# Problem Sheet 3

You should attempt all these questions and write up your solutions in advance of the workshop in week 7 where the answers will be discussed.

**9.** Consider the following dataset. Our aim is to predict \(y\) from the
variables \(x_1\), \(x_2\) and \(x_3\).

\(i\) | \(x_{1,i}\) | \(x_{2,i}\) | \(x_{3,i}\) | \(y_i\) |
---|---|---|---|---|

1 | -2.17 | -2.08 | -2.16 | 4.47 |

2 | -1.35 | -0.50 | -0.74 | 5.60 |

3 | -1.22 | -1.00 | -1.78 | 4.16 |

4 | -1.04 | -0.32 | -0.40 | 5.52 |

5 | -0.87 | -0.39 | -0.67 | 5.27 |

6 | -0.41 | 0.07 | -0.66 | 4.70 |

7 | 0.07 | 0.74 | 0.37 | 5.50 |

8 | 0.25 | 0.35 | 0.02 | 4.84 |

9 | 0.87 | 1.28 | 0.52 | 4.92 |

10 | 1.53 | 2.30 | 1.35 | 5.35 |

11 | 2.46 | 2.55 | 1.77 | 4.86 |

12 | 5.00 | 5.04 | 4.05 | 5.09 |

These data are also available in machine-readable form at

Using these data:

- Fit a linear model of the form \(y_i = \beta_0 + x_{1,i} \beta_1 + x_{2,i} \beta_2 + x_{3,i} \beta_3 + \varepsilon_i\).

To fit the required model we can use the following R commands:

```
url <- "https://www1.maths.leeds.ac.uk/~voss/2022/MATH3714/P03Q09.csv"
x <- read.csv(url)
m <- lm(y ~ x1 + x2 + x3, data=x)
```

The fitted model has the following coefficients:

```
Call:
lm(formula = y ~ x1 + x2 + x3, data = x)
Residuals:
Min 1Q Median 3Q Max
-0.104928 -0.032723 -0.001836 0.024802 0.142993
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.97725 0.06026 82.598 5.15e-13 ***
x1 -1.11954 0.07427 -15.074 3.71e-07 ***
x2 0.28176 0.12023 2.344 0.0472 *
x3 1.06621 0.09492 11.232 3.54e-06 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.07109 on 8 degrees of freedom
Multiple R-squared: 0.9817, Adjusted R-squared: 0.9748
F-statistic: 142.8 on 3 and 8 DF, p-value: 2.76e-07
```

- Determine a \(95\%\) confidence interval for the coefficient \(\beta_2\).

From lectures we know that a confidence interval for a single coefficient \(\beta_j\) is given by \[\begin{equation*} [U, V] = \Bigl[ \hat\beta_j - t_{n-p-1}(\alpha/2) \sqrt{\widehat{\sigma^2} C_{jj}}, \hat\beta_j + t_{n-p-1}(\alpha/2) \sqrt{\widehat{\sigma^2} C_{jj}} \Bigr], \end{equation*}\] where \(t_{n-p-1}\) is the \((1-\alpha/2)\)-quantile of the \(t\)-distribution, \(C_{jj} = (X^\top X)^{-1}_{jj}\), and \(X\) is the design matrix.

We can read off the required values for computing the confidence
interval from the output of `summary(m)`

: the centre of the
confidence interval can be found in the column `Estimate`

,
the standard error \(\sqrt{\widehat{\sigma^2} C_{jj}}\) is given in
column `Std. Error`

, and \(n-p-1 = 12 - 3 - 1 = 8\). Using
these values, we can find a \(95\%\)-confidence interval for
\(\beta_2\) as follows:

`[1] 0.004509123 0.559010877`

Thus, the confidence interval is \(\bigl[ 0.0045, 0.5590 \bigr]\).

- Perform a hypothesis test, at the \(95\%\)-level, for the hypothesis \(H_0\colon \beta_2 = 0\) with alternative \(H_1\colon \beta_2 \neq 0\).

The test statistic is
\[\begin{equation*}
T
= \frac{\bigl|\hat\beta_2\bigr|}{\sqrt{\widehat{\sigma^2} C_{jj}} }
= \frac{0.28176}{0.12023}
= 2.344.
\end{equation*}\]
(This value is also shown in the column `t value`

of the R
summary output.) The critical value is
\[\begin{equation*}
t
= t_{n-p-1}(\alpha/2)
= 2.306.
\end{equation*}\]
Since \(T > t\), we can reject the hypothesis \(\beta_2 = 0\) at the
\(95\%\)-level.

**10.** Let \(x_1, \ldots, x_n, y_1, \ldots, y_n \in \mathbb{R}\) be given. Assume
that it is known that the \(y\)-values satisfy
\(y \approx 1 - x + c x^2 - d x^3\), where \(c\) and \(d\) are unknown
constants.

- Explain how linear regression can be used to estimate the parameters \(c\) and \(d\) from the given data.

We can rewrite the equation for \(y\) as \[\begin{equation*} y + x - 1 = c x^2 - d x^3, \end{equation*}\] where \(c\) and \(d\) are the unknown quantities, and \(x\) and \(y\) are given. Thus we can estimate \(c\) and \(d\) by fitting a linear model with two inputs, \(x_i^2\) and \(x_i^3\) and one output \(y_i + x_i - 1\). The model does not include an intercept.

