\nabla g(\vec{x}) = \begin{bmatrix} 2x_1 -6 + 4x_1(x_1^2 - x_2) \\ -2(x_1^2 - x_2) \end{bmatrix}

We can find \nabla g(\vec{x}) by finding the partial derivatives of g(\vec{x}):

\frac{\partial g}{\partial x_1} = 2(x_1 - 3) + 2(x_1^2 - x_2)(2 x_1) \frac{\partial g}{\partial x_2} = 2(x_1^2 - x_2)(-1) \nabla g(\vec{x}) = \begin{bmatrix} 2(x_1 - 3) + 2(x_1^2 - x_2)(2 x_1) \\ 2(x_1^2 - x_2)(-1) \end{bmatrix} \nabla g\left(\begin{bmatrix} - 1 \\ 1 \end{bmatrix}\right) = \begin{bmatrix} 2(-1 - 3) + 2((-1)^2 - 1)(2(-1)) \\ 2((-1)^2 - 1) \end{bmatrix} = \begin{bmatrix} -8 \\ 0 \end{bmatrix}.

\vec x^{(1)} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}

Here’s the general form of gradient descent: \vec x^{(1)} = \vec{x}^{(0)} - \alpha \nabla g(\vec{x}^{(0)})

We can substitute \alpha = \frac{1}{2} and x^{(0)} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} to get: \vec x^{(1)} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} - \frac{1}{2} \nabla g(\vec x ^{(0)}) \vec x^{(1)} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} - \frac{1}{2} \begin{bmatrix} -8 \\ 0 \end{bmatrix}

\vec{x}^{(1)} = \begin{bmatrix} -1 \\ 1 \end{bmatrix} - \frac{1}{2} \begin{bmatrix} -8 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \end{bmatrix}

f(t) is not convex, but has a global minimum.

It is seen that f(t) isn’t convex, which can be verified using the second derivative test: f'(t) = 2(t - 3) + 2(t^2 - 1) 2t = 2t - 6 + 4t^3 - 4t = 4t^3 - 2t - 6 f''(t) = 12t^2 - 2

Clearly, f''(t) < 0 for many values of t (e.g. t = 0), so f(t) is not always convex.

However, f(t) does have a global minimum – its output is never less than 0. This is because it can be expressed as the sum of two squares, (t - 3)^2 and (t^2 - 1)^2, respectively, both of which are greater than or equal to 0.

Answer: Logistic regression

Logistic regression is a linear classification technique, meaning it creates a single linear decision boundary between classes. Among the five decision boundaries, only Decision Boundary 1 is linear, making it the correct answer.

Answer: Decision tree with \text{max depth} = 15

We know that Decision Boundaries 2 and 5 are decision trees, since the decision boundaries are parallel to the axes. Decision boundaries for decision trees are parallel to axes because the tree splits the feature space based on one feature at a time, creating partitions aligned with that feature’s axis (e.g., “Price ≤ 100” results in a vertical split).

Decision Boundary 2 is more complicated than Decision Boundary 5, so we can assume there are more levels.

Answer: k-nearest neighbors with k = 3

k-nearest neighbors decision boundaries don’t follow straight lines like logistic regression or decision trees. Thus, boundaries 3 and 4 are k-nearest neighbors.

A lower value of k means the model is more sensitive to local variations in the data. Since Decision Boundary 3 has more complex and irregular regions than Decision Boundary 4, it corresponds to k = 3, where the model closely follows the training data.

For example, if in our training data we had 3 points in the group represented by the light-shaded region at (300, 150), the decision boundary for k-nearest neighbors with k = 3 would be light-shaded at (300, 150), whereas it may be dark-shaded for k-nearest neighbors with k = 100.

Answer: k-nearest neighbors with k = 100

k-nearest neighbors decision boundaries don’t follow straight lines like logistic regression or decision trees. Thus, boundaries 3 and 4 are k-nearest neighbors.

k-nearest neighbors decision boundaries adapt to the training data, but increasing k makes the model less sensitive to local variations.

Since Decision Boundary 4 is smoother and has fewer regions than Decision Boundary 3, it corresponds to k = 100, where predictions are averaged over a larger number of neighbors, making the boundary less sensitive to individual data points.

Answer: \frac{35}{57}

There are 105 true positives and 66 false negatives. Hence, the recall is \frac{105}{105 + 66} = \frac{105}{171} = \frac{35}{57}.

Answer: 33

Let x be the number of true negatives. The number of correctly classified data points is 105 + x, and the total number of data points is 105 + 30 + 66 + x = 201 + x. Hence, this boils down to solving for x in \frac{69}{117} = \frac{105 + x}{201 + x}.

