Problem Sets
Click below to expand the problem sets that you should complete after each class. You should also watch the lecturer’s demonstration lectures on Canvas when they become available. These will mainly cover past exam questions.
Try each problem before revealing its solution. Each problem has its solution in a collapsed box directly beneath it. Problems 1–3 come from a previous year’s demonstration lecture. Problems 4–8 come from exams that I wrote for Econ 103 at the University of Pennsylvania and from past Oxford Prelims collections.
Prove, using the probability axioms
- The Complement Rule: \(P(A^c) = 1 - P(A)\).
- The Probability of the Union of Two Events Rule: \(P(A\cup B) = P (A) + P (B) - P (A\cap B)\)
- The Bounds on Probabilities Rule (Boole’s inequality): \(P (A \cup B) \leq P (A) + P (B)\)
- The Logical Consequence Rule: If \(B\) logically entails \(A\) then \(P (B) \leq P (A)\)
TipSolutionThe axioms are: (1) \(P(A) \geq 0\) for any event \(A\); (2) \(P(\Omega) = 1\); and (3) if \(A_1, A_2, \ldots\) are mutually exclusive then \(P(A_1 \cup A_2 \cup \cdots) = P(A_1) + P(A_2) + \cdots\).
Either \(A\) happens or it does not, so \(A \cup A^c = \Omega\). By Axiom 2, \(P(A \cup A^c) = P(\Omega) = 1\). Since \(A\) and \(A^c\) are mutually exclusive, \(P(A \cup A^c) = P(A) + P(A^c)\) by Axiom 3. Hence \(P(A) + P(A^c) = 1\), and rearranging gives \(P(A^c) = 1 - P(A)\).
Split \(A \cup B\) into two mutually exclusive pieces, according to whether \(A\) happens: \(A \cup B = A \cup (A^c \cap B)\), since an outcome in \(A \cup B\) that is not in \(A\) must be in \(B\). By Axiom 3, \(P(A \cup B) = P(A) + P(A^c \cap B)\). Now split \(B\) according to whether \(A\) happens: \(B = (A \cap B) \cup (A^c \cap B)\). These two pieces are also mutually exclusive. By Axiom 3, \(P(B) = P(A \cap B) + P(A^c \cap B)\), so \(P(A^c \cap B) = P(B) - P(A \cap B)\). Substituting into the first expression gives \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\).
By part (b), \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\). By Axiom 1, \(P(A \cap B) \geq 0\). Hence \(P(A \cup B) \leq P(A) + P(B)\).
If \(B\) logically entails \(A\), then \(A\) happens whenever \(B\) happens, so \(B \subseteq A\). Because \(B \subseteq A\), every outcome in \(A\) is either in \(B\) or in \(A \cap B^c\), so \(A\) splits into two mutually exclusive pieces: \(A = B \cup (A \cap B^c)\). By Axiom 3, \(P(A) = P(B) + P(A \cap B^c)\). By Axiom 1, \(P(A \cap B^c) \geq 0\), so \(P(A) \geq P(B)\).
Consider the experiment of tossing a coin repeatedly and counting the number of coin tosses required until the first head appears.
- Write down the sample space.
- Let \(A\) be the event that the number of coin tosses required for the first head to appear is even. Let \(B\) be the event that the number of coin tosses required until the first head appears is less than 5. Describe events \(A\) and \(B\) as sets.
TipSolutionThe basic outcomes are the possible numbers of tosses: the first head could appear on the first toss, the second, the third, and so on. Hence \(\Omega = \{1, 2, 3, \ldots\}\). This sample space is infinite, because there is no upper limit on the number of tosses needed for the first head.
\(A = \{2, 4, 6, \ldots\}\) and \(B = \{1, 2, 3, 4\}\).
Consider the experiment of tossing a coin 2 times. Consider the events: \[ \begin{aligned} A &= \text{There is at most one Head}\\ B &= \text{There is at least one Head and there is at least one Tail} \end{aligned} \]
- Are events \(A\) and \(B\) independent?
- What if you toss the coin 3 times?
- What if you toss the coin 4 times?
TipSolutionAssume that the coin is fair and that the tosses are independent, so that every sequence of heads and tails is equally likely. In each case, \(A \cap B\) is the event that there is exactly one head. “At most one head” rules out two or more heads, and “at least one head” rules out zero. With two or more tosses, exactly one head leaves at least one tail, so the requirement of a tail in \(B\) adds nothing. The events are independent if and only if \(P(A \cap B) = P(A)P(B)\).
