class: center, middle, inverse, title-slide # Growth Rate ## Macroeconomics (2018Q3) ### Kenji Sato ### Day 2 --- ## Growth Rate | Year | GDP | Symbol | |:-----:|:------:|:-------:| | 2011 | 15,591 | `\(Y_1\)` | | 2012 | 15,978 | `\(Y_2\)` | | 2013 | 16,274 | `\(Y_3\)` | | 2014 | 16,705 | `\(Y_4\)` | Growth rate between years 2011 and 2012 is `$$g_{2,1} = \frac{Y_2}{Y_1} - 1 \simeq 0.024822$$` --- ## Average growth rate The average annual growth rate can be computed by geometric average `$$\begin{aligned} \left( \frac{Y_4}{Y_1} \right)^{\frac{1}{4 - 1}} - 1 \simeq 0.0232714 \end{aligned}$$` Notice that `$$\begin{aligned} \left( \frac{Y_4}{Y_1} \right)^{\frac{1}{4 - 1}} - 1 = \underbrace{\sqrt[3]{\frac{Y_4}{Y_3} \frac{Y_3}{Y_2} \frac{Y_2}{Y_1}}}_{\text{geometric mean}} - 1 \end{aligned}$$` You don't use arithmetic mean to compute average growth rate. --- ## mean log difference `$$\begin{aligned} \frac{\ln Y_4 - \ln Y_1}{4 - 1} \simeq 0.0230048 \end{aligned}$$` Difference in log, devided by the length of the period, is close to the average growth rate. Geometric average corresponds to **effective rate**, while the mean log difference corresponds to **nominal rate**. --- ## Compounding Assume that a bank offers an annual, nominal interest rate of 6% compounded monthly and that you make a deposit of one thousand dollars ($1,000) at the bank today. How much do you expect to have in the bank account in one year from now? -- `$$1000 \times \left( 1 + \frac{0.06}{12} \right)^{12} \simeq 1061.6778119$$` The rate of return is about 6.17%, which is larger than 6%. --- ## Two annual interest rates 6% = Nominal annual interest rate 6.17% = Effective annual interest rate Be aware of the difference. As far as differential equations are concerned, nominal rates are more convenient. --- class: center, middle # Differential Equations --- ## More frequent compounding Compounded `\(N\)` times in one year. `$$\lim_{N\to \infty} \left( 1 + \frac{0.06}{N} \right)^{N} = e^{0.06} = \exp(0.06)$$` Alternatively, let `\(\Delta = 1/N\)` to get `$$\lim_{\Delta t \to 0} \left( 1 + 0.06 \Delta t \right)^{1/\Delta t} = e^{0.06} = \exp(0.06)$$` That is, * power 0.06 = nominal rate * `\(e^{0.06}\)` = effective rate Of course, log effective rate is nominal rate: `\(\ln e^{0.06} = 0.06\)` --- ## Relationship between nominal and effective rates `$$\begin{aligned} \left( \frac{Y_T}{Y_0} \right)^\frac{1}{T} &= \exp \left( \ln \left( \frac{Y_T}{Y_0} \right)^\frac{1}{T} \right) \\ &= \exp \left( \frac{\ln Y_T - \ln Y_0}{T} \right) \end{aligned}$$` `\(\frac{\ln Y_T - \ln Y_0}{T}\)` is the nominal rate. --- ## Constant nominal growth rate The most important case is when the nominal growth rate is constant. `$$\begin{aligned} \frac{\ln Y_t - \ln Y_0}{t} = g \end{aligned}$$` Rearranging, `$$\ln Y_t = \ln Y_0 + gt$$` which tells you why semilog plots exhibit linearity. By taking log of both sides: `$$Y_t = Y_0 e^{gt}$$` --- ## Continuous-time modeling Let `\(t\)` be any nonnegtive real number. To clarify that `\(Y\)` is a function of time, we rewrite as `\(Y(t) = Y(0) e^{gt}\)` <img src="day02_files/figure-html/unnamed-chunk-2-1.png" width="350px" style="display: block; margin: auto;" /> --- ## Differential equation Differentiate `$$Y(t) = Y(0) e^{gt}$$` with respect to time to get `$$\frac{dY}{dt} (t) = g Y(0) e^{gt} = gY(t).