Learning, Central Bank Conservatism, and Stock Price Dynamics

A.1 The Optimal Targeting Rule under Learning

Substituting E t * π t + 1 = a t , E t * x t + 1 = b t and E t * q t + 1 = s t into (1), (3) and (4), we write the Lagrangian of the CB’s minimization problem as follows:

where λ_i,t, with i = 1, 2, …, 6, are Lagrange multipliers associated with (1), (3), (4) and (13)–(15), respectively. Differentiating the Lagrangian with respect to π _t , x _t , r _t , q _t , a_t+1, b_t+1 and s_t+1 yields the first-order conditions:

Using (A.3), we get λ 3 , t = − 1 1 + ψ λ 2 , t . Substituting λ_3,t in (A.4) leads to λ_2,t = −λ_6,tγ_t+1 and λ 3 , t = 1 1 + ψ λ 6 , t γ t + 1 . Replacing λ_2,t, λ_3,t and β = 1 + ψ β ̃ into (A.6) and (A.7) yields:

Equation (A.9) implies that the only bounded forward-looking solution is λ_6,t = E _t λ_6,t+1 = 0. It follows that λ_2,t = λ_3,t = 0. Substituting E _t λ_6,t+1 = 0 into (A.8) similarly yields λ_5,t = E _t λ_5,t+1 = 0. Given λ_2,t = λ_5,t = 0, we get from (A.2) that λ 1 , t = − α x x t 1 + χ κ . Substituting λ_1,t into (A.1) gives the optimal inflation targeting rule when private agents are learning:

Rearranging terms in (A.10) yields λ 4 , t = − α x κ 1 + χ γ t + 1 x t − ( 1 + τ ) γ t + 1 π t . Substituting the solution of λ_1,t+1, λ_4,t and λ_4,t+1, and λ_2,t+1 = λ_3,t+1 = 0 into (A.5), we obtain:

which, after rearrangement of terms and forward iterations, gives rise to (16) for γ_t+i = γ. Notice that the Lagrange multipliers on (3), (4), (14) and (15) are zero, i.e. λ_2,t = λ_3,t = λ_5,t = λ_6,t = 0, implying that b _t , s _t , z _t and v _t have no effect on social welfare and are not true state variables for inflation and the output gap.

A.2 The ALM for Inflation

The CB’s rational inflation expectations, E t π t + 1 , are a function of π _t , a _t and z _t . From the Philips curve (1), we extract the value of x _t and x_t+1 while implementing the learning algorithm (13) for period t + 1:

Using (A.12) and (A.13) and the fact that β = 1 + ψ β ̃ , we rewrite (A.11) as

with

It follows from proposition 1 in Blanchard and Kahn (1980) that the solution of the ALM for inflation takes the following form:

Forwarding (13) and (A.18) by one period, taking expectations and eliminating a_t+1, we obtain:

Using (A.19) to eliminate E t π t + 1 in (A.14) and arranging terms give:

Comparing (A.20) with (A.18) yields

Substituting γ = 0 and γ = 1 into (A.15)A.17)–(A.17) and using the results in (A.21) and (A.22) lead straightforwardly to (20)–(22).

We can easily show, following Molnár and Santoro (2014), that the dynamic system formed by (13) and (A.14), as a _t is predetermined and π _t non-predetermined, is stable and there is a unique non-explosive solution among infinite stochastic sequences of c π c g satisfying (A.21) under two boundary conditions: a₀ and lim s → ∞ | E t π t + s | < ∞ , for ν ≥ 0 and 0 ≤ γ ≤ 1.

A.3 The Non-Explosive Solution of the ALM for Inflation

Rewriting (A.21) as c π c g c π c g γ − c π c g A 11 − A 12 + c π c g ( 1 − γ ) = 0 and substituting A₁₁ and A₁₂ by their expressions respectively given by (A.15) and (A.16), we obtain:

with

We can rewrite p₁ as

Then, it follows straightforwardly that the discriminant of the polynomial in (A.23) is positive and (A.23) admits two real solutions.

To find the nature of the solutions of c π c g in (A.23), we rewrite the latter as:

The function f ( c π c g ) is characterized by f ( 0 ) = − p 0 p 1 > 0 and f ( 1 ) = p 0 + p 2 − p 1 < 1 because (A.24) yields −p₁ > p₀ + p₂ > 0. This implies f ( c π c g ) : 0,1 → ( 0,1 ) . Given that f ′ c π c g = − 2 p 2 p 1 c π c g > 0 for c π c g ∈ 0,1 , f ( c π c g ) is strictly increasing in this interval. Applying the theorem of Brouwer, we deduce that there is one unique solution of c π c g inside the unit interval:

The other solution c π c g = − p 1 + p 1 2 − 4 p 2 p 0 2 p 2 is greater than unity and is to be excluded to avoid inflation following an explosive path.

Substituting A₁₁ and P₁ respectively given by (A.15) and (A.17) into (A.22) leads to the solution of d π c g given in (19).