It may be tempting to cross-multiply here, but that’s not necessary (in fact, we picked the numbers specifically so you would not have to)! Multiply \frac{69}{117} by \frac{2}{2} to yield \frac{138}{234}. Then, conveniently, setting x = 33 in \frac{105 + x}{201 + x} also yields \frac{138}{234}, so x = 33 and hence the number of true negatives our classifier has is 33.

Answer: True

Remember that \text{precision} = \frac{TP}{TP + FP} and \text{recall} = \frac{TP}{TP + FN}. In order for precision to be the same as recall, it must be the case that FP = FN, i.e. that our classifier makes the same number of false positives and false negatives. The only kinds of “errors" or”mistakes" a classifier can make are false positives and false negatives; thus, we must have

\text{mistakes} = FP + FN = FP + FP = 2 \cdot FP

2 times any integer must be an even integer, so the number of mistakes must be even.

Answer: Option B: low precision and high recall.

Our classifier is good at identifying when the input stream is “Kiss Me thru the Phone”, i.e. it is good at identifying true positives amongst all positives. This means it has high recall.

Since our classifier makes many false positive predictions – in other words, it often incorrectly predicts “Kiss Me thru the Phone” when that’s not what the input stream is – it has many false positives, so its precision is low.

Thus, our classifier has low precision and high recall.

Answer: Option 2

Conceptually, we’re looking for a combination of a and b such that when a > b, it’s true that in more than 50% of states, the "Math" value is larger than the "Verbal" value. Let’s look at all four options through this lens:

• Option 1: sat['Math'] > sat['Verbal'] is a Series of Boolean values, containing True for all states where the "Math" value is larger than the "Verbal" value and False for all other states. The mean of this series, then, is the proportion of states satisfying this criteria, and since b is 0.5, a > b is True only when the bolded condition above is True.
• Option 3 is the same as Option 1 – note that x > y is equivalent to x - y > 0.
• Option 4: sat['Math'] / sat['Verbal'] is a Series that contains values greater than 1 whenever a state’s "Math" value is larger than its "Verbal" value and less than or equal to 1 in all other cases. As in the other options that work, (sat['Math'] / sat['Verbal']) > 1 is a Boolean Series with True for all states with a larger "Math" value than "Verbal" values; a > b compares the proportion of True values in this Series to 0.5. (Here, p - 0.5 > 0 is the same as p > 0.5.)

Then, by process of elimination, Option 2 must be the correct option – that is, it must be the only option that doesn’t work. But why? sat['Math'] - sat['Verbal'] is a Series containing the difference between each state’s "Math" and "Verbal" values, and .mean() computes the mean of these differences. The issue is that here, we don’t care about how different each state’s "Math" and "Verbal" values are; rather, we just care about the proportion of states with a bigger "Math" value than "Verbal" value. It could be the case that 90% of states have a larger "Math" value than "Verbal" value, but one state has such a big "Verbal" value that it makes the mean difference between "Math" and "Verbal" scores negative. (A property you’ll learn about in future probability courses is that this is equal to the difference in the mean "Math" value for all states and the mean "Verbal" value for all states – this is called the “linearity of expectation” – but you don’t need to know that to answer this question.)

Answer: \frac{2}{3}

An accuracy of \frac{5}{9} means that the model is such that out of 9 values, 5 are labeled correctly. By extension, this means that 4 out of 9 are not labeled correctly as 0 or 1.

Remember, RMSE is defined as

\text{RMSE} = \sqrt{\frac{1}{n} \sum_{i = 1}^n (y_i - H(x_i))^2}

where y_i represents the ith actual value and H(x_i) represents the ith prediction. Here, y_i is either 0 or 1 and $H(x_i) is also either 0 or 1. We’re told that \frac{5}{9} of the time, y_i and H(x_i) are the same; in those cases, (y_i - H(x_i))^2 = 0^2 = 0. We’re also told that \frac{4}{9} of the time, y_i and H(x_i) are different; in those cases, (y_i - H(x_i))^2 = 1. So,

\text{RMSE} = \sqrt{\frac{5}{9} \cdot 0 + \frac{4}{9} \cdot 1} = \sqrt{\frac{4}{9}} = \frac{2}{3}

Answer: Option 2

Since the accuracy of Classifier 1 is much higher on the dataset used to train it than the dataset it was tested on, it’s likely Classifer 1 overfit to the training set because it was too complex. To fix the issue, we need to decrease its complexity, so that it focuses on learning the general structure of the data in the training set and not too much on the random noise in the training set.