$$` For notational simplicity, we write the time-derivative as `$$\dot Y = \frac{dY}{dt}$$` --- ## Differential equation (cont'd) The following equation is an example of differential equations, in which a function is the unknown. `$$\dot Y = gY \quad \text{or}\quad \frac{\dot Y}{Y} = g$$` This is the most important differential equation we use. How can we get `\(Y(t) = Y(0)e^{gt}\)` from this differential equation? In other words, how can we **solve** this differential equation? --- ## Solving simple differential equation Notice that `$$\frac{d}{dt}\left[ \ln Y(t) \right] = \frac{\dot Y(t)}{Y(t)}$$` We thus have `$$\frac{d}{dt}\left[ \ln Y(t) \right] = g \Longrightarrow \int d\left[ \ln Y(t) \right] = \int gdt$$` Then `$$\ln Y(t) = gt + \text{const.}$$` --- ## Solving simple differential equation (cont'd) At `\(t = 0\)`, `$$\ln Y(0) = \text{const.}$$` We therefore have `$$\ln Y(t) = \ln Y(0) + gt \Longrightarrow Y(t) = Y(0) e^{gt}$$` To sum up, ** `$$\frac{\dot Y}{Y} = g \Longleftrightarrow Y(t) = Y(0) e^{gt}$$` ** --- ## Continuous-time modeling We will use **differential equations** to model the evolution of the economy over time (i.e., dynamics of the model). Why is this strategy helpful? --- ## Force driving the economy `\(K(t)\)` is the amount of physical capital at time `\(t\)` * machines * buildings * etc. Capital is used for current and future production and is a fundamental source of wealth of the economy. Consider an increase in `\(K(t)\)` as one form of economic growth. --- ## Force driving the economy (cont'd) **Why does `\(K(t)\)` increase?** **Why does `\(K(t)\)` decrease?** --- ## Force driving the economy (cont'd) **Why does `\(K(t)\)` increase?** → Investment * Let `\(I(t)\)` denote the speed of investment ($/year) at time `\(t\)`. **Why does `\(K(t)\)` decrease?** → Depreciation * Let `\(\delta\)` denote the annual rate of depreciation. * `\(\delta K(t)\)` is the speed of depreciation at time `\(t\)`. The balance between the two forces determines the path of capital accumulation. --- ## Investment Let `\(K(t)\)` be given. Let's observe how `\(K(t + \Delta t)\)` is determined. The speed of investment is `\(I(t)\)`. If the economy keeps that speed from `\(t\)` to `\(t + \Delta t\)`, then the increment of capital between the two dates would be `$$I(t)\Delta t$$` --- ## Capital depreciation The annual rate of depreciation is `\(\delta\)`. In `\(\Delta t\)` years, capital is reduced by `$$(\delta \Delta t) K(t)$$` --- ## Capital accumulation Combining these forces, `$$K(t + \Delta t) - K(t) = I(t)\Delta t - (\delta \Delta t) K(t)$$` -- Deviding by `\(\Delta t\)`, we get `$$\frac{K(t + \Delta t) - K(t)}{\Delta t} = I(t) - \delta K(t)$$` -- By taking limit `\(\Delta t \to 0\)`, `$$\dot K(t) = I(t) - \delta K(t)$$` --- ## Differential equation, too mathematical? Instead of directly modeling the variable of interest `\(K\)`, we start by describing increment/decrement of `\(K\)`. This modeling strategy gives a clear picture of how the economy evolves over time. * When `\(I(t) > \delta K(t)\)`, i.e., when investment exceeds depreciation, then `\(\dot K > 0\)` and thus `\(K\)` is growing. * When `\(I(t) < \delta K(t)\)`, i.e., when depreciation exceeds investment, then `\(\dot K < 0\)` and thus `\(K\)` is decreasing. * When `\(I(t) = \delta K(t)\)`, `\(K(t)\)` stays steadily.