There are four equally likely outcomes: HH, HT, TH, TT. \(A = \{HT, TH, TT\}\), so \(P(A) = 3/4\). \(B = \{HT, TH\}\), so \(P(B) = 1/2\). \(A \cap B = \{HT, TH\}\), so \(P(A \cap B) = 1/2\). Since \(P(A)P(B) = (3/4)(1/2) = 3/8 \neq 1/2 = P(A \cap B)\), \(A\) and \(B\) are not independent. In fact \(B \subseteq A\), so if \(B\) happens then \(A\) must happen, and \(P(A|B) = 1 \neq 3/4 = P(A)\).
There are eight equally likely outcomes. \(A\) contains TTT and the three outcomes with exactly one head, so \(P(A) = 4/8 = 1/2\). \(B\) contains every outcome except HHH and TTT, so \(P(B) = 6/8 = 3/4\). \(A \cap B\) contains the three outcomes with exactly one head, so \(P(A \cap B) = 3/8\). Since \(P(A)P(B) = (1/2)(3/4) = 3/8 = P(A \cap B)\), \(A\) and \(B\) are independent.
There are sixteen equally likely outcomes. \(A\) contains TTTT and the four outcomes with exactly one head, so \(P(A) = 5/16\). \(B\) contains every outcome except HHHH and TTTT, so \(P(B) = 14/16 = 7/8\). \(A \cap B\) contains the four outcomes with exactly one head, so \(P(A \cap B) = 4/16 = 1/4\). Since \(P(A)P(B) = (5/16)(7/8) = 35/128 \neq 32/128 = P(A \cap B)\), \(A\) and \(B\) are not independent.
Events with the same descriptions are independent with three tosses but not with two or four. Whether two events are independent depends on their probabilities, so it has to be checked by computing whether \(P(A \cap B) = P(A)P(B)\), or equivalently whether \(P(A|B) = P(A)\).
Let \(A\) and \(B\) be two events where \(0<P(B)<1\). Prove that \[P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)\] You may use any rule except the one you are being asked to prove. Name each rule that you use.
TipSolutionFirst, note that \(A\cap B\) and \(A \cap B^c\) are mutually exclusive and \[A = (A\cap B) \cup (A \cap B^c)\] Therefore, by the third axiom of probability, \[P(A) = P(A\cap B) + P(A \cap B^c)\] Finally, by the multiplication rule, \[P(A\cap B) = P(A|B)P(B)\] and \[P(A \cap B^c) = P(A|B^c)P(B^c)\] Combining these, \[P(A) = P(A|B)P(B) + P(A|B^c)P(B^c)\]
From the HT 2022 collection.
Ruth wants to know the fraction of undergraduates who smoke marijuana. Because she worries that her subjects may be unwilling to admit to breaking the law, Ruth conducts her interviews as follows:
Hi, I’m carrying out a survey on marijuana use. In a moment I’m going to ask if you smoke marijuana, but before I do I’d like you to flip this fair coin in secret. If you get heads, answer my question truthfully. If you get tails, answer YES regardless of whether you smoke marijuana. Because I don’t know the outcome of your coin flip, if you answer YES I’ll never know whether you really smoke marijuana or just flipped a tails.
Ruth interviews a large number of subjects who all follow her instructions exactly: 85% answer YES. Calculate the fraction of undergraduates who smoke marijuana.
TipSolutionLet \(Y\) be the event that someone answers YES, \(M\) be the event she smokes marijuana, and \(T\) be the event that she flips a tails. Since \(Y = M \cup T\), by the Probability of the Union of Two Events Rule we have \[P(Y) = P(M) + P(T) - P(M\cap T)\] Since the coin is fair, \(P(T) = 0.5\). Since the outcome of a coin toss doesn’t depend on whether someone does or doesn’t smoke marijuana, \(M\) and \(T\) are independent which implies \(P(M\cap T) = P(M)P(T)\). Substituting this information along with \(P(Y) = 0.85\), \[0.85 = P(M) + 0.5 - 0.5 \times P(M)\] and solving we see that \(P(M) = 0.7\).
