Fix typos (#266)

* Update simple_linear_regression.md

* Update time_series_with_matrices.md

* Update input_output.md

Author: pitmonticone

lectures/input_output.md

@@ -118,7 +118,7 @@ A basic framework for their analysis is
 
 
 
-After  introducing the input-ouput model, we describe some of its connections to {doc}`linear programming lecture <lp_intro>`.
+After  introducing the input-output model, we describe some of its connections to {doc}`linear programming lecture <lp_intro>`.
 
 
 ## Input output analysis
@@ -307,7 +307,7 @@ L
 ```
 
 ```{code-cell} ipython3
-x = L @ d   # solving for gross ouput
+x = L @ d   # solving for gross output
 x
 ```
 
@@ -434,7 +434,7 @@ $$
 
 The primal problem chooses a feasible production plan to minimize costs for delivering a pre-assigned vector of final goods consumption $d$.
 
-The dual problem chooses prices to maxmize the value of a pre-assigned vector of final goods $d$ subject to prices covering costs of production.
+The dual problem chooses prices to maximize the value of a pre-assigned vector of final goods $d$ subject to prices covering costs of production.
 
 By the [strong duality theorem](https://en.wikipedia.org/wiki/Dual_linear_program#Strong_duality),
 optimal value of the primal and dual problems coincide:
@@ -482,7 +482,7 @@ plt.show()
 
 ## Leontief inverse
 
-We have discussed that gross ouput $x$ is given by {eq}`eq:inout_2`, where $L$ is called the Leontief Inverse.
+We have discussed that gross output $x$ is given by {eq}`eq:inout_2`, where $L$ is called the Leontief Inverse.
 
 Recall the {doc}`Neumann Series Lemma <eigen_II>` which states that $L$ exists if the spectral radius $r(A)<1$.
 
@@ -551,7 +551,7 @@ The above figure indicates that manufacturing is the most dominant sector in the
 
 ### Output multipliers
 
-Another way to rank sectors in input output networks is via outuput multipliers.
+Another way to rank sectors in input output networks is via output multipliers.
 
 The **output multiplier** of sector $j$ denoted by $\mu_j$ is usually defined as the
 total sector-wide impact of a unit change of demand in sector $j$.

lectures/simple_linear_regression.md

@@ -297,7 +297,7 @@ Calculating $\beta$
 ```{code-cell} ipython3
 df = df[['X','Y']].copy()  # Original Data
 
-# Calcuate the sample means
+# Calculate the sample means
 x_bar = df['X'].mean()
 y_bar = df['Y'].mean()
 ```
@@ -393,7 +393,7 @@ df
 Sometimes it can be useful to rename your columns to make it easier to work with in the DataFrame
 
 ```{code-cell} ipython3
-df.columns = ["cntry", "year", "life_expectency", "gdppc"]
+df.columns = ["cntry", "year", "life_expectancy", "gdppc"]
 df
 ```
 
@@ -415,10 +415,10 @@ It is always a good idea to spend a bit of time understanding what data you actu
 
 For example, you may want to explore this data to see if there is consistent reporting for all countries across years
 
-Let's first look at the Life Expectency Data
+Let's first look at the Life Expectancy Data
 
 ```{code-cell} ipython3
-le_years = df[['cntry', 'year', 'life_expectency']].set_index(['cntry', 'year']).unstack()['life_expectency']
+le_years = df[['cntry', 'year', 'life_expectancy']].set_index(['cntry', 'year']).unstack()['life_expectancy']
 le_years
 ```
 
@@ -453,13 +453,13 @@ df = df[df.year == 2018].reset_index(drop=True).copy()
 ```
 
 ```{code-cell} ipython3
-df.plot(x='gdppc', y='life_expectency', kind='scatter',  xlabel="GDP per capita", ylabel="Life Expectency (Years)",);
+df.plot(x='gdppc', y='life_expectancy', kind='scatter',  xlabel="GDP per capita", ylabel="Life Expectancy (Years)",);
 ```
 
 This data shows a couple of interesting relationships.
 
