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proof in a time-invariant fixed effects model
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Daniele RinaldoThe Graduate Institute, Geneva20 November
2015
Proof of equivalence of the within estimation and the dummy
variablesspecification (time-invariant fixed effects case)
Consider the following panel data model with a time-invariant
fixed effect component i:
yit = i+x
it+it, i= 1 . . . N , t= 1 . . . T
where each xit has k covariates. By writing this model in matrix
form stacking only the obser-vations in t we obtain:
yi1yi2
...yiT
(T1)
=
i...
i
(T1)
+
xi1xi2
...xiT
(Tk)
(k1)
+
i1i2
...iT
(T1)
Now, expanding the i individual observations and stacking them
yields the following form, usingfor notation clarity the horizontal
dots only for visual separation of the blocks of the matrix
(forexample, there is no data between y1T andy21):
y11|
y1T. . .y21
|y2T. . .
.... . .
yN1|
yNT
(NT1)
=
1 0 0 . . . 0| | | . . . |
1 0 0 . . . 0
0 1 0 . . . 0| | | . . . |0 1 0 . . . 0
......
... . . .
...
0 0 0 . . . 1| | | . . . |0 0 0 . . . 1
(NTN)
12
...N
(N1)
+
x11|
x
1T. . .x21
|x2T. . .
.... . .
xN1|
xNT
(NTk)
12...
k
(k1)
+
11|
1T. . .21
|2T. . .
.... . .N1
|NT
(NT1)
which is the full matrix form of a panel data model. We can then
give its compact form:
y= D +X+
where D = IN T, where T is a column vector of ones of size T: T
= [1 1 . . . 1]. Thesymbol denotes the Kronecker product, which is
an operation on two matrices of arbitrarysize (technically, the
generalization of the outer product) which multiplies the matrix T
to eachelement of the matrix IN. We can also see the (N T 1) matrix
Dto be precisely
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D=
|1T
|. . .
|2T
|. . .
.... . .
|NT
|
Let us now write the model in the even more compact form:
y= Z +
whereZ= [D X] and= [ ]. Using the Frisch-Waugh-Lovell
partitioned regression theorem,we can obtain consistent estimates
ofby projecting the variables on the orthogonal complementof the
matrix D and then regressing them. In other words, we multiply both
sides by theannihilator matrixMD = INT D(DD)1D, which gives us
MDy= MD D =0
+MDX+MD.
We know that OLS on this transformed model is consistent, and
using the symmetry and idem-potency ofMD we get the estimator
WG = (XMDMDX)
1(XMDMDy)
= (XMDX)1(XMDy)
But what is MD exactly? Its a linear combination of matrices
made of only 0s and 1s, and itwill be shown in two absolutely
equivalent ways, one algebraic and one explicit, that
multiplyingsuch matrix to any variable is equivalent to the within
transformation, which is subtracting thetime trend from each
observation of the variable:
MDy=
y11 1T
Ty=1y1t
...
y1T 1T Ty=1y1t. . .
y21 1T
Ty=1y2t
...
yNT 1T
Ty=1yNt
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proof in a time-invariant fixed effects model
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1) Algebraic way
We have the annihilator matrix MD = INT D(DD)1D. Let us analyze
the contents of
such matrix:
(DD) = (IN T)(IN T)
= (IN
TT) (property of the Kronecker product)
= (IN T) (because
T= [1 1 . . . 1])
(DD)1 = (IN 1/T) = 1
TIN
D(DD)1D = 1
T(IN T) IN (IN T)
= 1
T(IN T
T) =PD (the projection matrix)
Now, note that TT = T (a scalar), but TT is a T T matrix made
only of 1s and that
1/T (T
T) is a T Tmatrix made only of 1/Ts. In the variabley f you take
the individual 1,for example, and let it vary in t you have the
vector ofT observationsy1t. You can then see that1T
(TT)y1t = y1, which is a column ofT times the time mean of the
first observation. Then,
(INT
T
T ) y= yi, which is the (N T 1) column vector that contains all
the Nobservations
time means. We can now see that the it-th element of the
transformed dependent variable is
MD y
it
= yit yi
which is the within trasformation.
2) Explicit way (exactly the same)
(DD) =
T 0 . . . 0
0 T . . . 0
......
. . . ...
0 0 . . . T
(NNT)
T 0 . . . 0
0 T . . . 0
......
. . . ...
0 0 . . . T
(NTN)
=
T 0 . . . 00 T . . . 0...
... . . .
...0 0 . . . T
(NN)
(DD)1 =
1
T
0 . . . 00 1
T . . . 0
......
. . . ...
0 0 . . . 1T
(NN)
= 1
TIN
D(DD)1D = 1
T(DD)
Now we have the following:
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(DD) =
T 0 . . . 0
0 T . . . 0
......
. . . ...
0 0 . . . T
(NTN)
T 0 . . . 0
0 T . . . 0
......
. . . ...
0 0 . . . T
(NNT)
=
= (notice thatT
T is aT Tmatrix of ones)
=
1 . . . 1...
. . . ...
1 . . . 1(TT)
0 . . . 0
01 . . . 1...
. . . ...
1 . . . 1
. . . 0
. . .
0 0 . . .
1 . . . 1...
. . . ...
1 . . . 1
(NTNT)
1
T(DD) =
1/T . . . 1/T...
. . . ...
1/T . . . 1/T(TT)
0 . . . 0
0
1/T . . . 1/T...
. . . ...
1/T . . . 1/T
. . . 0
. . .
0 0 . . .
1/T . . . 1/T...
. . . ...
1/T . . . 1/T
(NTNT)
The annihilator matrix MD =INT 1T(DD) can be then seen to be
exactly equivalent to
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1 1/T . . . 1/T...
. ..
...1/T . . . 1 1/T
(TT)
0 . . . 0
0
1 1/T . . . 1/T...
. . . ...
1/T . . . 1 1/T
. . . 0
. . .
0 0 . . .
1 1/T . . . 1/T...
. . . ...
1/T . . . 1 1/T
(NTNT)
where one should notice that (1 1/T) appears only in the main
diagonal and all the other
nonzero elements are 1/T. The first row of the (N T 1) matrixMDy
is then y11 1T
Tt=1y1t
because all the rest of the observations iny after individual 1
will be multiplied by 0: if we repeatthis for all the observations
in y we get
MDy=
y11 1T
Tt=1y1t
...
y1T 1T
Tt=1y1t
y21 1T
Tt=1y2t
...
y2T 1T
Tt=1y2t
...
yN1 1T
Tt=1yNt
...
yNT 1T
Tt=1yNt
(NT1)
=
y11 y1...
y1T y1
y21 y2...
y2T y2
...
yN1 yN...
yNT yN
(NT1)
which is exactly the within transformation.
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