Assignment1Sol

Assignment 1 Solution: ECON-UA 266 - Intro to Econometrics Sahar Parsa, Xiaotong Wu, Anne Schick, and Odhrain McCarthy Spring 2024 IMPORTANT DISCLAIMER: The homework is NOT graded. The points are only to give you information about the weight assigned to each questions. The first assignment solution will be released on Friday 2nd of February 2024 5PM – you have until then to try to solve the homework on your own (or with the help of the TA during recitation). It covers the material related to the first week of class. 1) You are encouraged to discuss the problems with others, but 2) you must write up your own results. This question will cover statistical inference, the relationship between a population parameter and the sample average. This question will cover statistical inference, the relationship between a population parameter and the sample average. Suppose you flip a coin and you are interested in the probability of getting a head. Assuming the coin is not rigged, the random variable X defined as 1 if it is a head and 0 if it is a tail is described by a Bernoulli distribution which takes the value 1 (head) with the probability p = 0 . 5 . Question [100 points] This question will cover statistical inference, the relationship between a population parameter and the sample average. Suppose you flip a coin and you are interested in the probability of getting a head. Assuming the coin is not rigged, the random variable X defined as 1 if it is a head and 0 if it is a tail is described by a Bernoulli distribution which takes the value 1 (head) with the probability p = 0 . 5 . 1. [5 points] What is the probability of getting a tail? What is the expected value (= population mean) of the random variable X ? What about the population variance? Write down the formula first and then calculate the value. Solution : The probability of getting a tail is equal to: 1 − p = 1 − 0 . 5 = 0 . 5 [ X ] = µ X = Pr ( X = 1) × 1 + Pr ( X = 0) × 0 = 0 . 5 × 1 = 0 . 5 = p V ar ( X ) = σ 2 X = (1 − µ X ) 2 Pr ( X = 1) + (0 − µ X ) 2 Pr ( X = 0) = (1 − 0 . 5) 2 0 . 5 + (0 − 0 . 5) 2 0 . 5 = 0 . 5 2 = 0 . 25 2. [5 points] Generate a sample of N = 10 observations by randomly drawing 10 times from the coin experiment, { X 1 , X 2 , . . . , X N } , in R. 1

Solution : We’re going to use the function rbinom in R, which randomly draws values from a Binomial distribution in R. A binomial distribution with one trial is a Bernouilli distribution. The function rbinom has three arguments rbinom ( a, b, c ) , where a is the sample size, b is the number of trials which is set to 1, and c is the probability of getting a value equal to 1. library (tidyverse) ## -- Attaching packages --------------------------------------- tidyverse 1.3.2 -- ## v ggplot2 3.4.0 v purrr 1.0.1 ## v tibble 3.1.8 v dplyr 1.0.10 ## v tidyr 1.3.0 v stringr 1.5.0 ## v readr 2.1.3 v forcats 0.5.2 ## -- Conflicts ------------------------------------------ tidyverse_conflicts() -- ## x dplyr::filter() masks stats::filter() ## x dplyr::lag() masks stats::lag() set.seed ( 123 ) sample <- rbinom ( 10 , 1 , 0.5 ) %>% as.data.frame () colnames (sample) <- c ( ' S1 ' ) sample ## S1 ## 1 0 ## 2 1 ## 3 0 ## 4 1 ## 5 1 ## 6 0 ## 7 1 ## 8 1 ## 9 1 ## 10 0 In order to replicate the values drawn in your sample, you need to define a seed at the beginning of your code, using “set.seed(123)” command. In our sample, we obtain 6 “heads” or 1. We obtain 4 “tails”, or 0. We are calling on the package tidyverse. tidyverse is a package that allows us to manipulate and vizualize dataset easily in R. We will define ggplot2 in the next question. We named the column S1 for Sample 1. 3. [5 points] Plot the histogram of the sample you generated in 2 using the package ggplot. Solution : ggplot (sample %>% mutate ( S1 = as.factor (S1)), aes ( x = S1)) + geom_bar ( aes ( y = (..count..) / sum (..count..))) + labs ( title = "Histogram Sample 1" , x = "Outcome" , y = "Frequency" ) ## Warning: The dot-dot notation (‘..count..‘) was deprecated in ggplot2 3.4.0. ## i Please use ‘after_stat(count)‘ instead. ## This warning is displayed once every 8 hours. ## Call ‘lifecycle::last_lifecycle_warnings()‘ to see where this warning was ## generated. 2

