3.5. TESTING MONETARY MODEL PREDICTIONS

103

Problems

Let the fundamentals have the permanent—transitory components representation

ft = f¯t + zt,		(3.27)
iid
where f¯t = f¯t−1 + ²t is the permanent part with ²t N(0, σ²2) and zt =
iid
ρzt−1+ut is the transitory part with ut N(0, σu2), and 0 < ρ < 1. Note that
the time-t expectation of a random walk k periods ahead is E	(f¯	) = f¯,
t	t+k	t

and the time-t expectation of the AR(1) part k periods ahead is Etzt+k =

ρkz . (3.27) implies the k-step ahead prediction formula E			(f	) = f¯	+ρkz .
t		t	t+k	t	t
1. Show that	1
st = f¯t +		zt.			(3.28)
	1 + λ(1 − ρ)

2.Suppose that the fundamentals are stationary by setting σ² = 0. Then the permanent part f¯t drops out and the fundamentals are governed by a stationary AR(1) process. Show that

Var(st) = µ	1	¶2	Var(ft),	(3.29)
	1 + λ(1 − ρ)

3.Let’s restore the unit root component in the fundamentals by setting σ²2 > 0 but turn o the transitory part by setting σu2 = 0. Now the fundamentals follow a random walk and the exchange rate is given

exactly by the fundamentals

st = ft.

(3.30)

The exchange rate inherits the unit root from ft. Since unit root processes have inÞnite variances, we should take Þrst di erences to induce stationarity. Doing so and taking the variance, (3.30) predicts that the variance of the exchange rate is exactly equal to the variance of the fundamentals.

Now re-introduce the transitory part σu2 > 0. Show that depreciation of the home currency is

∆st = ²t +	(ρ − 1)zt−1 + ut	.	(3.31)
	1 + λ(1 − ρ)

104	CHAPTER 3. THE MONETARY MODEL
	where	2(1 − ρ)
	Var(∆st) = σ²2 +		Var(zt).
		[1 + λ(1 − ρ)]2

Why does the variance of the depreciation still not exceed the variance of the fundamentals growth?