Furthermore, we can show that f ( c π c g ) : 0 , α x β ̃ Υ → 0 , α x β ̃ Υ . Knowing that f(0) > 0 and substituting c π c g by α x β ̃ Υ into (A.25), we find

Using p 0 = α x β ̃ − Υ α x β ̃ p 0 + Υ α x β ̃ p 0 and the definition of p₀, p₁ and p₂ given above, we find after some rearrangements that

Substituting the above expression of −p₁ into (A.27) and using Υ = α x + κ 2 1 + χ 2 1 + τ , we obtain:

Since f ′ c π c g = − 2 p 2 p 1 c π c g > 0 for c π c g ∈ [ 0,1 ] , f ( c π c g ) is strictly increasing in 0 , α x β ̃ Υ . This characteristic and the fact that f ( c π c g ) : 0 , α x β ̃ Υ → 0 , α x β ̃ Υ prove the existence of a unique non-explosive solution for c π c g such that 0 < c π c g < α x β ̃ Υ . Using the last result and Υ = α x + κ 2 1 + χ 2 1 + τ in (19) implies that 0 < α x Υ < d π c g < 1 .

A.4 The Effect of Learning Gain

Differentiating the non-explosive solution of c π c g given by (A.26) with respect to γ gives

Using (A.28) and ∂ p 1 ∂ γ = − β ̃ ∂ p 2 ∂ γ − Y α β ̃ ∂ p 0 ∂ γ obtained using (A.28) to transform the above derivative, and after some rearrangements of terms, we finally obtain

where H ( γ ) ≡ p 0 ∂ p 1 ∂ γ − p 1 ∂ p 0 ∂ γ . The fact that c π c g < α x β ̃ Υ yields α x β ̃ − Υ c π c g > α x β ̃ − Υ α x β ̃ Υ = 0 . To show the conditions below which ∂ c π c g ∂ γ < 0 , we consider the case where H(γ) < 0 for γ = 1, and show that, in this case, we can have ∂ H ( γ ) ∂ γ > 0 , ∀γ ∈ (0, 1).

For γ = 1, we have ∂ p 0 ∂ γ = α x β ̃ 3 1 + ψ > 0 , ∂ p 1 ∂ γ = − 2 α x β ̃ 3 1 + ψ < 0 , p 1 = − α x β ̃ 3 1 + ψ − Υ , and p 0 = α x β ̃ . For γ = 1, we get

We have H(1) < 0 if

Differentiating H(γ) with respect to γ yields

Twice differentiating p₀ and p₁ with respect to γ, ∀γ ∈ (0, 1), leads to

Substituting these second derivatives as well as the definition of p₀ and p₁ into (A.31), we obtain that, ∀γ ∈ (0, 1),

Under condition (A.30), since H(1) < 0 for γ = 1 and ∂ H ( γ ) ∂ γ > 0 , ∀γ ∈ (0, 1), it follows that H(γ) < 0, ∀γ ∈ (0, 1). Using this result, we deduce from (A.29) that, if (A.30) holds,

Differentiating d π c g given by (19) with respect to γ and denoting Θ ≡ 2 α x γ β ̃ β ̃ − c π c g + 1 − 2 γ α x β ̃ − c π c g Υ yield

Using c π c g < α x β ̃ Υ and the definition of ϒ, we find that if β ̃ > 1 2 , we have

Associating the new expression of Θ with (A.32) leads to (27). If ∂ c π c g ∂ γ < 0 , it follows that

Using the definition of c x c g , d x c g , c q c g , and d q c g , c r c g and d r c g , it is straightforward to obtain the sign of their partial derivative with respect to γ.

A.5 The Effect of the Inflation Penalty

Differentiating c π c g given by (A.26) with respect to τ, using p₁ < 0 which implies p 1 − p 1 2 − 4 p 2 p 0 < 0 and p 1 p 1 2 − 4 p 2 p 0 < − 1 , and knowing that p₂ > 0, p₀ > 0, ∂ p 1 ∂ τ = − κ 2 1 + χ 2 [ 1 − β ̃ ( 1 − γ ) 2 1 + ψ ] < 0 , and ∂ p 2 ∂ τ = γ β ̃ κ 2 1 + ψ 1 + χ 2 ( 1 − γ ) > 0 , we obtain

Using this result and differentiating d π c g given by (19) with respect to τ, we get

Using the relationship between the feedback coefficients in the ALM for inflation and those in other ALMs, it is straightforward to obtain the sign of other feedback coefficients’ partial derivatives with respect to τ: ∂ c x c g ∂ τ = ∂ c q c g ∂ τ = − ∂ c r c g ∂ τ = 1 κ 1 + χ ∂ c π c g ∂ τ < 0 , ∂ d x c g ∂ τ = ∂ d q c g ∂ τ = − ∂ d r c g ∂ τ = 1 κ 1 + χ ∂ d π c g ∂ τ < 0 . We then compare the sign of a partial derivative with that of its corresponding feedback coefficient: if they have the same (different) sign, a higher inflation penalty rate has an amplification (attenuation) effect on the feedback coefficient. Notice that since κ is very small in the baseline calibration, the impact of the penalty weight on the feedback coefficients on inflation expectations and the cost-push shock in the ALMs for the output gap, stock prices and the interest rate is substantially greater than that on the corresponding coefficients in the ALM for inflation.