Conservatism, growth, and return on investment

Abstract

Return on Investment (ROI) is widely regarded as a key measure of firm profitability. The accounting literature has long recognized that ROI will generally not reflect economic profitability, as determined by the internal rate of return (IRR) of a firm’s investment projects. In particular, it has been noted that accounting conservatism may result in an upward bias of ROI, relative to the underlying IRR. We examine both theoretically and empirically the behavior of ROI as a function of two variables: past growth in new investments and accounting conservatism. Higher growth is shown to result in lower levels of ROI provided the accounting is conservative, while the opposite is generally true for liberal accounting policies. Conversely, more conservative accounting will increase ROI provided growth in new investments has been “moderate” over the relevant horizon, while the opposite is true if new investments grew at sufficiently high rates. Taken together, we find that conservatism and growth are “substitutes” in their joint impact on ROI.

Appendices

Appendix A

Proof of Lemma 1

Direct substitution yields

$$ \begin{aligned} \sum^{t}_{i=1} RI_{i} \cdot \gamma^{i}=& \gamma \cdot[c^{0}_{1} -(d_{1}+ r)\cdot b^{o}] +\gamma^{2}\cdot [c^{0}_{2}-(d_{2}+r \cdot(1-d_{1}))\cdot b^{o}]+\cdots \\ &+\gamma^{t}\cdot[c^{0}_{t}-(d_{t}+r \cdot(1 - d_{1} - \ldots d_{t-1}))\cdot b^{o}]. \end{aligned} $$

(10)

Collecting coefficients for each of the variables ${(d_{1},\ldots, d_{t})}$ on the right-hand side of the above equation, the coefficient for d _i is:

$$ \gamma^{i} - r \cdot[\gamma^{i+1}+\cdots + \gamma^{t}] = \gamma ^{i+1} - r\cdot [\gamma^{i+2}+\cdots + \gamma^{t}] = \cdots = \gamma^{t}. $$

Thus,

$$ \sum^{t}_{i=1} RI^{0}_{i} \cdot \gamma^{i} = \sum^{t}_{i=1} \gamma^{i} \cdot c^{0}_{i} -[(1-\gamma^{t}) + \gamma^{t} \cdot (d_{1} + \cdots + d_{t})] \cdot b^{0} $$

By the definition of neutral accounting,

$$ \sum^{t}_{i=1} \gamma^{i} \cdot c^{0}_{i} - [(1-\gamma^{t}) + \gamma^{t} \cdot (d_{1}^{\ast} + \cdots + d^{\ast}_{t})] \cdot b^{0} = 0. $$

Therefore ${\sum^{t}_{i=1} d_{i} \geq \sum^{t}_{i=1} d^{\ast}_{i}}$ is equivalent to ${\sum^{t}_{i=1} RI^{0}_{i} \cdot \gamma^{i} \le 0.}$   □

Proof of Proposition 1

The expression for ROI at date T is given in (1). Therefore ${ROI_{T}(\vec{\lambda})\ge r}$ is equivalent to:

$$ RI^{o}_{T}+(1+\lambda_{1})\cdot RI^{o}_{T-1}+(1+\lambda_{1})\cdot(1+\lambda_{2})\cdot RI^{o}_{T-2} + \cdots +\prod^{T-1}_{i=1}(1+\lambda_{i}) \cdot RI^{o}_{1} \ge 0, $$

(11)

where ${RI^{o}_{t} \equiv c^{o}_{t} - d_{t} \cdot b^{o} - r \cdot b^{o}(1-\sum^{t-1}_{i=1}d_{j})}$ denotes the residual income of the representative project ${\mathcal{P}=(b^{o}, c^{o}_{1}, \ldots, c^{o}_{T})}$. Lemma 1 shows that for any conservative depreciation schedule ${(d_{1}, \ldots, d_{T}):}$

$$ \sum^{t}_{i=1} \; RI^{o}_{i} \cdot \gamma^{i} \le 0 $$

(12)

for all ${1 \le t \le T-1}$. The claim therefore amounts to showing that for any T−1 tuple ${(RI^{o}_{1}, \ldots , RI^{o}_{T-1})}$ satisfying (12), inequality (11) will be met if ${\lambda_{t} \le r}$. Since ${\Sigma^{T}_{i=1} RI^{o}_{i} \cdot \gamma^{i} = 0}$, inequality (11) can equivalently be written as:

$$ \begin{aligned} RI_{T-1}& \cdot [(1+\lambda_{1})-\gamma^{-1}]+ RI_{T-2}\cdot [(1+\lambda_{1})(1+\lambda_{2})-\gamma^{-2}]+\cdots \\ &+RI_{1}\cdot \left[\prod^{T-1}_{i=1}(1+\lambda_{i}) - \gamma^{-(T-1)}\right] \ge 0. \end{aligned} $$