Answer: Option 1

Note that the columns of conf sum to 1 – 0.9 + 0.1 = 1, and 0.4 + 0.6 = 1. To create a Series with just the value 1, then, we need to sum the columns of conf, which we can do using conf.sum(axis=0). conf.sum(axis=1) would sum the rows of conf.

Answer: recall

The number 0.6 appears in the bottom-right corner of conf. Since conf is column-normalized, the value 0.6 represents the proportion of values in the second column that were predicted to be 1. The second column contains values that were actually 1, so 0.6 is really the proportion of values that were actually 1 that were predicted to be 1, that is, \frac{\text{actually 1 and predicted 1}}{\text{actually 1}}. This is the definition of recall!

If you’d like to think in terms of true positives, etc., then remember that: - True Positives (TP) are values that were actually 1 and were predicted to be 1. - True Negatives (TN) are values that were actually 0 and were predicted to be 0. - False Positives (FP) are values that were actually 0 and were predicted to be 1. - False Negatives (FN) are values that were actually 1 and were predicted to be 0.

Recall is \frac{\text{TP}}{\text{TP} + \text{FN}}.

Answer: \frac{5}{6}

Here is one way to solve this problem:

accuracy = \frac{TP + TN}{TP + TN + FP + FN}

Given the values from the confusion matrix:

accuracy = \frac{0.6 \cdot \alpha + 0.9 \cdot (1 - \alpha)}{\alpha + (1 - \alpha)}
accuracy = \frac{0.6 \cdot \alpha + 0.9 - 0.9 \cdot \alpha}{1}
accuracy = 0.9 - 0.3 \cdot \alpha

Therefore:

0.65 = 0.9 - 0.3 \cdot \alpha
0.3 \cdot \alpha = 0.9 - 0.65
0.3 \cdot \alpha = 0.25
\alpha = \frac{0.25}{0.3}
\alpha = \frac{5}{6}

R_\text{sq}(h) = \dfrac{1}{4}\sum_{i=1}^{4}(y_i-h)^2

Recall the equation for R_\text{sq}(h) = \frac{1}{n}\sum_{i=1}^{n}(y_i-h)^2, so we simply need to replace n with 4 because there are 4 elements in our dataset.

\frac{dR_\text{sq}(h)}{dh} = \frac{1}{2}\sum_{i=1}^{4}(h-y_i)

Since we have the equation for R_\text{sq}(h) we can calculate the derivative:

\begin{align*} \frac{dR_\text{sq}(h)}{dh} &= \frac{dR_\text{sq}(h)}{dh}(\frac{1}{4}\sum_{i=1}^{4}(y_i-h)^2) \\ &= \frac{1}{4}\sum_{i=1}^{4}\frac{dR_\text{sq}(h)}{dh}((y_i-h)^2) \\ \end{align*} We can use the chain rule to find the derivative of (y_i-h)^2. Recall the chain rule is: \frac{df(x)}{dx}[(f(x))^n] = n(f(x))^{n-1} \cdot f'(x).

\begin{align*} &= \frac{1}{4}\sum_{i=1}^{4}2(y_i-h) \cdot -1 \\ &= \frac{1}{4}\sum_{i=1}^{4} 2(h - y_i) \\ &= \frac{1}{2}\sum_{i=1}^{4} (h-y_i) \end{align*}

h^{(0)} = -\frac{5}{4}

The gradient descent equation is given by: \begin{aligned} h^{(t)} &= h^{(t-1)} - \alpha \frac{dR_{sq}(h^{(t-1)})}{dh} \end{aligned}

Recall the learning rate is equal to \alpha. We can plug in \alpha = \frac{1}{4}, h^{(2)} = \frac{1}{4}, and i=2, in that case we obtain: \begin{aligned} \frac{1}{4} &= h^{(1)} - \alpha \frac{dR_{sq}(h^{(1)})}{dh}\\ &= h^{(1)} - \alpha \frac{1}{2}\sum_{i=1}^{4}(h^{(1)}-y^{(i)})\\ &= h^{(1)} - \frac{1}{4} \cdot \frac{1}{2}(h^{(1)} + 2 + h^{(1)} + 1 + h^{(1)} - 2 + h^{(1)} - 4)\\ &= h^{(1)} - \frac{1}{8}(4h^{(1)} - 3)\\ \end{aligned}

Solving this equation, we obtain that h^{(1)} = -\frac{1}{4}. We can then repeat this step once more to obtain h^{(0)}: \begin{aligned} -\frac{1}{4} &= h^{(0)} - \alpha \frac{dR_{sq}(h^{(0)})}{dh}\\ &= h^{(0)} - \frac{1}{4} \cdot \frac{1}{2}(h^{(0)} + 2 + h^{(0)} + 1 + h^{(0)} - 2 + h^{(0)} - 4)\\ &= h^{(0)} - \frac{1}{8}(4h^{(0)} - 3)\\ \end{aligned}

Solving this equation, we obtain that h^{(0)} = -\frac{5}{4}.