Alternatively, you could solve this using the Law of Total Probability: \[\begin{aligned} P(Y) &= P(Y|T)P(T) + P(Y|T^c)P(T^c) \\ 0.85 &= 1 \times 1/2 + P(Y|T^c)\times 1/2\\ 0.7 &= P(Y|T^c) \end{aligned}\] using the fact that \(P(Y|T^c) = P(M)\) since anyone who flips heads answers truthfully, and the outcome of the coin flip has no bearing on whether someone smokes marijuana.
Suppose that 1 out of \(N\) coins is defective and I chose my coin at random. Whereas a regular coin is equally likely to come up heads or tails, a defective coin always comes up heads. I choose a coin at random and flip it ten times: I get ten heads. Let \(D\) be the event that my coin is defective and \(A\) be the event that I get ten heads in ten flips of my coin.
What is \(P(A|D)\)?
TipSolution\(P(A|D) = 1\) since a defective coin always comes up heads.
What is \(P(A|D^c)\)?
TipSolution\(P(A|D^c) = 1/2^{10} = 1/1024\) since a regular coin is equally likely to come up heads or tails, and each flip is independent.
Expressed as a function of \(N\), what is \(P(D)\)?
TipSolution\(P(D) = 1/N\) since 1 out of \(N\) coins is defective and I chose a coin at random.
Expressed in terms of \(N\), what is \(P(D^c)\)?
TipSolutionBy the complement rule \(P(D^c) = 1 - P(D) = 1 - 1/N = (N-1)/N\).
Expressed in terms of \(N\), what is \(P(A)\)?
TipSolutionBy the law of total probability: \[\begin{aligned} P(A) &= P(A|D) \times P(D) + P(A|D^c) \times P(D^c)\\ &= 1/N \times 1 + (N-1)/N \times 1/1024 = (1024 + N - 1) / (N \times 1024) \end{aligned}\]
Expressed in terms of \(N\), what is \(P(D|A)\)?
TipSolutionBy Bayes’ rule \[P(D|A) = \frac{P(A|D)P(D)}{P(A)} = \frac{1/N}{(1024 + N - 1)/(N \times 1024)} = \frac{1024}{1024 + N - 1}\]
Approximately 80% of all emails sent over the internet are spam. About 10% of spam emails contain the word “viagra” compared to 1% of non-spam emails. About 5% of spam emails contain the word “herbal” compared to 3% of non-spam.
Assume that the occurrences of words in emails are independent for both spam and non-spam. If an email contains both the words “viagra” and “herbal” what is the probability that it is spam?
TipSolutionUse Bayes’ Rule: \[P(\text{spam}|\text{herbal}\cap\text{viagra}) = \frac{P(\text{herbal}\cap\text{viagra}|\text{spam})P(\text{spam})}{P(\text{herbal}\cap\text{viagra})}\] By the Law of Total Probability, \[\begin{aligned} P(\text{herbal}\cap\text{viagra}) &= P(\text{herbal}\cap\text{viagra}|\text{spam})P(\text{spam})\\ &\quad + P(\text{herbal}\cap\text{viagra}|\text{non-spam})P(\text{non-spam})\\ &= 1/10 \times 1/20 \times 4/5 + 1/100 \times 3/100 \times 1/5\\ &= 1/1000 \times (4 + 3/50) \end{aligned}\] Hence, \[P(\text{spam}|\text{herbal}\cap\text{viagra}) = \frac{4}{4 + 3/50} \approx 99\%\]
After completing your calculations, you learn an additional piece of information: approximately 14.5% of spam emails contain the word “herbal” or “viagra.” Does this new information support or contradict the assumption that words appear independently in emails? Explain.
TipSolutionThis question is only for spam emails, so we won’t explicitly write the conditioning on spam. Let \(H\) be the event that a spam email contains at least one occurrence of “herbal” and \(V\) be the event that it contains at least one occurrence of “viagra.” We are given that \(P(V) = 0.1\), \(P(H) = 0.05\) and \(P(V\cup H) = 0.145\). By the Probability of the Union of Two Events Rule \(P(V\cup H) = P(V) + P(H) - P(V\cap H)\) Substituting the known probabilities, we see that \(0.145 = 0.15 - P(V\cap H)\). Hence, \(P(V\cap H) = 0.15 - 0.145 = 0.005\). Independence requires that \(P(V\cap H) = P(H)P(V)= 0.1 \times 0.05 = 0.005\) so this piece of information tells us that, at least for spam emails, the words “herbal” and “viagra” occur independently.