(13)

The inequalities in (12) can be represented in matrix form as:

$$ \Upgamma \cdot \vec{RI}^{0} \le 0 $$

(14)

where

$$ \Upgamma = \left[ \begin{array}{ccccc} \gamma &&&& \\ \gamma & \gamma^2 &&& \\ \gamma & \gamma^2 & \gamma^3 && \\ \cdot & \cdot & \cdot & \ddots &\\ \cdot & \cdot & \cdot & \cdot\\ \gamma & \gamma^2 & \gamma^3 & \cdots & \gamma^{T-1}\end{array}\right] $$

and ${\vec{RI}^{0}=(RI^{0}_{1}, \ldots ,RI^{0}_{T-1})}$. By Farkas Lemma (see Rockafellar 1970), (13) holds for any vector satisfying (14) if and only if there exists a non-negative row-vector ${v' = (v_1, \ldots, v_{T-1})}$ such that:

$$ v'\cdot \Upgamma = \left( \begin{array}{c} \gamma^{1-T} - \prod^{T-1}_{i=1}(1+\lambda_{i})\\ \vdots \\ \gamma^{-1} - (1 + \lambda_1)\end{array}\right). $$

(15)

Since ${\Upgamma}$ is a diagonal matrix, the system of equations in (15) can be solved explicitly, yielding:

$$ \begin{array}{lll} v_{T-1}&=& \gamma^{-(T-1)} [ \gamma^{-1} - (1 + \lambda_1)]\\ v_{T-2} &=& \gamma^{-(T-2)} (1 + \lambda_1) [\gamma^{-1}-(1+\lambda_2)]\\ &\vdots &\\ v_1 &=& \gamma^{-1} \cdot \prod^{T-1}_{i=1}(1+\lambda_{i}) [ \gamma^{-1}-(1 + \lambda_{T-1})] \end{array}. $$

It follows that v _t ≥ 0 if and only if ${\gamma^{-1} \ge (1 + \lambda_t)}$, or equivalently, ${ r \ge \lambda_t}$.

If the annual growth rates all exceed r, the claim is that inequality (13) reverses. The same line of arguments as before applies with the vector w′ ≥ 0, required by Farkas Lemma, given by:

$$ w_t = - v_t \ge 0. $$

Finally, for liberal accounting, the vector of residual income numbers satisfies:

$$ \Upgamma \cdot \vec{RI}^{0} \ge 0. $$

Therefore the inequalities are “flipped,” such that (13) holds provided λ _t ≥ r and the opposite is true for λ _t ≤ r.   □

Proof of Proposition 2

We show that:

$$ ROI_{T}(\vec{\lambda},\vec{d}+ \vec{u}) \ge ROI_{T}(\vec{\lambda},\vec{d}), $$

(16)

provided that ${\sum^{t}_{i=1} u_{i} \ge 0}$ and λ _t ≤ r for ${1 \le t \le T-1}$. It will be convenient to denote ${S_{t} \equiv \sum^{t}_{i=1} u_{i}}$. Referring back to the definition of ROI _T in (1), the inequality in (16) is equivalent to:

$$ \begin{array}{l} ROI_{T}(\vec{\lambda}, \vec{d}) \cdot \left[S_{T-1} + (1+ \lambda_{1}) \cdot S_{T-2} + \cdots \prod \limits^{T-2}_{i=1} (1 +\lambda_{i}) \cdot S_{1} \right] \ge \cr -S_{T-1} + (S_{T-1} - S_{T-2}) (1+\lambda_{1})+(S_{T-2} - S_{T-3})(1+\lambda_{1}) (1+\lambda_{2}) + \cdots (S_{1} - S_{0}) \prod\limits^{T-1}_{i=1} (1+\lambda_{i}), \end{array} $$

