, there are many methods and strategies
to counteract overfitting. In this section we follow a simple approach.
First, we split the data set into two parts. We call one part the
**training set** and the other part the **validation
set**. Then we use all the data in the training set to learn the
model parameters \(\beta_i\), in the
same fashion as we did above. Thereafter, we apply the trained model to
predict the data of the validation set and evaluate the model
performance by computing the RMSE. Thus, we use the validation set to
optimize the modelâ€™s complexity, given by \(k\).

Unfortunately, if we want to learn from data we ultimately need data to learn from. So far we worked with 25 observations. That is not much. In real life applications we would probably have to obtain new observations by conducting a new measurement campaign. In our exercise, however, we may generate more data fairly easily. Thus, we continue this example with a new data set of 150 observations.

Let us plot the data!

`plot(new_poly_data$x, new_poly_data$y)`

Now, we are ready to build our training and validation set. For this
we use the `sample.split()`

function from the
`caTools`

package. We split the data in such a way, that 65 %
of the data is assigned to the training set and the rest, 35 % of the
data, is assigned to the validation set.

```
library(caTools)
set.seed(100) # set seed for reproducibility
# define split vector
split <- sample.split(new_poly_data$y, SplitRatio = 0.65)
# split data set
train_set <- new_poly_data[split == TRUE, ]
val_set <- new_poly_data[split == FALSE, ]
# check dimensions of the training and validation set
dim(train_set)
```

`## [1] 97 2`

`dim(val_set)`

`## [1] 53 2`

In this exercise we are going to build 14 models with \(k = 1,2,...,14\). We evaluate each of the 14 polynomial regression models by calculating the RMSE on the training set. Further, after building the model and obtaining the model parameters \(\beta_i\), we use the model to predict the response variable in the validation set. Again, we rely on the RMSE in order to evaluate the predictions based on the validation set. Finally, we plot the RMSE of each model for both, the training set and the validation set. Based on the RMSE we assess the generalization of the model.

We already developed a nice piece of code above. Thus, we only modify and extend the existing code:

```
# setup
RMSE <- data.frame("kth_order" = NA, "RMSE_train" = NA, "RMSE_val" = NA) # empty data frame to store RMSE
vals <- list("x" <- seq(min(train_set$x), max(train_set$y), by = 0.01)) # set up vector used for prediction
# run loop
k <- seq(1, 14) # k-th order
for (i in k) {
# build models
model <- lm(y ~ poly(x, k[i]), data = train_set)
# calculate RMSE and store it for further usage
RMSE[i, 1] <- k[i] # store k-th order
RMSE[i, 2] <- sqrt(sum((fitted(model) - train_set$y)^2) / length(train_set$y)) # calculate RMSE of the training set
# predict
predictions <- predict(model, newdata = val_set)
RMSE[i, 3] <- sqrt(sum((predictions - val_set$y)^2) / length(val_set$y)) # calculate RMSE of the validation set
}
# plot RMSE for training and validation set
plot(RMSE[, 1], RMSE[, 2],
xlab = "k-th order",
ylab = "RMSE",
ylim = c(min(RMSE[, c(2, 3)]), max(RMSE[, c(2, 3)])),
type = "b",
col = "blue",
pch = 16
)
lines(RMSE[, 3],
type = "b",
col = "red",
pch = 16
)
legend("topright",
legend = c("training set", "validation set"),
lty = c(1, 1),
col = c("blue", "red")
)
```

The figure shows, that the error on the training data (blue line) is
constantly decreasing. This makes perfect sense, the more complex the
model becomes by increasing \(k\), the
better the model fits the training data. We observed the same behavior
in the section above, when we trained our model with just 25
observations. If we take a look at the RMSE for the validation set (red
line), we see that with increasing \(k\), and thus increasing model complexity,
the error decreases. However, there is a sweet spot, indicated by the
lowest RMSE, where the model is just complex enough to generalize well
on the so far unseen validation data. If the model complexity increases
further, the RMSE is starting to increase, too. This indicates that the
model is overfitting. Thus, the model memorizes the data of the training
set well, however, the model becomes less predictive for so far unseen
data, such as the data of the validation set. Take a look at the figure
above to see that the lowest error on the validation set is obtained for
a regression model of 5^{th}-order.

In the previous section we found a 5^{th}-order polynomial
regression model to perform best on the validation set. Now, we plot
that model, including the 95%-confidence intervals, on the data set in
order to visually assess its quality. In addition, we plot the
underlying function for the data generation process.

```
best_order <- RMSE[which.min(RMSE[, 3]), 1]
final_model <- lm(y ~ poly(x, best_order), data = train_set)
# predictions
vals <- list("x" <- seq(min(new_poly_data$x), max(new_poly_data$y), by = 0.01))
predictions <- predict(final_model, newdata = vals, interval = "confidence", level = 0.95)
# plot data
plot(new_poly_data$x, new_poly_data$y, pch = 16, ylab = "", xlab = "")
# plot data generation function
lines(vals[[1]], sin(2 * pi * vals[[1]]), lwd = 3, col = "green")
# plot fit and confidence levels
lines(vals[[1]], predictions[, "fit"], lwd = 2, col = "blue")
lines(vals[[1]], predictions[, "upr"], lwd = 2, lty = 2, col = "blue")
lines(vals[[1]], predictions[, "lwr"], lwd = 2, lty = 2, col = "blue")
legend("topright",
legend = c("observed data", "prediction", "signal"),
lty = c(NA, 1, 1),
pch = c(16, NA, NA),
cex = 0.75,
col = c(1, "blue", "green"),
lwd = c(NA, 2, 2)
)
```

The figure shows that our model does a decent job at fitting the data and therefore we can be quite satisfied with it.

The input values \(x_n\) for the underlying function are generated uniformly in range \(U(0, 1)\). The corresponding target values \(y\) are obtained by first computing the corresponding values of the function \(sin(2\pi x)\) and then adding random noise with a Gaussian distribution with a standard deviation of \(0.35\).

```
##### Data generating function #####
set.seed(123) # set seed for reproducibility
n <- 25
x <- runif(n, 0, 1)
y <- sin(2 * pi * x) + rnorm(n, 0, 0.35)
poly_data <- data.frame("x" = x, "y" = y)
```

`## Warning: Paket 'htmltools' wurde unter R Version 4.3.1 erstellt`

**Citation**

The E-Learning project SOGA-R was developed at the Department of Earth Sciences by Kai Hartmann, Joachim Krois and Annette Rudolph. You can reach us via mail by soga[at]zedat.fu-berlin.de.

You may use this project freely under the Creative Commons Attribution-ShareAlike 4.0 International License.

Please cite as follow: *Hartmann,
K., Krois, J., Rudolph, A. (2023): Statistics and Geodata Analysis
using R (SOGA-R). Department of Earth Sciences, Freie Universitaet Berlin.*