In general, the update rule for gradient descent is: x^{(t+1)} = x^{(t)} - \alpha \nabla f(x^{(t)}) = x^{(t)} - \alpha \frac{df}{dx}(x^{(t)}), where \alpha is the learning rate or step size. We know that the derivative of f is as follows: \frac{df}{dx} = f'(x) = 3x^2 + 2x, thus the update rule can be written down as: x^{(t+1)} = x^{(t)} - \alpha(3x^{(t)^2} + 2x^{(t)}) = -\frac{3}{4}x^{(t)^2} + \frac{1}{2}x^{(t)}.

We have f'(x^{(0)}) = f'(-1) = 3(-1)^2 + 2(-1) = 1 > 0, so we go left, and x^{(1)} = x^{(0)} - \alpha f'(x^{(0)}) = -1 - \frac{1}{4} = -\frac{5}{4}. Intuitively, the gradient descent cannot converge in this case because \text{lim}_{x \rightarrow -\infty} f(x) = -\infty,

We need to find all local minimums and local maximums. First, we solve the equation f'(x) = 0 to find all critical points.

We have: f'(x) = 0 \Leftrightarrow 3x^2 + 2x = 0 \Leftrightarrow x = -\frac{2}{3} \ \ \text{and} \ \ x = 0.

Now, we consider the second-order derivative: f''(x) = \frac{d^2f}{dx^2} = 6x + 2.

We have f''(x) = 0 only when x = -1/3. Thus, for x < -1/3, f''(x) is negative or the slope f'(x) decreases; and for x > -1/3, f''(x) is positive or the slope f'(x) increases. Keep in mind that -1 < -2/3 < -1/3 < 0 < 1.

Therefore, f has a local maximum at x = -2/3 and a local minimum at x = 0. If the gradient descent starts at x^{(0)} = -1 and it always goes left then it will never meet the local minimum at x = 0, and it will go left infinitely. We say the gradient descent cannot converge, or is divergent.

We have f'(x^{(0)}) = f'(-1) = 3 \cdot 1^2 + 2 \cdot 1 = 5 > 0, so we go left, and x^{(1)} = x^{(0)} - \alpha f'(x^{(0)}) = 1 - \frac{1}{4} \cdot 5 = -\frac{1}{4}.

From the previous part, function f has a local minimum at x = 0, so the gradient descent can converge (given appropriate step size) at this local minimum.

h^* = -1

The minimum possible value of the exponent is 0, since anything squared is non-negative. The exponent is 0 when (x+1)^2 = 0, i.e. when x = -1. Since e^{(x+1)^2} gets larger as (x+1)^2 gets larger, the minimizing input h^* is -1.

h^{(1)} = -\alpha \cdot 2e

First, we find \frac{dR}{dh}(h):

\frac{dR}{dh}(h) = 2(h+1) e^{(h+1)^2}

Then, we know that

h^{(1)} = h^{(0)} - \alpha \frac{dR}{dh}(h^{(0)}) = 0 - \alpha \frac{dR}{dh}(0)

In our case, \frac{dR}{dh}(0) = 2(0 + 1) e^{(0+1)^2} = 2e, so

h^{(1)} = -\alpha \cdot 2e

\alpha = \frac{1}{2e}

We know from the part 2 that h^{(1)} = -\alpha \cdot 2e, and we know from part 1 that h^* = -1. If gradient descent converges in one iteration, that means that h^{(1)} = h^*; solving this yields

-\alpha \cdot 2e = -1 \implies \alpha = \frac{1}{2e}

True.