(17)

where, by definition, S ₀ = 0. We first note that:

$$ ROI_{T}(\vec{\lambda}, \vec{d}) \cdot S_{T-1} \ge - S_{T-1} + (1+\lambda_{1}) \cdot S_{T-1} $$

because S _T-1 ≥ 0, ${\lambda_{1} \le r}$ and ${ ROI_{T}(\vec{\lambda}, \vec{d}) \ge r}$. The latter claim is implied by Proposition 1, since the accounting is conservative and ${\lambda_{t} \le r}$ for all ${1 \le t \le T-1}$. Turning next to S _T-2, the same arguments apply since:

$$ ROI_{T}(\vec{\lambda}, \vec{d})(1+\lambda_{1}) \cdot S_{T-2} \ge -(1+\lambda_{1}) \cdot S_{T-2} + (1+\lambda_{1})(1+\lambda_{2})\cdot S_{T-2}. $$

Proceeding inductively, we conclude that for any ${1 \le t \le T-2}$:

$$ ROI_{T}(\vec{\lambda}, \vec{d})\cdot \prod^{T-t-1}_{i=1}(1+\lambda_{i}) \cdot S_{t}\ge - \prod^{T-t-1}_{i=1} (1+\lambda_{i}) \cdot S_{t} + \prod^{T-t}_{i=1} (1 +\lambda_{i}) \cdot S_{t}, $$

provided ${S_{t} \ge 0, \ \lambda_{t} \le r}$ and ${ ROI_{T}(\vec{\lambda}, \vec{\delta}) \ge r}$; thus, (17) holds, as was to be shown. Finally, when growth is aggressive, i.e., ${\lambda_{t} \ge r}$, note that all inequalities are reversed since by Proposition 1 ${ROI_{T}(\vec{\lambda}, \vec{d}) \le r}$ whenever the accounting is conservative and ${\lambda_{t} \ge r}$.  □

Proof of Proposition 3

To demonstrate that ${ ROI_{T}(\vec{\lambda})}$ is monotone decreasing in each λ _t , we first establish the following technical result.

Claim

The function

$$ f(\vec{\lambda}) = \frac{\sum^{T-1}_{i=0} a_i(\vec \lambda)}{\sum^{T-1}_{i=0} \omega_i \cdot a_i (\vec \lambda)} $$

is monotone decreasing in λ _t provided: ${(i) \ \omega_{i+1}\ge \omega_i}$ and

$$ (ii)\ \frac{\frac{\partial}{\partial \lambda_t} a_i (\vec \lambda)}{a_i (\vec \lambda)} \le \frac{\frac{\partial}{\partial \lambda_t} a_{i+1} (\vec \lambda)}{a_{i+1} (\vec \lambda)} $$

(18)

for all 0 ≤ i ≤ T−2.

Proof of Claim

The numerator of the derivative of ${f (\vec \lambda)}$ with respect to λ _t is given by:

$$ \sum^{T-1}_{i=0} \sum^{T-1}_{j=i+1} (\omega_j - \omega_i) \left[ \frac{\partial}{\partial \lambda_t} a_i (\vec \lambda)\cdot a_j (\vec \lambda)-\frac{\partial}{\partial \lambda_t} a_j(\vec \lambda) \cdot a_i (\lambda)\right]. $$

Therefore ${\frac{\partial}{\partial \lambda_t} f(\vec \lambda) \le 0}$ whenever conditions (i) and (ii) are met.

To apply the above Claim, we recall that ${Inc_t^{o}= c_t^{o} -d_{t}\cdot b^{o}}$ and set:

$$ \begin{array}{lll} a_0 (\vec \lambda) & = & Inc^0_T \\ a_1 (\vec \lambda) & = & Inc^0_{T-1} \cdot (1 + \lambda_1)\\ & \vdots &\\ a_{T-1} (\vec \lambda) & = & Inc^0_1 \cdot \prod^{T-1}_{i=1}(1+\lambda_{i}).\end{array} $$

Recalling that ${BV^{o}_t= (1- \sum^{t-1}_{i=1} d_{i})\cdot b^{o}}$, we also set:

$$ \omega_0 = \frac{BV^0_{T-1}}{Inc^0_T}, \quad \omega_1 = \frac{BV^0_{T-2}}{Inc^0_{T-1}}, \ldots, \quad \omega_{T-1} = \frac{BV^0_0}{Inc^0_1}. $$