From the HT 2022 collection.
A plane has crashed in one of three possible locations: the mountains (\(M\)), the desert (\(D\)), or the sea (\(S\)). Based on its flight path, experts have calculated the following prior probabilities that the plane is in each location: \(P(M) = 0.5\), \(P(D) = 0.3\) and \(P(S) = 0.2\). If we search the mountains then, given that the plane is actually there, we have a 30% chance of failing to find it. If we search the desert then, given that the plane is actually there, we have a 20% chance of failing to find it. Finally, if we search the sea then, given that the plane is actually there, we have a 90% chance of failing to find it. Naturally if the plane is not in a particular location but we search for it there, we will not find it. You may assume that searches in each location are independent given the location of the plane. Let \(F_M\) be the event that we fail to find the plane in the mountains. Define \(F_D\) and \(F_S\) analogously.
We started by searching the mountains. We failed to find the plane. What is the conditional probability that the plane is nevertheless in the mountains? Explain.
TipSolutionBy Bayes’ Rule: \(P(M|F_M) = P(F_M|M)P(M)/P(F_M)\). We first calculate the denominator using the Law of Total Probability: \[\begin{aligned} P(F_M) &= P(F_M|M)P(M) + P(F_M|M^c)P(M^c) \\ &= 0.3 \times 0.5 + 1 \times 0.5 = 0.15 + 0.5 = 0.65 \end{aligned}\] Hence, the desired probability is \(15/65 = 3/13 \approx 0.23\).
After failing to find the plane in the mountains, we searched the desert, and the sea. We did not find the plane in either location. After this more exhaustive search what is the conditional probability that the plane is in the mountains? Explain.
TipSolutionWe are asked to calculate \(P(M|F_M \cap F_D \cap F_S)\). By Bayes’ rule, \[P(M|F_M \cap F_D \cap F_S) = \frac{P(F_M \cap F_D \cap F_S|M)P(M)}{P(F_M \cap F_D \cap F_S)}\] Define the shorthand \(A = F_M \cap F_D \cap F_S\). By the Law of Total Probability \[\begin{aligned} P(A) &= P(A|M)P(M) + P(A|D)P(D) + P(A|S)P(S) \\ &= (0.3 \times 1 \times 1) \times 0.5 + (1 \times 0.2 \times 1) \times 0.3 + (1 \times 1 \times 0.9 ) \times 0.2 \\ &= 0.15 + 0.06 + 0.18 = 0.39 \end{aligned}\] using independence. Hence, the desired probability is \(15/39 \approx 0.38\).
- Consider a random variable \(X\) which is uniformly distributed \(U(0, 1)\).
- Draw the probability density function (pdf), \(f(x)\). What is the formula for the pdf?
- Draw the cumulative distribution function (cdf), \(F(x)\). What is the formula for the cdf?
- Consider a random variable \(X\) which is uniformly distributed \(U(0, 1)\).
- What is the distribution of \(Z = 2 + X\)?
- What is the distribution of \(Z = 3X\)?
- What is the distribution of \(Z = 2 + 3X\)?
- Consider a random variable \(X\) which is uniformly distributed \(U(-2, 3)\).
- What is the distribution of \(Z = 2 + X\)?
- What is the distribution of \(Z = 3X\)?
- What is the distribution of \(Z = 2 + 3X\)?
- For a random variable \(X \sim U (2, 12)\) (read the \(\sim\) as “that is distributed”)
- What is \(P(3 \leq X \leq 8)\)?
- What is \(P(X = 9)\)?
- Consider a random variable \(X\) which is normally distributed with mean 0 and variance of 1.
- What is the distribution of \(Z = 5 + X\)?
- What is the distribution of \(Z = 2X\)?
- What is the distribution of \(Z = 2 + 3X\)?
- Consider a random variable \(X\) which is normally distributed with mean 2 and variance of 9.
- What is the distribution of \(Z = -2 + X\)?
- What is the distribution of \(Z = X/\sqrt{9}\)?
- What is the distribution of \(Z = (-2+X)/\sqrt{9}\)?
- If \(X \sim N (\mu=-1, \sigma^2 =2)\) what is
- \(P(X = 0)\)?
- \(P(X \leq -1)\)?
- \(P(-1 \leq X \leq 1)\)?