R(h) is convex, since the graph is bowl shaped. (It can also be proved that R(h) is convex using the second derivative test.) It is also differentiable, as we saw in part 2. As a result, since it’s both convex and differentiable, gradient descent is guaranteed to be able to minimize it given an appropriate choice of step size.

h^{(1)} = 4, h^{(2)} = 2

The updating rule for gradient descent in the one-dimensional case is: h^{(t+1)} = h^{(t)} - \alpha \cdot \frac{dR}{dh}(h^{(t)})

We can find \frac{dR}{dh} by taking the derivative of R(h): \frac{d}{dh}R(h) = \frac{d}{dh}(\sqrt{(h - 3)^2 + 1}) = \dfrac{h-3}{\sqrt{\left(h-3\right)^2+1}}

Now we can use \alpha = 2\sqrt{2} and h^{(0)} = 2 to begin updating:

\begin{align*} h^{(1)} &= h^{(0)} - \alpha \cdot \frac{dR}{dh}(h^{(0)}) \\ h^{(1)} &= 2 - 2\sqrt{2} \cdot \left(\dfrac{2-3}{\sqrt{\left(2-3\right)^2+1}}\right) \\ h^{(1)} &= 2 - 2\sqrt{2} \cdot (\dfrac{-1}{\sqrt{2}}) \\ h^{(1)} &= 4 \end{align*}
\begin{align*} h^{(2)} &= h^{(1)} - \alpha \cdot \frac{dR}{dh}(h^{(1)}) \\ h^{(2)} &= 4 - 2\sqrt{2} \cdot \left(\dfrac{4-3}{\sqrt{\left(4-3\right)^2+1}}\right) \\ h^{(2)} &= 4 - 2\sqrt{2} \cdot (\dfrac{1}{\sqrt{2}}) \\ h^{(2)} &= 2 \end{align*}

No, this algorithm will not converge to the minimizer because if we do more iterations, we’ll keep oscillating back and forth between predictions of 2 and 4. We showed the first two iterations of the algorithm in part 1, but the next two would be exactly the same, and the two after that, and so on. This happens because the learning rate is too big, resulting in steps that are too big, and we keep jumping over the true minimizer at h^* = 3.

Answer:

Let’s first identify all elements of the confusion matrix by comparing predictions with true labels:

1. Accuracy: The proportion of correct predictions (TP + TN) among the total number of predictions. \text{Accuracy} = \frac{TP + TN}{TP + TN + FP + FN} = \frac{6 + 1}{10} = \frac{7}{10} = 0.7

2. Precision: The proportion of true positives among all positive predictions. \text{Precision} = \frac{TP}{TP + FP} = \frac{6}{6 + 3} = \frac{6}{9} = \frac{2}{3} \approx 0.67

3. Recall: The proportion of true positives among all actual positives. \text{Recall} = \frac{TP}{TP + FN} = \frac{6}{6 + 0} = \frac{6}{6} = 1

Answer: Model B

Accuracy is defined as \frac{\text{number\;of\;correct\;predictions}}{\text{number\;of\;total\;predictions}}, so we sum the (Yes, Yes) and (No, No) cells to get the number of correct predictions and divide that by the sum of all cells as the number of total predictions. We see that Model B has the highest accuracy of 0.9 with that formula. (Note that for simplicity, the confusion matrices are such that the sum of all values is 100 in all three cases.)

Answer: Model C

Precision is defined as \frac{\text{number of correctly predicted yes values}}{\text{total number of yes predictions}}, so the number of correctly predicted yes values is the (Yes, Yes) cell, while the total number of yes predictions is the sum of the Yes column. We see that Model C has the highest precision, \frac{80}{85}, with that formula.

Answer: Model B

Recall is defined as \frac{\text{number of correctly predicted yes values}}{\text{total number of values actually yes}}, so the number of correctly predicted yes values is the (Yes, Yes) cell, while the total number of values that are actually yes is the sum of the Yes row. We see that Model B has the highest recall, 1, with that formula.

Answer: \frac{8}{13}

Precision is the proportion of predicted positives that actually were positives. So, given this confusion matrix, that value is \frac{8}{8 + 5}, or \frac{8}{13}.

Answer:

1. B
2. 5

We already know that the precision is \frac{8}{13}. Recall is the proportion of true positives that were indeed classified positives, which in this matrix is \frac{8}{B + 8}. So, in order for precision to equal recall, B must be 5.

Answer:

1. A \cdot B
2. 40

We can solve this problem by simply stating recall and accuracy in terms of the values in the confusion matrix. As we already found, recall is \frac{8}{B+8}. Accuracy is the sum of correct predictions over total number of predictions, or \frac{A + 8}{A + B + 13}. Then, we simply set these equal to each other, and solve.

\frac{8}{B+8} = \frac{A + 8}{A + B + 13} 8(A + B + 13) = (A + 8)(B + 8) 8A + 8B + 104 = AB + 8A + 8B + 64 104 = AB + 64 AB = 40