Straightforward differentiation shows that the functions ${a_t (\vec{\lambda})}$ in (18) satisfy the elasticity conditions in part (ii) of the Claim. Finally, by neo-conservatism, RI ⁰ _t is increasing in t. It follows that

$$ ROI^0_{t+1}-r=\frac{RI^0_{t+1}}{BV^0_t} > ROI^0_t-r= \frac{RI^0_t}{BV^0_{t-1}} $$

since ${BV^0_{t-1} \ge BV^0_t}$. We conclude that

$$ ROI^0_1 \equiv \frac{1}{\omega_{T-1}} \le ROI^0_2 \equiv \frac{1}{\omega_{T-2}} \le \cdots ROI^0_T = \frac{1}{\omega_0} $$

implying that ${\omega_0 \le \omega_1 \le \cdots \le \omega_{T-1}}$, as required by condition (i) in the Claim.   □

Proof of Proposition 4

We first show that

$$ RI_T^{\lambda} \equiv Inc_{T} -\lambda \cdot BV_{T-1} $$

is independent of the depreciation schedule ${(d_1,\ldots,d_T)}$ , where

$$ Inc_{T} \equiv Inc^{o}_{T}+ Inc^{o}_{T-1}\cdot (1+\lambda)+\cdots + Inc^{o}_{1} \cdot (1+\lambda)^{T-1} $$

and

$$ BV_{T-1}\equiv BV^{o}_{T-1}+ BV^{o}_{T-2}\cdot (1+\lambda)+\cdots + BV^{o}_0 \cdot (1+\lambda)^{T-1}. $$

Pulling out the factor (1 + λ)^T, we observe that:

$$ RI_T^{\lambda}= (1+\lambda)^T \cdot \left[\sum^{T}_{t=1} RI^{o}_{t}(\lambda) \cdot (1+\lambda)^{-t}\right], $$

(19)

where ${RI^{o}_{t}(\lambda) \equiv Inc^{o}_{t} -\lambda \cdot BV^{o}_{t-1}}$. Since the residual income numbers ${RI^{o}_{t}(\lambda)}$ have the same present value as the underlying cash flows, the right hand side of (19) is equal to:

$$ (1+\lambda)^T \cdot \left[\sum^{T}_{t=1} c^{o}_{t} \cdot (1+\lambda)^{-t} -b^o\right], $$

which obviously is independent of ${(d_1,\ldots,d_T)}$. If one chooses the depreciation schedule ${(d_1^{\ast},\ldots,d_T^{\ast})}$, which is neutral for ${\mathcal{P}}$, then by definition:

$$ Inc_{T}^{\ast}= r \cdot MV_{T-1}, $$

and

$$ BV_{T-1}^{\ast} = MV_{T-1}. $$

Using the above observation that ${RI_T^{\lambda}}$ is invariant to the depreciation schedule, we conclude that for any ${(d_1,\ldots,d_T)}:$

$$ (r-\lambda) \cdot MV_{T-1}= Inc_{T} -\lambda \cdot BV_{T-1}, $$

or equivalently:

$$ ROI_{T}(\lambda, \delta) = \lambda + (r-\lambda) \cdot \frac{MV_{T-1}(\lambda, r)}{BV_{T-1}(\lambda)}. $$

(20)

Proof of Lemma 2

Differentiating the function h(s), we obtain:

$$ h'(s)= \frac{(1+s)^{T-1}}{[(1+s)^{T}-1]^2} \cdot \left[(1+s)^{T+1}-(1+s+sT)\right] > 0; $$

(21)

$$ h''(s) = T \cdot (1+s)^{T-2} \cdot \left[\frac{(2+s) \cdot [1-(1+s)^{T}]+sT \cdot (1+(1+s)^T)}{[(1+s)^{T}-1]^3}\right]. $$

(22)

