Housing Dynamics
Technical Appendix
Edward L. Glaeser, Joseph Gyourko,
Eduardo Morales and Charles G. Nathanson
∗January 7, 2014
1
Definitions of Trend Variables
We write the housing demand equation as
Ht−
E(Ht+1)
1 +r −
rC
1 +r =x+qt+xt−αNt
and the housing supply equation as
E(Ht+1) = C+c0t+c1It+c2Nt.
Our third equation is the relationship between city population and construction:
Nt+1=Nt+It.
∗Glaeser: Department of Economics, Harvard University, and NBER; Gyourko: The Wharton
We define ˆHt, ˆNt, and ˆIt to be the unique solutions to the non-stochastic elements of
these three equations, which are linear. They are the unique solutions to
ˆ
Ht−
ˆ
Ht+1
1 +r −
rC
1 +r = x+qt−α
ˆ
Nt
ˆ
Ht+1 = C+c0t+c1Iˆt+c2Nˆt
ˆ
Nt+1 = Nˆt+ ˆIt
and are therefore given by
ˆ
Ht =
c2
2(x+r(C+x)) +α(1 +r)(c2C−c0c1)
c2(rc2+α(1 +r))
+
(1 +r)(αc0+qc2)(c22+α(1 +r)(c1−c2))
c2(rc2+α(1 +r))2
+
(1 +r)(αc0+qc2)
rc2+α(1 +r)
t;
ˆ
Nt =
rc0c1+ (1 +r)c2x
c2(rc2+α(1 +r))
− (1 +r)(r(c1−c2)−c2)(αc0+qc2)
c2(rc2+α(1 +r))2
+
q(1 +r)−rc0
rc2+α(1 +r)
t;
ˆ
It =
q(1 +r)−rc0
rc2+α(1 +r)
.
2
Proof of Lemma 1
Let h = H −Hˆ, n = N −Nˆ, and i = I −Iˆ, where H, N, I constitute a solution to
The three key equations reduce to
ht−
E(ht+1)
1 +r = xt−αnt
E(ht+1) = c1it+c2nt
nt+1 = nt+it.
Note that because E(xt+j) = δj−1E(xt+1), limj→∞E(xt+j) = 0. We can therefore
de-duce from the first key equation that
lim
j→∞
E(nt+j)
(1 +r)j = 0. (1)
By combining the three key equations, we obtain the following difference equation:
(1 +r)E(xt+1) = (1 +r)(c2−c1)E(nt) + ((1 +r)(α+c1) +c1−c2)E(nt+1)−c1E(nt+2) = −c1(φL−I)(φL−I)E(nt+2).
Here L is the lag operator, I is the identity operator, and φ < φ are the two roots to the characteristic equation
0 =−c1y2+ ((1 +r)(α+c1) +c1−c2)y+ (1 +r)(c2−c1).
Because limj→∞E(xt+j) = 0, one of the following two equations must hold:
lim
j→∞(φL−I)E(nt+j) = 0
lim
We claim only the first holds. To prove this, we show that 0 < φ < 1 < 1 +r < φ. Recall that α >0 and c2 < c1. Note that
φ = (1 +r)(α+c1) +c1−c2+ p
((1 +r)(α+c1) +c1−c2)2−4(1 +r)c1(c1−c2) 2c1
> (1 +r)(α+c1) +c1−c2+
p
((1 +r)c1+c1−c2)2−4(1 +r)c1(c1−c2) 2c1
= (1 +r)(α+c1) +c1−c2+rc1+c2 2c1
= 1 +r+α(1 +r) 2c1
> 1 +r.
We also have
φ+φ= (1 +r)(α+c1) +c1−c2
c1
>0
and
φφ= (1 +r)
1− c2
c1
∈(0,1 +r).
Therefore 0< φ <1. Because of equation (1) and the fact that φ >1 +r, the limit
lim
j→∞(φL−I)E(nt+j) = 0
cannot hold. Therefore,
lim
j→∞(φL−I)E(nt+j) = 0
holds, and because 0 < φ < 1, this limit implies the limit limj→∞E(nt+j) = 0.
The equation nt+1 = nt +it shows that limj→∞E(it+j) = 0, and then the equation
condition. We can therefore simplify the above equation to
(φL−I)nt+1 =− 1 +r
c1
φ−1(I−φ−1L−1)−1E(xt+1) =−
1 +r c1(φ−δ)
E(xt+1).
Therefore
it=nt+1−nt=−(φL−I)nt+1−(1−φ)nt=
1 +r c1(φ−δ)
E(xt+1)−(1−φ)nt,
which proves the second equation of the lemma. For the first equation of the lemma,
note that
ht = xt−αnt+
E(ht+1)
1 +r
= xt−αnt+
c1 1 +rit+
c2 1 +rnt
= xt+
E(xt+1)
φ−δ +
c2−(1−φ)c1
1 +r −α
nt
= xt+
E(xt+1)
φ−δ −
α(1 +r) 1 +r−φnt.
