Transcript What is the MPC?
What is the MPC?
Learning Objectives
1. Use linear regression to establish the relationship between two variables 2. A more formal approach to hypothesis testing
Consumption Function
• • • • Keynesian Consumption function income today, consumption today
C=a+b*Y
Econometrics : quantify economic relationships – What are “a” and “b”
Look at some data
• • • Look at individual level data: individual.dta
Stata: scatter cons nmwage This gives a scatter plot with the first variable on the vertical axis and the second variable on the x axis
Look at data
0 500 1000 net monthly wage 1500 2000
Two Obvious facts
1. Observe many households at different income levels – There is clearly a positive relationship 2. cons depends on income but households with same income will not have same consumption – other factors influence consumption
How do we Calculate the MPC?
• • • • Draw a line Many possible lines Intuition tells us that an “average” line would be a better estimate – We will show why this intuition is correct later Any line we draw (even the “best”) will not go through all the points – There will be deviations from the line
Conditional Expectation
• • • As an alternative to the line we could follow the logic of the gender example from the pervious section and look at conditional expectation Recall we answered the question of gender discrimination by comparing the average wage of two groups – The expected waged conditional on being a man or woman – we used the “summ if” command Formally – E(hwage|gender==1)=6.701875 – E(hwage|gender==2)= 5.451302
Conditional Expectation
• • • We can apply the same logic to the consumption function.
Divide in two groups – Rich: nmwage>1000 – Poor:nmwage<1000 – generate rich=(nmwage>1000) Compare the average consumption of each using summ if
Conditional Expectation
• • We get average consumption conditional on being rich or poor – E(Cons|Rich)= 1297.3 – E(Cons|Poor)= 599.89
We can measure the marginal propensity of consume by taking the average income of each group – E(nmwage|Rich)= 1611.698
– E(nmwag|Poor)= 711.9268
Conditional Expectation
• • • • As you move from “poor” to “rich” your income rises by: – 1611 -711=900 – And consumption rises by: 1297-599=698 So an estimate of the MPC would be 698/900 which is 0.77
This is a simple and intuitive method that builds on the logic of the gender example But…..
• • • • •
Obvious Problem
The division between risk and poor was entirely arbitrary – Not natural like gender We throw away information by forcing individuals into one group or another Why not have 3 groups or any number of groups you like Intuitively the more the better – 10 group example But large numbers of groups would make calculations tedious and would always leave out some information
10 Income Groups
0 500 1000 group_w 1500 2000
Compromise
• • • Imagine there are an infinity of groups but the conditional means are all related Specifically they have a linear relationship – E(cons|nmwage)=a+b*nmwage From now on we will write in more general notation – E(Y|X)= b 1 + b 2 X
• • • •
Comment
Note this is a restriction and it may not be true in the real world We impose it on the model – Looks reasonable in the consumption example If it isn't true then there might be a problem – – Linear approx GIGO Relationship doesn’t have to be linear but it does have to be parametric – We will see more on this later
So to Recap…
• • • • We have data that appears to illustrate a relationship between two variables Intuitively we will put a line through the data that represents the data in some way What way? Two ways: 1. the line links all the conditional means 2. We choose the particular line that is closest to the data in a defined way These turn out to be the same
Draw a line to represent the data
Y E(Y|X)= b 1 + b 2
X
Y 1 u 1 Y 3 Y 2 b 1 u 2 X 2 X 1 u 3 X 3 b 2
E Y X
X
) : slope coefficient: Change in
E(Y|X)
for a change in
x
. X
An Explanation
• • • • Change in notation to be more general – – Y is the LHS or dependent variable X is the RHS or independent variable E(Y|X i ) = conditional mean i.e. does not describe every observation – – Y i = E(Y|X i ) + u i u i represents the deviation of each individual observation from the conditional mean Y i Y i = E(Y|X i ) + u i = b
1+
b
2 X i
+ u i
What is U
i
?
• • • • Any factor other than income (X) which influences consumption (Y) – individual tastes and unpredictability approximation error because of assumption of linear relationship Later we will model this a random variable Perhaps with a normal distribution – Remember our warnings about the bell curve
OLS Estimation
• • • • Find line of “best fit” Method of Ordinary Least Squares (OLS) to estimate b 1 b 2 Objective: find estimates of minimize the error terms b
1
b
2
that minimizes the distance between the regression line and the actual data points, i.e. Minimise the sum of squared deviations i.e. – Aside: why not absolute deviation or others?
Algebra of OLS
• • • • • min i
Y i =
i b
2
u i 2 u i 2 = i
min S(
b
1
i.e. min (u 1 2 + u 2 2 +u 3 2 +…+u i 2 ) b
1 +
b
2 X i +u i
(Y
i => u i = Y i
b
1 +
b
2
X ) 2
,
b
2 ) = min
i (Y
i -
b = S(
1
b
+ 1
b
, 2 X
b
2 ) => sum of squared errors is a function of
b
1
b
1 +
b
2
X ) 2
,
• To find minimum of any function: differentiate with respect to the arguments and set derivative = 0 i.e. find the point where the slope with respect to the argument = 0.