For our sample, the sample variable is equal to s 2 x = (0 − 0 . 6) 2 + (1 − 0 . 6) 2 + (0 − 0 . 6) 2 + (1 − 0 . 6) 2 + (1 − 0 . 6) 2 + (0 − 0 . 6) 2 + (1 − 0 . 6) 2 + (1 − 0 . 6) 2 + (1 − 0 . 6) 2 + (0 − 0 . 6) 2 10 = 0 . 36 + 0 . 16 + 0 . 36 + 0 . 16 + 0 . 16 + 0 . 36 + 0 . 16 + 0 . 16 + 0 . 16 + 0 . 36 10 = 2 . 4 10 = 0 . 24 As expected, the sample average and variance are similar to, but they differ from the the population average and the population variance of the Bernouilli random variable given in question 1. 5. [10 points] Generate another sample of size N = 10 and repeat 3 and 4. Do you observe a difference in the histogram, the sample average, and the sample variance in this new sample with the sample generated in 2? Explain. Solution : set.seed ( 234 ) sample $ S2 <- rbinom ( 10 , 1 , 0.5 ) sample ## S1 S2 ## 1 0 1 ## 2 1 1 ## 3 0 0 ## 4 1 1 ## 5 1 0 ## 6 0 1 ## 7 1 1 ## 8 1 1 ## 9 1 1 ## 10 0 0 ggplot (sample %>% mutate ( S2 = as.factor (S2)), aes ( x = S2)) + geom_bar ( aes ( y = (..count..) / sum (..count..))) + labs ( title = "Histogram Sample 2" , x = "Outcome" , y = "Frequency" ) 4

0.0 0.2 0.4 0.6 0 1 Outcome Frequency Histogram Sample 2 Notice that to obtain a different sample, we need to change the value inside set.seed . In this sample, we obtain 7 heads and 3 tails. The sample average is: ¯ x = QQQQQQQ 10 i =1 x i 10 = 1 + 1 + 0 + 1 + 0 + 1 + 1 + 1 + 1 + 0 10 = 7 10 = 0 . 7 and the sample variance is: s 2 x = QQQQQQQ 10 i =1 ( x i − ¯ x ) 2 10 = (1 − 0 . 7) 2 + (1 − 0 . 7) 2 + (0 − 0 . 7) 2 + (1 − 0 . 7) 2 + (0 − 0 . 7) 2 + (1 − 0 . 7) 2 + (1 − 0 . 7) 2 + (1 − 0 . 7) 2 + (1 − 0 . 7) 2 + (0 − 0 . 7) 2 10 = 0 . 09 + 0 . 09 + 0 . 49 + 0 . 09 + 0 . 49 + 0 . 09 + 0 . 09 + 0 . 09 + 0 . 09 + 0 . 49 10 = 2 . 1 10 ≈ 0 . 21 From the class notes, we know that different samples will generate different values for our statistics. It is worth pausing to discuss the formula we are using for the sample variance. You might have seen two formula for the sample variance. In the first formula, one divides the numerator by N and in the other one by N − 1 . These two formulas are the same when the sample size is very large. They will differ when the sample size is small. The main difference between the two formula is that the sample variance measured by dividing by N − 1 is an unbiased estimator of the population variance, while the sample variance measured by dividing by N will be a biased estimator. Why do we need to divide by N − 1 ? Intuitively, when we are estimating the population variance σ 2 X = [ X − [ X ]] 2 , one needs to estimate two unknown objects:the 5

Solution : Standard Deviation: sd (sampleaverage $ average) ## [1] 0.1671478 Sampling Average: mean (sampleaverage $ average) ## [1] 0.479 8. [10 points] Write down the formulas for the standard deviation as well as the mean of the sample average? Solution : The population standard deviation of the sample average is: σ ¯ X = rrrrrrr V ar ( ¯ X ) = sssssss σ 2 X N = qqqqqqq 0 . 25 / 10 The mean of the sample average is: [ ¯ X ] = µ X = 0 . 5 9. [15 points] Repeat 6, 7 and 8 but with a sample of size N = 100 . Solution : set.seed ( 111 ) sample <- replicate ( 100 , rbinom ( 100 , 1 , 0.5 )) %>% as.data.frame () sample.numcols <- sample[, sapply (sample, is.numeric)] # Use only the numeric columns -- all the columns sampleaverage <- apply (sample.numcols, 2 , function (x) mean (x, na.rm = TRUE )) %>% as.data.frame () colnames (sampleaverage) <- c ( ' average ' ) ggplot (sampleaverage, aes ( x = average)) + geom_histogram ( aes ( y = ..ncount..)) + labs ( title = "Histogram Sample Average" , x = "Sample Average" , y = "Frequency" ) ## ‘stat_bin()‘ using ‘bins = 30‘. Pick better value with ‘binwidth‘. 7

0.00 0.25 0.50 0.75 1.00 0.40 0.45 0.50 0.55 Sample Average Frequency Histogram Sample Average Solution : Standard Deviation: sd (sampleaverage $ average) ## [1] 0.0444454 Sampling Average: mean (sampleaverage $ average) ## [1] 0.5006 The population standard deviation of the sample average is: V ar ( ¯ X ) = sssssss σ 2 X N = qqqqqqq 0 . 25 / 100 The mean of the sample average is: [ ¯ X ] = µ X = 0 . 5 10. [10 points] How does the distribution (look at the histogram), the standard deviation, and the mean value of the sample averages differ when we increase the sample size from N = 10 to N = 100 ? How do the values compare to the values when you are using formulas derived in 8? Explain your results and the lesson learned (Hint: The sampling distribution changes with the sample size.) 8