The monotonicity of h(·) in (21) follows from Bernoulli’s inequality. To establish convexity, first note that as ${s \rightarrow 0, (22) \rightarrow \frac{T^{2}-1}{6T}> 0.}$ Assume henceforth that s ≠ 0. We can ignore the positive ${T \cdot(1+s)^{T-2}}$ expression in front of (22). Now, for s ≠ 0, the denominator of (22) has the same sign as s. The numerator equals (1−T) < 0 as s → −1, and its value as well as its derivative equal 0 at s → 0. Further, its second derivative is given by the expression ${T(T-1)(T+1)s(1+s)^{T-2}}$, which has the same sign as s. Combining these facts, the numerator of (22) is monotone increasing and has an inflection point at s = 0, i.e., it has the same sign as s everywhere, thereby proving that (22) > 0 for all s. To prove that the ratio of h′(·) to h′′(·) increases in s, we use (21) and (22) to obtain the simplified ratio:

$$ \frac{h'(s)}{h''(s)} = \frac{(1+s) \cdot [(1+s)^{T}-1] \cdot [(1+s)^{T+1}-s(1+T)-1]}{(2+s) \cdot [1-(1+s)^{T}]+s \cdot T \cdot [1+(1+s)^T]}. $$

(23)

Letting z =  (1 + s) > 0, (23) can be re-expressed as:

$$ \frac{z \cdot (z^{T}-1) \cdot[z^{T+1}-z-T(z-1)]}{(1+z) \cdot (1-z^T)+T \cdot (z-1) \cdot(1+z^T)} $$

(24)

The key to the proof is to recognize that (z−1)³ is a factor of both the numerator and denominator of (24). Dividing through by this term, (24) can then be reduced to the following ratio of polynomials with positive coefficients:

$$ \frac{\left(\sum_{i=0}^{T-1} \; z^{i+1}\right) \cdot \left(\sum_{i=0}^{T-1}\; (T-i)z^{i}\right)}{\sum^{T-1}_{i=1} \; i(T-i)z^{i-1}} $$

(25)

$$ =\frac{\sum_{i=1}^{T} \; i(2T-i+1)z^{i} \;+ \sum_{i=T+1}^{2T-1} \; (2T-i+1)(2T-i)z^{i}}{2 \cdot \sum^{T-2}_{i=0} \; (i+1)(T-i-1)z^{i}}. $$

(26)

We ignore the constant and differentiate (26) with respect to z. The resulting numerator, which we must show is positive, is the following polynomial of order (3T−4):

$$ \begin{aligned} \left(\sum_{i=0}^{T-2}\; (i+1)(T-i-1)z^{i}\right) \cdot \left[\sum_{i=0}^{T-1} \; (i+1)^2(2T-i)z^{i} + \sum_{i=T}^{2T-2} \; (i+1)(2T-i)(2T-i-1)z^{i}\right]\\ - \left(\sum_{i=0}^{T-3} \; (i+1)(i+2)(T-i-2)z^{i}\right) \cdot \left[\sum_{i=1}^{T} \; i(2T-i+1)z^{i} + \sum_{i=T+1}^{2T-1} \; (2T-i+1)(2T-i)z^{i}\right]. \end{aligned} $$

(27)

For ${1 \le i \le (3T-4)}$ , the coefficient of z ⁱ in (27) can then be simplified as follows:

Range of i	Coefficient of i
i ≤ (T−1)	${\frac{1}{60}\cdot (1+i)\cdot (2+i) \cdot (3+i)\cdot [i^{2}(T+3)+i(7-11T)+20T(T-1)]}$
i =  T	${\frac{1}{60}\cdot T\cdot(T+1)\cdot(T-1)\cdot[T^{3}+18T^2+71T-162]}$
(T + 1) ≤ i ≤ (2T−3)	${\frac{1}{60}\cdot\left({\begin{array}{l}-84i-190i^{2}-150i^ {3}-50i^{4}-6i^{5}+246t +784iT+760i^{2}T\cr +270i^3T+20i^4T-4i^5t-665T^2-1005iT^2-300i^2T^2 \cr +100i^3T^2+40i^4T^2+215T^3-335iT^3-540i^2T^3-150i^3T^3 \cr +485T^4+785iT^4+260i^2T^4-341T^5-205iT^5+60T^6 \end{array}}\right)}$
i ≥ (2T−2)	${\begin{array}{l}\frac{1}{60}\cdot(T+1)\cdot(3T-i)\cdot[3T-1-i]\cdot[3T-2-i]\cr \quad\cdot[3T-3-i]\cdot[3i+7-4T]\end{array}}$

It can be verified that each of these coefficients is positive for any suitable values of T and i (details are available on request). As the polynomial in (27) is defined over the positive real line, we thus obtain the desired result that h′(·) is (strictly) log-concave everywhere.  □