The last equality comes verifying that the equality
c2−(1−φ)c1
1 +r −α=−
α(1 +r) 1 +r−φ
4
Proof of Proposition 1
We have
E(nt+j) = φjnt+ j−1 X
k=0
φk(E(nt+j−k)−E(nt+j−k−1))
= φjnt+
1 +r c1(φ−δ)
j−1 X
k=0
φkE(xt+j−k)
= φjnt+
1 +r c1(φ−δ)
E(xt+1)
j−1 X
k=0
φkδj−1−k
= φjnt+
1 +r c1(φ−δ)
φj −δj
φ−δ E(xt+1).
Note that ˆNt+j−Nˆt=jIˆ. Therefore
E(Nt+j −Nt) = jIˆ+
1 +r c1(φ−δ)
φj−δj
φ−δ E(xt+1)−(1−φ
j)n t.
Next, we have
E(It+j) = E(Nt+j+1)−E(Nt+j)
= Iˆ+ 1 +r
c1(φ−δ)
δj(1−δ)−φj(1−φ)
φ−δ
E(xt+1)−φj(1−φ)nt.
Finally,
E(Ht+j −Ht) = Hˆt+j −Hˆt+E(ht+j)−ht
that prices and construction fall below trend att+j for largej, we must by Proposition 1 show that
δj(1−δ)−φj(1−φ)
φ−δ =δ
j1−δ−(1−φ)(φ/δ) j
φ−δ
is negative for large j. If φ > δ, then the numerator is negative for large j, while the denominator is positive. The opposite occurs when φ < δ. In either case, the fraction is negative for largej, which is what we wanted to show.
6
Proof of Proposition 3
From Lemma 2, we have
I0 = ˆI+
1 +r c1(φ−δ)δ0. Therefore
N1 = ˆN1+
1 +r c1(φ−δ)
δ0.
It follows again from Lemma 2 that
I1 = ˆI+
1 +r
c1(φ−δ)δ(1+δ0)−(1−φ)
1 +r c1(φ−δ)δ0.
Recall that
ˆ
I = q(1 +r)−rc0
rc2+α(1 +r)
.
Therefore
Cov(I0, I1) =
1 +r rc2+α(1 +r)
2
Var(q) +
δ(1 +r)
c1(φ−δ) 2
(δ+φ−1)Var(0),
which is positive as long as
Var(q) Var(0)
>(1−δ−φ)
δ(rc2+α(1 +r))
c1(φ−δ)
2
.
Because δ >1−φ by assumption, this inequality must hold. We now turn to price growth. From Lemma 2, we have
H0 = ˆH0+
We also have
H1 = ˆH0+
φ φ−δ1+
φ φ−δ −
α(1 +r) 1 +r−φ
1 +r c1(φ−δ)
δ0.
To obtainH2, we first note that
N2 =N1 +I1 = ˆN2+
δ(1 +r)
c1(φ−δ)(1+ (δ+φ−1)0).
Therefore
H2 = Hˆ2+
φ φ−δ2+
φ φ−δ −
α(1 +r) 1 +r−φ
1 +r c1(φ−δ)
δ1
+
δφ φ−δ −
α(1 +r) 1 +r−φ
(1 +r)(φ+δ)
c1(φ−δ)
δ0.
Recall that
ˆ
Ht+1−Hˆt =
(1 +r)(αc0+qc2)
rc2+α(1 +r)
.
Much algebra produces
Cov(H2−H1, H1−H0) =
(1 +r)c2
rc2+α(1 +r) 2
Var(q)− Ω
c1(φ−δ)2
Var(),
where Var() is the common variance of0 and 1, and Ω is given by
Ω =
α(1 +r)2δ
1 +r−φ +c1(1−δ)φ
(1−δ−φ)α(1−r)2δ
1 +r−φ +c1(1−δ+δ
2 )φ
.
7
Calculation of Volatilities in Table 5
We showed above that
(φL−I)nt+1 =−
1 +r c1(φ−δ)
E(xt+1).
It follows from decomposing nt into a linear combination of the terms that
Std(nt) =
(1 +r)(θ+δ)
c1(φ−δ)
∞
X
j=1
φj −δj
φ−δ σ.
This expression allows us to compute the congestion externality volatility, as it is a
constant times this expression.
The other volatility is of a sum of xt and E(xt+1), which can be written as a sum of xt and t. The variance of x is obtained from taking the recursive equation
xt=δxt−1+t+θt−1
and then computing the variance of both sides. The result is
σx2 = 1 + 2δθ+θ 2
1−δ2 σ
2
.
The covariance ofxtandtis justσ2. Therefore, we can readily compute the volatility of
the direct wage term, using the standard formula for the variance of a sum of correlated