- the value of \(t\) such that \(P(X \leq t) = 0.25\)
- the value of \(t\) such that \(P(X > t) = 0.80\)
- the value of \(t\) such that \(P(-|t| \leq X \leq |t|) = 0.9\)
- Suppose that students’ marks on the economics prelims paper are normally distributed with mean 61 and standard deviation 9.5.
- What is the probability that a student scores (i) less than 50? (ii) 70 or more?
- What score is exceeded by only 10% of students?
- Find the median, and the upper and lower quartiles.
- What proportion of students have scores within 5 marks of the mean?
- Consider the following bivariate probability mass function \(f(X, Y )\) \[
\begin{array}{l|ccc}
& X=0 & X=2 & X=3 \\
\hline
Y=0&1/36 & 1/18 & 1/12\\
Y=1& 1/9 & 2/9 & 1/3\\
Y=2& 1/36 &1/18 & 1/12
\end{array}
\] what is
- the marginal PMF of \(X\)?
- the marginal CDF of \(Y\)?
- the conditional PMF of \(Y\): \(f(Y |X = 3)\)?
- the conditional CDF of \(X\): \(F(X|Y = 2)\)?
- Are \(X\) and \(Y\) independent?
- Let \(X\) and \(Y\) be random variables and \(a,b,c\) be constants. Use the linearity of expectation to derive the following results:
- \(\text{Cov}(X,Y) = \mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)\)
- \(\text{Var}(a + bX + cY) = b^2 \text{Var}(X) + c^2 \text{Var}(Y) + 2bc \text{Cov}(X,Y)\)
- Let \(X\) and \(Y\) be random variables such that \(\mathbb{E}[Y] \neq 0\). Show that, provided that all of the relevant expectations exist and are finite, \[\frac{\mathbb{E}[X]}{\mathbb{E}[Y]} - \mathbb{E}\left[ \frac{X}{Y}\right] = \frac{\text{Cov}(X/Y,\, Y)}{\mathbb{E}[Y]}.\]
- Let \(X\) be a random variable and \(a, b\) be constants. Suppose that \(Y = a + b X\). Calculate the correlation between \(X\) and \(Y\) in each of the following cases:
- \(b > 0\)
- \(b < 0\)
- \(b = 0\)
- Let \(X\) and \(U\) be random variables and \(a, b\) be constants. Suppose that \(Y = a + b X + U\) where \(\mathbb{E}(U) = 0\) and \(\text{Cov}(X, U) = 0\).
- Calculate \(\mathbb{E}(Y)\).
- Calculate \(\text{Var}(Y)\).
- Show that \(\text{Cov}(X, Y) = b \text{Var}(X)\)
- Show that \(\text{Corr}(X,Y) = b \sqrt{\frac{\text{Var}(X)}{b^2 \text{Var}(X) + \text{Var}(U)}}\).
- Let \(X\) be a \(\text{Bernoulli}(p)\) random variable, i.e. \(X \in \{0, 1\}\) with \(\mathbb{P}(X=1) = p\). Show that, for any other random variable \(Y\) such that the relevant expectations exist and are finite \[ \frac{\text{Cov}(X,Y)}{\text{Var}(X)} = \mathbb{E}(Y|X=1) - \mathbb{E}(Y|X=0). \]
- Define the terms “estimand”, “estimator”, and “estimate” using the median as an example.
- Let \(X_1, X_2, \dots, X_N \sim \text{iid}\) with mean \(\mu\) and variance \(\sigma^2\) and let \(\bar{X}_N \equiv \frac{1}{N} \sum_{i=1}^N X_i\).
- Is \(\bar{X}_N\) an unbiased estimator of \(\mu\)?
- Is \((0.1 X_1 + 0.9 X_2)\) an unbiased estimator of \(\mu\)? Why or why not?
- Which is a more efficient estimator of \(\mu\): \((0.1 X_1 + 0.9 X_2)\) or \((0.5 X_1 + 0.5 X_2)\)? Explain.
- Suppose that \(\mu\) is known. Is \(\widehat{\sigma}^2 \equiv \frac{1}{N-1}\sum_{i=1}^N (X_i - \mu)^2\) an unbiased estimator of \(\sigma^2\)? Explain.
- Consider a fictional university, Camford University, which is composed of a large number of colleges. Each college has the same number of first-year economics students. Suppose that students’ marks on the economics prelims paper are normally distributed with mean 61 and standard deviation 9.5.
