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This unit provides an overview of the key concepts and tools of modern economics, covering topics such as market behavior, institutional analysis, and policy formation. The course will begin with a comparison of historical and cross-country economic trends to provide context for the study of economic analysis. The course will then focus on the behavior of economic actors in the goods, labor, and credit markets, exploring how institutions and policy shape economic outcomes.
The course will examine both successful and failed market systems and study key economic variables such as GDP, unemployment, inequality, and inflation. Students will also learn about monetary and fiscal policy and explore how these policies affect the global economy. To analyze these topics, students will use empirical data, graphical and mathematical models, and historical and methodologically informed narrative.
Through their coursework, students will develop the skills to communicate economic concepts and ideas to both specialist and non-specialist audiences. This will include the ability to analyze and interpret economic data and use economic models to explain economic phenomena. Overall, this course provides a foundation in modern economics and prepares students to apply economic analysis to a wide range of issues and questions.
Let $X$ and $Y$ be random variables with finite variances. (i) Show that $$ \min _{g(\cdot)} E(Y-g(X))^2=E(Y-E(Y \mid X))^2 $$ where $g(\cdot)$ ranges over all functions.
(i) Let $g^(x) = E(Y \mid X=x)$, then we have \begin{align} E(Y – g^(X))^2 &= E((Y – E(Y \mid X)) + (E(Y \mid X) – g^(X)))^2 \ &= E((Y – E(Y \mid X))^2) + E((E(Y \mid X) – g^(X))^2) \ &\qquad + 2E((Y – E(Y \mid X))(E(Y \mid X) – g^(X))) \ &= E((Y – E(Y \mid X))^2) + E((E(Y \mid X) – E(Y \mid X))^2) \ &= E((Y – E(Y \mid X))^2), \end{align*} where we have used the fact that $E(E(Y \mid X)) = E(Y)$ and the law of iterated expectations.
(ii) Assume $m(X)=E(Y \mid X)$ and write $Y=m(X)+e$. Show that $\operatorname{Var}(Y)=$ $\operatorname{Var}(m(X))+\operatorname{Var}(e)$
(ii) We have \begin{align*} \operatorname{Var}(Y) &= E(Y – E(Y))^2 \ &= E((m(X) + e – E(m(X) + e))^2) \ &= E((m(X) – E(m(X)))^2) + E(e^2) + 2E((m(X) – E(m(X)))e) \ &= \operatorname{Var}(m(X)) + \operatorname{Var}(e) + 2E((m(X) – E(m(X)))e), \end{align*} where we have used the fact that $E(e) = 0$ and $\operatorname{Var}(m(X)) = E((m(X) – E(m(X)))^2)$.
To show that $E((m(X) – E(m(X)))e) = 0$, note that \begin{align*} E((m(X) – E(m(X)))e) &= E(E((m(X) – E(m(X)))e \mid X)) \ &= E((m(X) – E(m(X)))E(e \mid X)) \ &= 0, \end{align*} where we have used the fact that $E(e \mid X) = 0$.
(iii) If $E(Y \mid X=x)=a+b x$ find $E(Y X)$ as a function of moments of $X$.
(iii) We have \begin{align*} E(YX) &= E((a+bX)X) \ &= aE(X) + bE(X^2). \end{align*}
