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Is the Non-disclosure Policy of Audit Intensity Always Effective? A Theoretical Exploration

  • Yong Ma ORCID logo , Wanlin Deng and Hao Jiang EMAIL logo
Published/Copyright: August 22, 2022

Abstract

This study theoretically explores the effectiveness of the non-disclosure policy of audit intensity using the portfolio choice approach. In our setting, audit intensity follows a two-state Markov chain, which is not disclosed by the tax authority, and agents will exploit the available information to learn the state and accordingly make tax evasion decisions. We find that the effectiveness of the non-disclosure policy in reducing tax evasion and increasing tax revenues depends on the proportion of time in the high-intensity state. Interestingly, when this proportion is high during a period, the disclosure policy is more effective.


Corresponding author: Hao Jiang, College of Finance and Statistics, Hunan University, Changsha 410006, China, E-mail:

Award Identifier / Grant number: 71971077

Appendix A: Proof of Proposition 1

According to (12), the first order conditions for the optimal consumption C* and investment proportion π * , π e * are

(A.1) f C ( C * , J ) J W = 0 ,
(A.2) π * ( 1 τ ) + π e * = J W W J W W μ ( 1 τ ) r G ( 1 τ ) σ 2 ,
(A.3) π * ( 1 τ ) + π e * = ( μ r G ) J W σ 2 W J W W + J ( W θ ( τ ) π e * W , s ̂ ) ( W θ ( τ ) π e * W ) θ ( τ ) [ ( 1 s ̂ ) λ N + s ̂ λ H ] σ 2 W J W W .

According to (8), we have

f C = ρ C ψ 1 | ( 1 γ ) J | 1 β ,

and then

C * = ρ ψ J W ψ | ( 1 γ ) J | 1 γ ψ 1 γ .

If we equate the left-hand sides of the (A.2) and (A.3), π e * must solve

J ( W θ ( τ ) π e * W , s ̂ ) ( W θ ( τ ) π e * W ) = τ r G J W ( 1 τ ) θ ( τ ) [ ( 1 s ̂ ) λ N + s ̂ λ H ] .

We conjecture the value function has the following form:

(A.4) J W , s ̂ = W 1 γ 1 γ G s ̂ .

Thus, the optimal consumption and tax evasion can be written as

(A.5) C * W = ρ ψ G 1 ψ 1 γ ,
(A.6) π e * = 1 θ ( τ ) 1 ( 1 τ ) θ ( τ ) [ ( 1 s ̂ ) λ N + s ̂ λ H ] τ r G 1 γ ,
(A.7) π * = μ ( 1 τ ) r G γ ( 1 τ ) 2 σ 2 π e * 1 τ .

using

f ( C * , J ) = ρ 1 ψ 1 C * 1 ψ 1 ( ( 1 γ ) J ) β ( ( 1 γ ) J ) β 1 = 1 γ 1 ψ 1 ρ ψ G 1 ψ 1 γ ρ J ,

We obtain the ODE (14).

Appendix B: Numerical Method for Solving G(⋅)

To solve G ( s ̂ ) , we approximate it by using an exponential function as follows:

(B.1) G ( s ̂ ) e c 0 + c 1 s ̂

where c0 and c1 are the constant coefficients to be solved.

Substituting (B.1) into the ODE (14), we have.

0 = A 1 ρ ψ G 1 ψ 1 γ + A 2 [ ( 1 s ̂ ) λ N + s ̂ λ H ] 1 γ [ ( 1 s ̂ ) λ N + s ̂ λ H ] + [ ( 1 s ̂ ) p 01 s ̂ p 10 ] G ( s ̂ ) G
(B.2) 1 2 κ 2 ( λ H λ N ) 2 s ̂ 2 ( 1 s ̂ ) 2 G ( s ̂ ) G + A 3 ,

where

A 1 = 1 γ 1 ψ , A 2 = γ τ r G θ ( τ ) ( 1 τ ) 1 1 γ ,

and

A 3 = ( 1 γ ) r G + τ r G θ ( τ ) ( 1 τ ) ρ 1 ψ 1 + ( μ ( 1 τ ) r G ) 2 2 γ ( 1 τ ) 2 σ 2 .

Let

(B.3) T E ( s ̂ ) = A 1 ρ ψ G 1 ψ 1 γ + A 2 [ ( 1 s ̂ ) λ N + s ̂ λ H ] 1 γ 1 2 κ 2 ( λ H λ N ) 2 s ̂ 2 ( 1 s ̂ ) 2 G ( s ̂ ) G .

Then, (B.2) gives

(B.4) T E ( s ̂ ) = λ N p 01 c 1 A 3 + [ ( λ H λ N ) + ( p 01 + p 10 ) c 1 ] s ̂ .

In addition, we approximate TE by its first-order Taylor expansion around s ̂ = s ̄ (We in this study set s ̄ = 0.5 .), that is,

(B.5) T E ( s ̂ ) = T E ( s ̄ ) T E ( s ̄ ) s ̄ + T E ( s ̄ ) ( s ̂ ) .

By matching the coefficients of (B.4) and (B.5), we have the following two equations for the unknowns c0 and c1:

T E ( s ̄ ) + λ N λ H ( p 01 + p 10 ) c 1 = 0 , T E ( s ̄ ) T E ( s ̄ ) s ̄ λ N + p 01 c 1 + A 3 = 0 .

Finally, following Chacko and Viceira (2005), we can solve the system numerically.

Acknowledgments

We are grateful to two anonymous referees for their helpful comments. This research is supported by the National Natural Science Foundation of China (71971077).

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Received: 2022-04-30
Accepted: 2022-08-08
Published Online: 2022-08-22

© 2022 Walter de Gruyter GmbH, Berlin/Boston

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