- What is the distribution of the sample mean in a random sample of size \(N\)?
- In a random sample of 10 students, what is the probability that their average mark exceeds 63?
- Suppose that you have a sample of 10 students that is selected by choosing a college at random, and then choosing 10 students at random from the college. What can you say about the expected value and variance of their average mark? Explain.
- In a random sample of 1165 Oxford PPE applicants, the average score on the TSA test was 60.86, with a standard deviation of 8.02. Construct a 95% confidence interval for the population mean score.
- Suppose you are studying the impact of COVID-19 on the labor market. In a random sample of 1000 workers, you find that 354 had been “furloughed” in July 2020. Construct a 90% confidence interval for the population mean furlough rate.
- Suppose that we observe hourly wages for a random sample of 150 university graduates: the sample mean is £31 with a standard deviation of £15. In contrast, the sample mean wage is £17 with a standard deviation of £10 for a sample of 225 non-university graduates. Construct an approximate 95% confidence interval for the difference in population mean wages \((\mu_X - \mu_Y)\) between university graduates (X) and non-university graduates (Y).
- You are interested in whether the average rent for a room in Oxford is £500 per month. In a random sample of 300 rooms to rent in Oxford, the average rent is £525 per month with a standard deviation of £200. Test the hypothesis that the average Oxford room rent per month is £500.
- Suppose you have a large random sample \(X_1, \ldots, X_n \sim \text{iid} \, N(\mu, \sigma^2)\), and you test \(H_0: \mu = \mu_0\) vs. \(H_1: \mu \neq \mu_0\) where \(\mu_0\) is some hypothesized value of \(\mu\). You decide to set a significance level, \(\alpha\), at which to conduct the hypothesis test.
- Say you choose \(\alpha = 0.05\) and reject. Would you still have rejected if you had instead chosen \(\alpha = 0.1\)? Explain.
- Say you choose \(\alpha = 0.01\) and fail to reject. Would you have failed to reject if you had instead chosen \(\alpha = 0.05\)? Explain.
- A random sample of 35 Netflix users are asked to rate two shows: The Queen’s Gambit and Bridgerton. You can access the dataset as a .csv file from this url. Use the data to test the null hypothesis that both shows are equally highly rated.
- You are interested in whether the Oxford unemployment rate is the same as the national unemployment rate (5%). In a random sample of 1,000 Oxford residents, you find that 63 are unemployed. Test the hypothesis that the Oxford unemployment rate is the same as the national rate.
- Consider two groups A and B composed of 100 people who are suffering from a disease. A drug is given to group A. A placebo is given to group B. It is found that in group A 75 people recover whereas in group B only 65 people recover.
- Test the hypothesis that the drug helped cure the disease.
- Now suppose that each group contains 300 people, of whom 225 recover in group A and 195 recover in group B. What do you conclude?
- At this url you can find a plot illustrating the change in global surface temperatures over time, relative to the 1951-1980 average. With reference to the plot, discuss the key differences between time series and cross-sectional data.
- If \(X_t\) is white noise with mean \(\mu\) and variance \(\sigma^2\), what process does its first difference (\(\Delta X_t = X_t - X_{t-1}\)) follow? What is the mean, variance, first order autocovariance, and second order autocovariance of the process?
- Consider the following time series process: \[X_t = \epsilon_t + 0.5\epsilon_{t-1} + 0.5\epsilon_{t-2}\] where \(\epsilon_t\) is identically and independently distributed with \(\epsilon_t \sim N(0, \sigma^2)\).
- What is the mean and variance of \(X_t\)?
- What are the first, second, and third order autocovariances?
- Let \(\epsilon_t\) be identically and independently distributed with \(\epsilon_t \sim N(0, \sigma^2)\). Consider the following time series process: \[X_t = \alpha + \beta X_{t-1} + \epsilon_t\]
- What is the name commonly given to this type of process? What if \(\alpha = 0\) and \(\beta = 1\)? What if \(\alpha = 0\) and \(\beta = 0\)?