S(
b
1 ,
b
2 ) =
N i (Y
i -
b
1 -
b
2 X
) 2
S
( , 1 b b b b 1 1 2 2 ) 2 ) 2
i N
1
Y i
2
i N
1
i
b b 2
X i i
b b 1 2
X i
)
To find the minimum set these equal to zero. b 1
,
b 2 are the solutions to these equations when they are set = 0: 2 2
i N
i
N
1 1
Y i i
1
b
2
X i i
0 1
b X
2
i b
1 2
b
2
i N
1 (
x i
i N
1 (
x i x
)(
y i x
) 2
y
)
An Explanation
• • •
b 1 , b 2
are the Ordinary Least Squares (OLS) estimators of the true population parameters b
1 ,
b
2
.
b 2
is the estimator of the slope coefficient: the slope coefficient measures the effect on y of a one unit change in x
b 1
is the estimator of the intercept: the value of Y which occurs if X=0;
OLS in stata
regress cons nmwage i u i 2 Source | SS df MS Number of obs = 1330 -------------+----------------------------- -------------+----------------------------- F( 1, 1328) = 605.97
Model | 98124170.1 1 98124170.1 Prob > F = 0.0000
Residual | 215041332 1328 161928.714 R-squared = 0.3133
Adj R-squared = 0.3128
Total | 313165502 1329 235639.956 Root MSE = 402.4
----------------------------------------------------------------------------- cons | Coef. Std. Err. t P>|t| [95% Conf. Interval] -------------+--------------------------------------------------------------- nmwage | .7562304 .0307205 24.62 0.000 .6959644 .8164964
_cons | 62.47876 25.9165 2.41 0.016 11.63701 113.3205
----------------------------------------------------------------------------- Estimated coef
The Answer
• • • The regression gives us a measure of the MPC The OLS estimate of the MPC is 0.756
What use is this – Prediction – Causation – Statistical inference
Prediction
• • • • We can use this to make predictions What would the consumption be if income were 2500 Cons= 62.47876 + 0.7562304*2500 – This is equal to 1953 Be careful this is the predicted conditional mean – It is the next point on the line – – What people with 2500 would consume on average What they actually will consume is unknown because we don’t observe their U i
Predicted Consumption
Y Y 4 Y 1 Y 3 Y 2 b 1 u 2 X 2 u 1 u 3 X 1 X 3 E(Y|X)= b 1 + b 2
X
u 4 X 4 X Actual Consumption Predicted Cons
Causation
• • • • • Remember all this only really identifies variables that move together It doesn’t show causation Need theory for that Obvious in the gender example (wages don’t cause changes in gender) Not obvious here causation can run both ways
Statistical Inference
• • • This estimate is generated from a sample Recall that the issue is whether we can use this fact about the sample to make statements about the world (“population”) The same issues of statistical inference arise in context of regression – OLS estimates are sample statistics just like the sample average wages in the gender example
More on the Residual (U
i
)
• • • • The residual is the difference between the line (conditional expectation) and the actual data Think of every individuals consumption as being made up of two bits – Conditional expectation – Residual The conditional expectation is that same for everyone with the same X (income) Residual is potentially different even for those with same income
Random Variable
• • • Residual is unknown in advance so we model it as a random variable Think of consumption being determined by systematic bit plus a roll of a dice See diagram – Actual consumption (expectation+residual) is distributed around the mean – All the means are linked
Each distribution is a slice in the data
0 500 1000 net monthly wage 1500 2000
Distribution of Y for two different “slices” of X
f(Y|X=900)
f(Y|X)
y|x=900 consumption Probability distribution of expenditure given income = 900 and income = 1200
f(Y|X=1200) f(Y|X=900)
Empirical Distribution
• • • • • • We can use the hist comand in stata to look at this Just as we got distribution of hwage for men and women hist cons, by(rich) norm We could do the same for any income group – hist cons if nmwage<1100 &nmwage >900, norm All OLS does is draw a line through all the means Imagine laying all these distributions side by side
0 1 -1000 0 1000 2000 3000 -1000 monthly consumption Density normal cons 0 1000 2000 3000 Graphs by rich
The “Slice” Around nmwage=1000
-500 0 500 1000 monthly consumption 1500 2000
1 2 3 4 5 .002
5 6 7 8 5 .002
-1000 0 1000 2000 3000 -1000 0 1000 2000 3000 9 10 5 .002
-1000 0 1000 2000 3000 -1000 0 1000 2000 3000 monthly consumption Density normal cons Graphs by group
Linking the Means
• • • We assume that there is a linear relationship between the means of all these distributions Imagine taking each and lining them up in order of their average Get the next diagram
Distribution of Y
f(Y|X i )
E(Y|X i )
X=600 X =900
X
X=1200
Putting it all together
• • • We usually assume that the residual is a normal random variable Seems reasonable in this case – But remember our concerns about normal So the full model is – – – Y i = b
1 +
b
2 X i
+ u i Where E(Y|X i )= b
1 +
b
2
And u i ~N(0, s 2 )
X i