Proof of Proposition 5

Direct substitution yields:

$$ ROI_{T}(\lambda, \delta) = \lambda + \frac{(\lambda+\delta) [1-(1-\delta)^T]}{[(1+r)^T-1]} \cdot \frac{r(1+r)^T [(1+\lambda)^T-1]-\lambda(1+\lambda)^T [(1+r)^T-1]}{\lambda(1+\lambda)^T [1-(1-\delta)^T]+\delta(1-\delta)^T [1-(1+\lambda)^T]}. $$

(28)

The monotonicity of ${ROI_{T}(\lambda, \delta)}$ in λ follows from Proposition 3 and the observation that geometric depreciation results in neo-conservative accounting when cash flows are uniform and δ > −r. To prove convexity in λ, we let p = −δ and express (28) in the following form:

$$ ROI_{T}(\lambda, p) = p + (p-\lambda) \cdot \left[\frac{h(p)-h(r)}{h(\lambda)-h(p)}\right]. $$

(29)

Differentiating (29) twice with respect to λ, we find that:

$$ \frac{\partial^2}{\partial\lambda^2}\left(ROI_{T}(\lambda,p)\right) =\frac{[h(r)-h(p)]}{[h(\lambda)-h(p)]^4} \cdot \left([h(\lambda)-h(p)] \cdot Q(\lambda,p)\right), $$

(30)

where ${Q(\lambda,p) = -h''(\lambda)(\lambda-p)[h(\lambda)-h(p)]-2h'(\lambda) \cdot [h(\lambda)-h(p)-h'(\lambda)(\lambda-p)]}$. We will demonstrate that for any ${\lambda \ne p,[h(\lambda)-h(p)] \cdot Q(\lambda,p) > 0}$. Since h(·) is an increasing function, (30) then immediately yields the desired results for conservative accounting, i.e., h(r) > h(p). Conversely, for liberal accounting ROI _T (·, p) is concave in λ since (30) is negative when h(r) < h(p).

Fix some λ > −1 and consider the behavior of Q(·) in p. Note that Q(λ,λ) = 0 and Q _p (λ,λ) =  0. In addition, we have:

$$ Q_{pp}(\lambda,p) = h''(\lambda)h''(p) \cdot \left[(\lambda-p)+2 \cdot \left(\frac{h'(\lambda)}{h''(\lambda)} - \frac{h'(p)}{h''(p)}\right)\right]. $$

(31)

For p < λ, ${\frac{h'(\lambda)}{h''(\lambda)} > \frac{h'(p)}{h''(p)}}$ by the log-concavity of h′(·), implying that Q _pp (·) > 0. In turn, this implies that Q _p (·) < 0 for p < λ and therefore that Q(·) > 0 for p < λ. We thus have ${[h(\lambda)-h(p)] \cdot Q(\lambda,p) > 0}.$

For ${p > \lambda}$, a similar analysis of (31) using the log-concavity of h′(·) yields Q _pp (·) < 0. This implies that Q _p (·) < 0 for p > λ and thus Q(·) < 0 for p > λ. Again, we get [h(λ)−h(p)]·Q(λ,p) > 0, thereby completing the proof of the first claim.

To prove decreasing differences in λ and δ, we find it convenient to let p =  −δ and to rewrite (28) as:

$$ ROI_{T}(\lambda, p) = \lambda + (p-\lambda) \cdot \left[\frac{h(\lambda)-h(r)}{h(\lambda)-h(p)}\right]. $$

(32)

We show that the cross-partial of (32) in λ and p is positive when the accounting is conservative (i.e., p <  r). Simplifying the cross-partial and ignoring its (positive) denominator, we find that the sign of ${\frac{\partial^2}{\partial \lambda \partial p} \left(ROI_{T}(\lambda, p)\right)}$ is given by the sign of:

$$ F(\lambda,r,p) = [h(\lambda)-h(p)] \cdot \left[\begin{array}{l} h'(\lambda) \cdot [h(r)-h(p)] \cdot [h(\lambda)-h(p)-h'(p)(\lambda-p)]\\ + h'(p) \cdot [h(r)-h(\lambda)] \cdot [h(\lambda)-h(p)-h'(\lambda)(\lambda-p)]\end{array}\right]. $$

(33)

