Test 3, St590 2015, Dickey
When exponentials or natural logarithms needed, use e0=1, e0.69 =2, e1.10=3, e1.39=4, e1.61=5 (so
ln(2)=0.69 etc.) and assume no priors are used.
1. (5 points) For what value of x, if any, is the hyperbolic tangent of x equal to 1.10? x=______
2. (20 pts.) Here is a neural network diagram for a continuous response Y.
In neural network terminology,
How many hidden units ____ are there?
How many hidden layers ____ are there?
How many total biases ____ and
weights____ are involved in going from
the inputs to the output Y?
X1
H1
X2
Y
H2
X3
3. (25 pts.) A logistic regression for the probability p that someone will pay off their loan (Y=1) rather
than default (Y=0) gives the logit L as L=0.02 -0.01*age + 0.025*income, where age is in years and
income is in thousands of dollars.
(a) find the income ($________ thousand) that would make L=0.69 for a person 50 years old.
(b) Using the list of exponentials at the top of the page, give the probability p=_______ and the odds
______ that a 50 year old with the income you computed in part (a) will pay off their loan.
(c) For that same 50 year old, what increase in income would triple the predicted odds of paying off the
loan?
Answer increase income by $_____ thousand
(d) According to this model are young people more or less likely to pay off their loan than older people
with the same salary?
4. (50 pts.) I ran a simple logistic regression for the probability p that Y=1 as a function of X, that is, I
assumed logit(p)=a+bX. I printed the X and Y values and predicted probability (p_hat), sorted in
increasing order of X, to get this:
Obs
1
2
3
4
5
6
7
8
9
10
X
2
3
7
9
10
12
13
15
15
19
Y
1
0
1
0
1
1
0
0
0
0
p_hat
0.77
0.73
0.55
0.45
0.40
0.31
0.27
0.20
0.20
0.10
Answer these questions if possible. If any is not possible mark it “NP”.
(a) Is the logit’s slope b positive or negative?
(b) How many concordant pairs _____ discordant pairs _____ and tied pairs ____ are there?
(c) Compute the cumulative lift at depth 25 _______
(d) If I predict that Y will be 1 whenever X is less than 11 and that Y will be 0 otherwise, what will be the
corresponding vertical ____ and horizontal ____ coordinates for the resulting point on the ROC curve?
What will be the specificity__________?
Answers with detailed explanation
1. Hyperbolic tangents range from -1 to 1 so no x will give H(x)=1.10.
2. Each X connects to each H so there are 3 weights for each of two H’s for 6 weights. Each of these two
H equations has an intercept (2 bias terms). Each H then connects to Y so we have 2 more weights and 1
more bias term. There is 1 hidden layer of 2 hidden units, 8 total weights and 3 bias terms.
3. We want to solve log(2)=0.69=0.02 -0.01*(50) + 0.025*income so 0.025(income)=0.69-0.02+0.50=1.17
and thus income = 1.17/0.025 = $46.8 thousand ($46,800). Since e0.69=2 from the list, the probability is
2/(2+1)=2/3 so the odds would be (2/3)/(1-2/3)=2 (i.e. “2 to 1 odds”). To make (new odds)/(old
odds)=3, we would have ln(new odds)-ln(old odds)=ln(3) or new logit – old logit = ln(3). The change in
logits is the change in income (delta) multiplied by 0.025 so 0.025delta = ln(3) and, using the list of
exponentials, delta=ln(3)/0.025 = 1.10/0.025=$44 so an extra 44 thousand would triple the odds. This is
the same logic as for the doubling amount discussed in class and our notes but using ln(3) rather than
ln(2). Because of the negative coefficient on age, young people of the same income as older people will
be more likely to pay off the loans. Perhaps younger people have to take out loans to get started and
then pay them off while an older person still carrying those loans may be in financial difficulty.
4. The probabilities decrease with increasing X so the slope must be negative. The first 1 is concordant
with all the 6 0s to its right on the X axis (below it in the list). The next 1 is concordant with the 5 0s to
its right and the other two 1s each give 4 concordant pairs for 6+5+4+4=19 concordant pairs in all. Even
though there are two observations at X=15, they both have Y=0 and are not counted in the pairs having
Y=1 for one element and Y=0 for the other. Thus there are no ties and with 4 1s and 6 0s there are 24
total pairs. With no ties, 24-19=5 of them must be discordant.
Column C: # 0s below each 1
Column D: # 1s below each 0
C
Obs
1
2
3
4
5
6
7
8
9
10
X
2
3
7
9
10
12
13
15
15
19
Y
1
0
1
0
1
1
0
0
0
0
p_hat
0.77
0.73
5
0.55
0.45
4
0.40
4
0.31
0.27
0.20
0.20
0.10
19
D
6
3
2
0
0
0
0
5
The 25th percentile cuts off the first two observations where the response rate is 0.50 whereas the
overall response rate is 0.40 so the lift is the ratio 0.50/0.40 = 1.25 (and the gain is 25%). Using a cutoff
at X=11, we correctly identify the first 3 ones (they have X<11) so we’ve identified 3 of the 4 Y=1 values
for a sensitivity of 0.75, the vertical coordinate. For X larger than 11, there are 4 of the 6 Y=0 values so
we have specificity 2/3 and thus the horizontal coordinate is 1/3.