第2章 多元回归分析(2)

∑ ( y y ) is the total sum of squares (SST) ∑ ( y y ) is the explained sum of squares (SSE) ∑ u is the residual sum of squares (SSR)2 2 i i 2 i

Then SST = SSE + SSR13

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Goodness-of-Fit (continued)How do we think about how well our sample regression line fits our sample data? Can compute the fraction of the total sum of squares (SST) that is explained by the model, call this the R-squared of regression R2 = SSE/SST = 1 – SSR/SST14

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Goodness-of-Fit (continued)We can also think of R 2 as being equal to the squared correlation coefficient between the actual yi and the values yi R2

(∑ ( y y )(y y )) = (∑ ( y y ) )(∑ (y y ) )2 i i 2 2 i i

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More about R-squaredR2 can never decrease when another independent variable is added to a regression, and usually will increase Because R2 will usually increase with the number of independent variables, it is not a good way to compare models16

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An Example: an crime model (w p82)What determines the person to commit crime? (the dependent variable is the number of times the man was arrested during 1986, narr86)pcnv, the proportion of arrests that led to conviction. avgsen, average sentence length served for prior convictions. ptime86, months spent in prison in 1986. qemp86, the number of quarters during which the man was employed in 1986.

The Regression model

narr86=β0+β1pcnv+β2avgsen+β3ptime86+β4qemp86+u

The Estimated equation

narr86=0.7120.150pcnv-0.034ptime86-0.104qemp86 n=2,725 R2=0.0413 narr86=

0.7070.151pcnv+0.0074avgsen-0.037ptime86-0.103qemp86 n=2,725 R2=0.0422 The fact the estimated model explain only about 4.2% of the variable in narr86 does not necessarily mean that the equation is useless. Generally, a low R2 indicates that it is hard to predict individual outcomes on y with much accuracy.17

Another word

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Assumptions for UnbiasednessPopulation model is linear in parameters: y = β0 + β1x1 + β2x2 +…+ βkxk + u We can use a random sample of size n, {(xi1, xi2,…, xik, yi): i=1, 2, …, n}, from the population model, so that the sample model is yi = β0 + β1xi1 + β2xi2 +…+ βkxik + ui E(u|x1, x2,… xk) = 0, implying that all of the explanatory variables are exogenousE(u|x)=0 Cov(ux)=0, E(ux)=0

None of the x’s is constant, and there are no exact linear relationships among themIt does allow the independent variables to be correlated; they just cannot be perfectly linear correlated. Student performance: avgscore=β0+β1expend+β2avginc+u Consumption function: consum=β0+β1inc+β2inc2+u But, the following is invalid: log(consum)=β0+β1inc+β2inc2+u

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Unbiasedness of OLS estimationUnder the four assumptions above, we can get

E β j = β j , j = 0,1,L , k

we prove the result only for β1 : the proof for the other parameters is virtually identical. we first write the β1 as following: n n n n 2 = r y r = r (β + β x + β x +L+ β x + u ) r2 β1

( )i n

∑i =1 n i =1 n

i1 i

∑i =1 n i =1

i1

∑i =1

i1

0

1 i1

2 i2 n

k ik n

∑i =1 2 i1

i1

= β 0 ∑ rii1 + β1 ∑ rii1xi1 + β 2 ∑ rii1xi 2 + L + β k ∑ rii1xik + ∑ rii1ui = β1 ∑ ri1 ( xi1 + ri1 ) + ∑ ri1uii =1 n i =1 i =1

n

∑ ri =1

n

2 i1

= β1 ∑ r + β1 ∑ ri1 xi1 + ∑ ri1uii =1 2 i1 i =1 i =1

n

i =1

n

i =1

∑ rn i =1

∑ ri =1

n

2 i1

= β1 + ∑ ri1uii =1

n

∑ ri =1

n

2 i1

therefore, n = Eβ + r u E β1 1 ∑ i1 i i =1

( )

n ∑ r = β1 + ∑ ri1E (ui ) i =1 i =1 n 2 i1

∑ ri =1

n

2 i1

E βj = βj19

( )

βj

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Too Many or Too Few VariablesWhat happens if we include variables in our specification that don’t belong?

suppose we specify the model as y = β 0 + β1 x1 + β 2 x2 + β 3 x3 + u , and this model satisfies the four assumptions. but the x3 has no effect on y after control x1 , x2 that is, the really model is E ( y | x1 , x2 , x3 ) = E ( y | x1 , x2 ) = β 0 + β1 x1 + β 2 x2 the estimated model including x3 is y = β 0 + β1 x1 + β 2 x2 + β 3 x3 the estimated parameters is unbiased, there is no effect.There is no effect on our parameter estimate, and OLS remains unbiased What if we exclude a variable from our specification that does belong? OLS will usually be biased20

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Omitted Variable Bias

Suppose the true model is given as y = β 0 + β1 x1 + β 2 x2 + u, but we ~ ~ ~ = β + β x + u, then estimate y

β1

~

∑ (x x ) y = ∑ (x x )i1 1 2 i1 1

0

1 1

i

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