We must show that F(λ,r,p) > 0 for p < r. Note that if p < r < λ, then, by the monotonicity and convexity of the h(·) function, (33) immediately yields F(·) > 0. Accordingly, we restrict attention to values of λ < r. Next, note that F(·) is symmetric in p and λ. With no loss of generality, we can thus assume that p < λ < r. Since h(·) is an increasing function, this implies that ${0 < [h(r)-h(\lambda)] < [h(r)-h(p)]}$, or:

$$ F(\lambda,r,p) > [h(\lambda)-h(p)] \cdot [h(r)-h(\lambda)] \cdot G(\lambda,p), $$

(34)

where ${G(\lambda,p) = h'(\lambda) \cdot [h(\lambda)-h(p)-h'(p)(\lambda-p)] + h'(p) \cdot [h(\lambda)-h(p)-h'(\lambda)(\lambda-p)]}$. It is thus sufficient for us to demonstrate that G(·)≥0 for all values of λ > p. To do so, fix λ > −1 and consider the behavior of G(·) in p. First, note that ${G(\lambda,-1) = h(\lambda)h'(\lambda) > 0}$ and G(λ,λ) = 0. Moreover, ${G_p(\lambda,\lambda) = G_{pp}(\lambda,\lambda) = 0}$, while ${G_{ppp}(\lambda,\lambda)=[h'(\lambda)h'''(\lambda)-3h''(\lambda)^2] < 0}$, by the log-concavity of h′(·). Together, these facts imply that G(·) is positive at small values of p and is tangent to the origin line from above as p → λ. Finally, to verify that G(·) does not cut the origin line at any point prior to p = λ, we show that it cannot have a local maximum in this region. Suppose not, i.e., assume there exists p ^* < λ such that G _p (λ,p ^*) = 0 and ${G_{pp}(\lambda,p^{\ast})\le 0}$. But, ${G_p(\lambda,p^{\ast})=0}$ implies:

$$ h(\lambda)-h(p^{\ast})-2h'(\lambda)(\lambda-p^{\ast})= \frac{h'(p^{\ast})}{h''(p^{\ast})} \cdot[h'(p^{\ast})-h'(\lambda)]. $$

(35)

In turn, (35) indicates that:

$$ \begin{aligned} G_{pp}(\lambda,p^{\ast})&=h'''(p^{\ast})\cdot[h(\lambda)-h(p^{\ast}) -2h'(\lambda)(\lambda-p^{\ast})]+3h''(p^{\ast})\cdot[h'(\lambda)-h'(p^{\ast})]\\ &=\frac{(h'(\lambda)-h'(p^{\ast}))}{h''(p^{\ast})} \cdot \left[3h''(p^{\ast})^2-h'(p^{\ast})h'''(p^{\ast})\right] > 0, \end{aligned} $$

(36)

where the final inequality in (36) follows from p ^* < λ, the convexity of h(·), and the log-concavity of h′(·). This contradicts the supposition that p ^* is a local maximum, thus verifying that G(·) > 0 for all λ > p.   □

Proof of Corollary 3

To establish the limit results (i)–(iii), we note that the limit result for λ → −1 is obvious. The result for λ → 0 follows from repeated applications of l’Hospital’s rule, as shown below:

$$ \begin{array}{lll}\lim_{\lambda \rightarrow 0} \frac{\lambda}{T \cdot h(\lambda) -1}& =& \lim_{\lambda \rightarrow 0} \frac{(1 + \lambda)^T - 1 + \lambda T(1 + \lambda)^{T-1}}{-T (1 + \lambda)^{T-1} + T (1 + \lambda)^T + \lambda T^2 (1 + \lambda)^{T-1}}\\ & = & \lim_{\lambda \rightarrow 0} \frac{2 (1 + \lambda) + \lambda (T-1)}{-T + 1 + 2T (1 + \lambda) + \lambda T (T-1)}\\ & = & \lim_{\lambda \rightarrow 0} \frac{2 + \lambda + T\lambda}{1 + T + T\lambda + \lambda T^2} = \frac{2}{1+ T}. \end{array} $$

The limit result as λ → ∞ follows from applying l’Hospital’s rule once more to the final expression above.