- What is the design matrix in this situation?

Since there is no intercept, the design matrix is \[\begin{equation*} X = \begin{pmatrix} x_1^2 & x_1^3 \\ x_2^2 & x_2^3 \\ \vdots & \vdots \\ x_n^2 & x_n^3 \\ \end{pmatrix} \in \mathbb{R}^{n\times 2}. \end{equation*}\]

- Why can
*linear*regression be used, despite the presence of the*non-linear*terms \(x^2\) and \(x^3\)?

A linear model is appropriate, because the response depends on the unknown coefficients \(c\) and \(d\) in a linear way: \(c\) and \(d\) are only multiplied by known constants, and there are no non-linear functions of \(c\) and \(d\) in the model.

**11.** In this question we consider four different datasets, given by
inputs \(x_i\) and responses \(y_i\) for \(i \in \{1, 2, 3, 4\}\).

- Based on the following plots, discuss model fit of each model. Describe all relevant features of the plots. Describe any problems with the model fit you find.

```
par(mfrow=c(4,2))
m1 <- lm(y1 ~ x1)
plot(fitted(m1), resid(m1), xlab="fitted values", ylab="residuals")
qqnorm(resid(m1), main=NULL)
m2 <- lm(y2 ~ x2)
plot(fitted(m2), resid(m2), xlab="fitted values", ylab="residuals")
qqnorm(resid(m2), main=NULL)
m3 <- lm(y3 ~ x3)
plot(fitted(m3), resid(m3), xlab="fitted values", ylab="residuals")
qqnorm(resid(m3), main=NULL)
m4 <- lm(y4 ~ x4)
plot(fitted(m4), resid(m4), xlab="fitted values", ylab="residuals")
qqnorm(resid(m4), main=NULL)
```

We first note that the given R commands generate the plots along
rows, so the first row of plots corresponds to model `m1`

, the
second row to model `m2`

, and so on. The plots shown are
residual plots in the left column, and Q-Q-plots in the right
column.

**m1:**Samples in the residual plot on the left seem to follow an upside down parabola indicating a non-linear relationship between \(x\) and \(y\). Samples in the QQ-plot on the right seem to (approximately?) follow a straight line, so residuals seem to be (approximately?) normally distributed. No outliers are visible.**m2:**Samples in the residual plot form a band centred at zero. Possibly very negative residuals are less frequent for fitted values \(\hat y_i < 4\)? No outliers are apparent. The samples in the QQ-plot seem to clearly fall on a straight line. Overall, model fit seems very good.**m3:**The samples in the residual plot are centred around \(\hat\varepsilon= 0\), but the width of the spread increases as the fitted values increase, leasing to a triangular shape. This indicates that the variance of the residuals is not constant but increases with \(y\). Possibly, a model with multiplicative noise would be appropriate for these data. The QQ-plot shows an approximately straight line, but fit is not as good as for model`m2`

.**m4:**The samples in the residual plot form a horizontal band centred at \(\hat\varepsilon= 0\), no outliers are visible. The QQ-plot shows clearly that the residuals are not normally distributed: the samples form an “S-shaped” line, indicating that the errors have lighter tails than we would expect for a normal distribution.

- Which of the four models has the best fit?

The data in `m1`

seems to
have a non-linear relationship between \(x\) and \(y\), the data
in `m3`

seems to be better modelled using multiplicative noise,
and the residuals in `m4`

seem not to be normally
distributed. Thus, the model with the best fit is `m2`

,
because it is the only model which does not show any obvious
problems.

**12.** Using singular value decomposition,
we can write a design matrix \(X\) as \(X = U D V^\top\),
where \(D\in\mathbb{R}^{(p+1)\times(p+1)}\) is a diagonal matrix
and \(U\in\mathbb{R}^{n\times(p+1)}\) and \(V\in\mathbb{R}^{(p+1)\times(p+1)}\) are matrices with orthonormal columns.
Denote the diagonal elements of \(D\) by \(\sigma_0(X), \ldots, \sigma_p(X) \geq 0\),
and the columns of \(V\) by \(v_0, \ldots, v_p\).
Show that \(\| X v_k \| = \sigma_k(X)\).
(We used this fact in example 15.3.)

Using the singular value decomposition of \(X\), we find \[\begin{equation*} \| X v_k \| = \| U D V^\top v_k \|. \end{equation*}\] For every vector \(x\) we have \(\| U x \|^2 = x^\top U^\top U x = x^\top x = \| x \|^2\). Thus, multiplying a vector \(x\) by \(U\) does not change its length. Using this rule we find \[\begin{equation*} \| X v_k \| = \| D V^\top v_k \|. \end{equation*}\] Since the columns of \(V\) are orthonormal, we have \(V^\top v_k = e_k\), where \(e_k = (0, \ldots, 1, \ldots, 0)\) is the \(k\)th standard basis vector. Thus we get \[\begin{equation*} \| X v_k \| = \| D e_k \| = \| \sigma_k(X) e_k \| = \sigma_k(X) \| e_k \| = \sigma_k(X) \end{equation*}\] as required. This completes the proof.