1.Let Y be an individual's income over one calendar year. Imagine a state with a flat 10% income tax, so the individual's after-tax income is 90% of Y. In the population, the mean of Y is $50,000 per year, and Var(Y)=4×10 ⁸ dollars ² . The population mean after-tax income is

Group of answer choices

$45,000 per year

$40,000 per year

[not enough information]

[none of these]

2.Assume that you have the following probabilities for your final grade in your econometrics course, with the standard GPA scale of A=4, B=3, C=2, D=1, F=0 applies (ignoring +/- for simplicity). Letting Y be your grade, P(Y=4)=0.2, P(Y=3)=0.5, and P(Y<3)=1-0.2-0.5=0.3. The expected value E(Y)=

Group of answer choices

2.0

2.3

[not enough information]

[none of these]

3.You're part of a team developing an algorithm to tag uploaded pet photos as either "cat" or "dog" automatically. For simplicity, assume all pet photos are cat or dog. Currently, your algorithm estimates the probability that the photo is a dog, p=P(dog); your job is to translate this probability to either a "cat" or "dog" label. From consumer research, you know that cat owners really hate when their photos are incorrectly labeled "dog," but dog owners only care a little when their dog photos are incorrectly labeled "cat." You consider the loss function L(actual, label). A reasonable loss function might be

Group of answer choices

L(cat,cat)=L(dog,dog)=0, L(cat,dog)=10, L(dog,cat)=10

L(cat,cat)=L(dog,dog)=0, L(cat,dog)=1, L(dog,cat)=1

L(cat,cat)=L(dog,dog)=0, L(cat,dog)=10, L(dog,cat)=1

L(cat,cat)=L(dog,dog)=0, L(cat,dog)=1, L(dog,cat)=10

4.It takes either 1, 2, or 3 minutes for you to get to work (Y). If your guess (g) is correct, then you'll be exactly on time, so in your loss function L(y,g) you let L(1,1)=L(2,2)=L(3,3)=0. If you leave yourself 3 minutes and it only takes 1 or 2, then you waste the extra time sitting at your desk (you're not paid overtime and have to stay till 5:00pm regardless), which is only mildly annoying. If instead you guess 1 minute but it takes 2 or 3 minutes, you'll arrive late and might get fired, which you really don't want to happen. If you guess 2 minutes, you could be early or late or on time. If P(Y=1)=P(Y=2)=P(Y=3)=1/3, you should

Group of answer choices

try to be on time on average, guessing g=E(Y)=2

err on the early side, g=3

err on the late side, g=1

randomize such that you guess 1, 2, or 3 with probability 1/2 each

5.You're trying to predict whether it will rain on the day of your child's birthday party. The party is at a park. If it doesn't rain, then you can sit in the grass to eat. If it does rain, sitting in the grass will be very unpleasant, and you'd prefer to sit at the sheltered tables. But, the shelter costs $100 to reserve, and you have to decide 1 week ahead you can't just wait to see the actual weather before deciding. Assume you can't use the shelter if you don't reserve it. You quantify your loss in different scenarios as the loss function L(y,g), where y=1 if it actually rains and y=0 if not, and g is your guess: L(1,1)=L(0,0)=0, and L(0,1)=100, L(1,0)=500. Let P(Y=1)=0.2. Then, your optimal prediction is g=

6.Let P(Y=0)=1/4, P(Y=1)=1/2, P(Y=2)=1/4. Given the quadratic loss function L(y,g)=(y-g) ² , the expected loss (risk) of predictions 0, 1, and 2 are E[L(Y,0)]=___, E[L(Y,1)]=___, E[L(Y,2)]=___, and the best prediction among these three options is ___.

Group of answer choices

3/4, 1/2, 3/4, g=1

1, 1/2, 1, g=1

1.5, 0.5, 1.5, g=1

[none of these]