Performance, Growth and Earnings Management

Abstract

We study the relationship between the amount of managed earnings and firms’ earnings performance and expected growth in a reporting model, where managers manipulate earnings to influence the valuation of firms’ equity while bearing a cost that is increasing and convex in the amount of managed earnings. In the unique revealing equilibrium to the model, firms with higher performance and growth over-report earnings by a larger amount because price responsiveness increases with earnings performance and growth. And earnings quality, defined as the proportion of true economic earnings in total reported earnings, increases with earnings performance but decreases with earnings growth. We conduct empirical tests on a large sample and a restatement sample using different proxies for earnings management. Results from the large sample tests support our predictions while results from the restatement sample tests are mixed. Our study provides an alternative explanation to the positive relationship between discretionary accruals estimated from the Jones model and firms’ performance and growth.

Appendices

Appendix A: Proofs

Proof of Proposition 1

Using Eqs. 1–4, we get the differential equation:

$$ap_r^{\ast\prime} +2cbp^\ast -2cr=0$$

(A.1)

where $r\ge 0$, and $b=\frac{d-g}{1+d}$. The general solution for Eq. A.1 is:

$$p^\ast (r)=Ge^{-\frac{2cb}{a}r}+\frac{r}{b}-\frac{a}{2cb^2}$$

(A.2)

Using the initial condition $p(r=0)=0$, we get $G=\frac{a}{2cb^2}$. We plug G back into Eq. A.2, and get:

$$p^\ast (r)=\frac{a}{2cb^2}e^{-\frac{2cb}{a}r}+\frac{r}{b}-\frac{a}{2cb^2}$$

(A.3)

Eq. A.3 gives the equilibrium pricing function when reported earnings is positive.

Plugging (A.3) back to the objective function of the manager, we could show that the utility of the manager is concave in r when $r\ge0$; i.e.,

$$ 2c\left(e^{-\frac{2cb}{a}r}-1\right)\le 0$$

For any $r\ge 0$. Therefore, the first order condition is sufficient for a global maximum.

Proof of $\frac{\partial m}{\partial \alpha} =1-\left({1+\frac{r}{\alpha}} \right)e^{-\frac{r}{\alpha}}\ge 0$where $\alpha =\frac{a}{2cb}$. Let $g(r)=\frac{\partial m}{\partial \alpha}$, then $g(r=0)=0$. And $g^{\prime}(r)=\frac{r}{\alpha }e^{-\frac{r}{\alpha}}$. When $r> 0,g^{\prime}(r)> 0$. So for all $r\ge 0,\frac{\partial m}{\partial \alpha} \ge 0$.

Proof of Corollary 2

Notice that the sign of $\frac{\partial q}{\partial \alpha}$ would be the same as the sign of $\frac{\partial m}{\partial \alpha}$ when $r\ge 0$, where $\alpha =\frac{a}{2cb}$, and $b=\frac{d-g}{1+d}$. Given $\frac{\partial m}{\partial \alpha} \ge 0$, it is obvious that when $r\ge 0,\frac{\partial q}{\partial \alpha} \ge 0$. Now we prove that $\frac{\partial q}{\partial r}\ge 0$: Take derivative of Eq. 11, we get:

$$\frac{\partial q}{\partial r}=\frac{a}{2cbr^2}e^{-\frac{2cbr}{a}}\times A$$

where

$$A=\left(e^{\frac{2cbr}{a}}-\frac{2cbr}{a}-1\right)$$

Because $\frac{a}{2cbr^2}e^{-\frac{2cbr}{a}}> 0$, $\frac{\partial q}{\partial r}$ has the same sign as A. Because when $r=0,\,A=0$. And when $r> 0$, $\frac{\partial A}{\partial r}> 0$, we get $A\ge 0$, when $r\ge 0$. For the above reasoning, $\frac{\partial q}{\partial r}\ge 0$.

Proof of Corollary 3

From Eq. A.3, we get:

$$\frac{{\partial ^{2} p}} {{\partial r^{2}}} = \frac{{2c}} {a}e^{{- \frac{{2cb}} {a}r}},\quad \frac{{\partial ^{2} p}} {{\partial r\partial a}} = - \frac{{2cr}} {{a^{3}}}e^{{- \frac{{2cb}} {a}r}},\quad \frac{{\partial ^{2} p}} {{\partial r\partial c}} = \frac{{2r}} {a}e^{{- \frac{{2cb}} {a}r}}$$

Obviously, $\frac{\partial ^2p}{\partial r^2}> 0$. And when $r\ge 0$, $\frac{\partial ^2p}{\partial r\partial a}\ge 0$ and $\frac{\partial^2p}{\partial r\partial c}\ge 0$ From Eq. A.3, we get:

$$\frac{\partial ^2p}{\partial r\partial b}=\frac{2c}{ab}e^{-\frac{2cbr}{a}}\times B$$

where

$$B=\left(r-\frac{a}{2bc}e^{\frac{2cbr}{a}}+\frac{a}{2bc}\right)$$

Since $\frac{2c}{ab}e^{-\frac{2cbr}{a}}> 0$, $\frac{\partial ^2p}{\partial r\partial b}$ has the same sign as B. Because when $r=0,\,B=0$, and when $r> 0$, $\frac{\partial B}{\partial r}< 0$. We get $B\le 0$, when $r\ge 0$. For the above reason, $\frac{\partial ^2p}{\partial r\partial b}\le 0$ and $\frac{\partial ^2p}{\partial r\partial g}\ge 0$.

Appendix B

• ACCL = Net Income (#172)-Operating Cash flow (#308)

• ART = Accounts receivable (#2) less Tax receivables (#161)

• EXP = expenses = Revenue (#12)- Operating income before depreciation (#13)

• DEP = Depreciation from Income Statement (#14)

• OCAL = other current assets and liabilities = Current assets (#4)-ART-Cash (#1)- Income tax refund (#161)-(Current liabilities (#5)- Income taxes payable (#71)

• GPPE = gross plant, property and equipment (#7)

• TA = Total assets (#6)

• REV = Revenues (#12)