- By recursing back one period, show that: \[X_t = \alpha + \beta \alpha + \beta^2 X_{t-2} + \epsilon_t + \beta \epsilon_{t-1}\]
- By using the results on sequences covered in Chapter 4 of the Maths Workbook, show that \[ X_t = \frac{1 - \beta^t}{1 - \beta} \alpha + \beta^t X_0 + \sum_{j=0}^{t-1} \beta^j \epsilon_{t-j}\] Suppose \(0 \leq \beta < 1\) and \(X_0\) is a random variable with mean \(\frac{\alpha}{1 - \beta}\) and variance \(\frac{\sigma^2}{1 - \beta^2}\), and is uncorrelated with \(\epsilon_t\) for all \(t > 0\).
- Find the mean, variance and autocovariance function of \(X_t\) for any \(t \geq 1\)? Is the process (weakly) stationary?
- In studies dating back over 100 years, it’s well established that “regression toward the mean” occurs between the heights of parents and the heights of their adult children. Indicate whether the following statement is true or false: “Parents of tall children will tend to be taller than their children.”
In “Long-Lasting Effects of Socialist Education” by Nicola Fuchs-Schundeln and Paolo Masella the authors study the effects of a socialist education among children educated in the former East Germany (GDR). Socialist teaching was an official school subject in the GDR from 7th grade on and aimed to provide a deep knowledge of Marxism-Leninism, and of the socialist system of the GDR. The authors also suggest that an important feature of this educational system was that “critical thinking was not incentivized and divergent opinions were suppressed”. The authors were interested in the causal effect of exposure to the GDR’s educational system on subsequent labor market outcomes. To study this they exploit the very rapid reorganization of the school system in the East after the fall of the Berlin Wall (November 9, 1989). The socialist elements of the curriculum were dropped almost immediately. The authors used the German Microcensus (2005-08) to look at individuals in date-of-birth cohorts born between 1974 and 1977 in the GDR. These date-of-birth cohorts were still in education at reunification. The outcome variable of interest was whether or not they were employed at the time of the Microcensus (2005-08). The study relies on the following institutional feature: in the GDR, children turning six on or before May 31 of a given year were automatically enrolled in the first grade by September 1 of the same year. Children born on or after June 1 waited another year before starting school. The authors define their control group as consisting of individuals born on or after June 1. The treatment group consisted of individuals born on or before May 31. Treated individuals within each date-of-birth cohort were exposed to a year more of socialist teaching (and one year less of Western education) than individuals in the control group. The authors find that, comparing males in the treatment and control groups, the proportion of males who were employed was 0.021 lower in the treatment than the control group with a standard error of 0.004. Comparing females in the treatment and control groups, the proportion of females who were employed was 0.009 lower in the treatment than the control group with a standard error of 0.019.
- Conduct suitable hypothesis tests to determine whether the difference in employment rates between the treatment and control groups for men and women respectively were statistically significant. What do you conclude?
- The authors interpret their results (subject to the results of the statistical tests above) as causal effects. Using the potential outcomes framework and the standard notation write down an expression describing this causal effect.
- What are the principal assumptions which would justify the authors’ claim that this effect is causal? Why might these assumptions not hold?
- Unfortunately the German Microcensus only records the current residence of respondents and not their residence at the time when they were at school. The authors therefore assumed that respondents currently residing in the East also received their education in the East. Why might this not be the case? If this is not the case how would it bias the results of the study?
You are interested in the effect of taking extra online classes (e.g. edX or Khan Academy courses) on whether teenagers apply to university. In a sample of 1000 young adults you observe whether or not they applied to university and whether or not they took at least one online class. You also observe whether they got at least five A*-C grades at GCSE. The tables show the numbers in the sample of the various types of young adult and the proportion who applied to university.
Numbers of young adults by GCSE grades and whether took at least one online course No. Online Courses 0 ≥1 GCSE Grades < 5 A*-C 150 100 ≥ 5 A*-C 250 500 Proportion applying to University No. Online Courses 0 ≥1 GCSE Grades < 5 A*-C 0.3 0.8 ≥ 5 A*-C 0.3 0.5 - Calculate the average difference in university application rates between those who did at least one online course and those who did not.
- Does this difference have a causal interpretation? Why or why not?
- A friend who has done a statistics course suggests that controlling for GCSE grades might help in this context. What does this mean and why might it help?
- Calculate the conditional treatment effects for those with less than 5 A*-C grades at GCSE and for those with at least 5 A*-C at GCSE.
- Calculate the Average Treatment Effect and the Average Effect of Treatment on the Treated and compare them to the difference you found in part (a).