 1. there are a number of countries with similar GDP per capita levels but a wide range in Life Expectancy
-2. there appears to be a positive relationship between GDP per capita and life expectancy. Countries with higher GDP per capita tend to have higher life expectency outcomes
+2. there appears to be a positive relationship between GDP per capita and life expectancy. Countries with higher GDP per capita tend to have higher life expectancy outcomes
 
 Even though OLS is solving linear equations -- one option we have is to transform the variables, such as through a log transform, and then use OLS to estimate the transformed variables
 
@@ -470,7 +470,7 @@ ln -> ln == elasticities
 By specifying `logx` you can plot the GDP per Capita data on a log scale
 
 ```{code-cell} ipython3
-df.plot(x='gdppc', y='life_expectency', kind='scatter',  xlabel="GDP per capita", ylabel="Life Expectancy (Years)", logx=True);
+df.plot(x='gdppc', y='life_expectancy', kind='scatter',  xlabel="GDP per capita", ylabel="Life Expectancy (Years)", logx=True);
 ```
 
 As you can see from this transformation -- a linear model fits the shape of the data more closely.
@@ -486,11 +486,11 @@ df
 **Q4:** Use {eq}`eq:optimal-alpha` and {eq}`eq:optimal-beta` to compute optimal values for  $\alpha$ and $\beta$
 
 ```{code-cell} ipython3
-data = df[['log_gdppc', 'life_expectency']].copy()  # Get Data from DataFrame
+data = df[['log_gdppc', 'life_expectancy']].copy()  # Get Data from DataFrame
 
 # Calculate the sample means
 x_bar = data['log_gdppc'].mean()
-y_bar = data['life_expectency'].mean()
+y_bar = data['life_expectancy'].mean()
 ```
 
 ```{code-cell} ipython3
@@ -499,7 +499,7 @@ data
 
 ```{code-cell} ipython3
 # Compute the Sums
-data['num'] = data['log_gdppc'] * data['life_expectency'] - y_bar * data['log_gdppc']
+data['num'] = data['log_gdppc'] * data['life_expectancy'] - y_bar * data['log_gdppc']
 data['den'] = pow(data['log_gdppc'],2) - x_bar * data['log_gdppc']
 β = data['num'].sum() / data['den'].sum()
 print(β)
@@ -513,13 +513,13 @@ print(α)
 **Q5:** Plot the line of best fit found using OLS
 
 ```{code-cell} ipython3
-data['life_expectency_hat'] = α + β * df['log_gdppc']
-data['error'] = data['life_expectency_hat'] - data['life_expectency']
+data['life_expectancy_hat'] = α + β * df['log_gdppc']
+data['error'] = data['life_expectancy_hat'] - data['life_expectancy']
 
 fig, ax = plt.subplots()
-data.plot(x='log_gdppc',y='life_expectency', kind='scatter', ax=ax)
-data.plot(x='log_gdppc',y='life_expectency_hat', kind='line', ax=ax, color='g')
-plt.vlines(data['log_gdppc'], data['life_expectency_hat'], data['life_expectency'], color='r')
+data.plot(x='log_gdppc',y='life_expectancy', kind='scatter', ax=ax)
+data.plot(x='log_gdppc',y='life_expectancy_hat', kind='line', ax=ax, color='g')
+plt.vlines(data['log_gdppc'], data['life_expectancy_hat'], data['life_expectancy'], color='r')
 ```
 
 :::{solution-end}

lectures/time_series_with_matrices.md

@@ -504,7 +504,7 @@ print("Sigma_y = ", Sigma_y)
 
 Notice that  the covariance between $y_t$ and $y_{t-1}$ -- the elements on the superdiagonal -- are **not** identical.
 
-This is is an indication that the time series respresented by our $y$ vector is not **stationary**.  
+This is is an indication that the time series represented by our $y$ vector is not **stationary**.  
 
 To make it stationary, we'd have to alter our system so that our **initial conditions** $(y_1, y_0)$ are not fixed numbers but instead a jointly normally distributed random vector with a particular mean and  covariance matrix.