Proof of Proposition 6

If a β-fraction of new investments is expensed, steady state depreciation is given by:

$$ (1-\beta)\cdot b^o\cdot\frac{1}{T}\sum_{j=1}^T(1+\lambda)^{j-1} =(1-\beta)\cdot b^o\cdot\frac{1}{T}\frac{[(1+\lambda)^T-1]}{\lambda}. $$

(37)

The starting book value is simply scaled by (1−β), while cash flows and the specification of the internal rate of return (r) are unaltered. Income in the numerator of ROI _T is therefore given by:

$$ c^o\cdot\left[\frac{(1+\lambda)^T -1}{\lambda}-\beta (1+\lambda)^T\frac{[1-(1+r)^{-T}]}{r}-\frac{(1-\beta) [1-(1+ r)^{-T}][(1+\lambda)^T-1]}{T\lambda r}\right], $$

(38)

while book value in the denominator becomes:

$$ \frac{1-\beta}{T}\cdot\frac{c^o[1-(1+ r)^{-T}]}{ r}\cdot\frac{[1-(1+\lambda)^T+\lambda T(1+\lambda)^T]}{\lambda^2}. $$

Simplification of the resulting ratio then yields ROI _T (λ,0,β):

$$ \begin{aligned} &=\lambda\cdot\left[\frac{[(1+\lambda)^T-1][r T-1+(1+r)^{-T}]+\beta[1-(1+r)^{-T}][(1+\lambda)^T-1-\lambda T(1+\lambda)^T]}{(1-\beta)[1-(1+r)^{-T}][1-(1+\lambda)^T+\lambda T(1+\lambda)^T]}\right]\\ &=\frac{\lambda\cdot[(1+\lambda)^T-1][1-(1+r)^T+ r T(1+r)^T]+\beta[(1+r)^{T}-1][(1+\lambda)^T-1-\lambda T(1+\lambda)^T]}{(1-\beta)\cdot[(1+r)^{T}-1][1-(1+\lambda)^T+\lambda T(1+\lambda)^T]}\\ &=\frac{\lambda\cdot[(1+\lambda)^T-1][1-(1+r)^T+ r T(1+r)^T]}{[1-(1+\lambda)^{T}+\lambda T(1+\lambda)^T][(1+r)^T-1]}\cdot\frac{1}{1-\beta}-\frac{\lambda\cdot\beta}{1-\beta}\\ &= \lambda\cdot \frac{[h(r) -\frac{1}{T}]}{[h(\lambda) -\frac{1}{T}]}\cdot\frac{1}{1-\beta}-\frac{\lambda\cdot\beta}{1-\beta}\\ &=\frac{1}{1-\beta} \cdot [ROI_{T}(\lambda,0)-\lambda \cdot \beta]. \end{aligned} $$

The claim regarding the negative cross-partial derivative of ROI _T (λ,0,β) follows immediately upon recalling that ROI _T (λ,0) is everywhere decreasing in λ because setting δ = 0 corresponds to conservative accounting.

Appendix B

2.1 Alternative methods for calculating the cost of capital

2.1.1 Method 1: Target Price Method (r _DIV )

The target price method, introduced in Botosan and Plumlee (2002), employs a short-horizon form where the infinite series of future cash flows is truncated at the end of year five by inserting a forecasted terminal value. This yields the equation below. The primary assumption underlying this method is that analysts’ forecasts of dividends per share during the forecast horizon and stock price at the end of the forecast horizon capture the market’s expectation of those values.

$$ P_0=\sum^5_{t=1}(1+r_{DIV})^{-t}(dps_t)+(1+ r_{DIV})^{-5} \cdot P_5 $$

where:

• P₀ =  price at time t = 0.

• P₅ =  price at time t = 5.

• r _DIV =  estimated cost of equity capital.

• dps _t =  dividends per share.

Dividend forecasts for the current fiscal year (t = 1), the following fiscal year (t = 2), and the long run (t = 5), as well as maximum and minimum long-run target price estimates are collected from forecasts published by Value Line during the third quarter of the calendar year. Since Value Line does not provide dividend forecasts for years t = 3 and t = 4, we interpolate between the year t = 2 and t = 5 dividend forecasts using an implied straight-line rate of growth in dividends from year t = 2 to year t = 5.

Our forecast of terminal value (P ₅) is the 25th percentile of Value Line’s forecasted long-run price range, although our conclusions are robust to the use of the 50th percentile or the minimum value. We use the 25th percentile to adjust for an apparent optimistic bias in analysts’ forecasts of target price. Current stock price (P ₀) equals the stock price reported on CRSP on the Value Line publication date or closest date thereafter within 3 days of publication.

2.1.2 Method 2: Industry Method (r _GLS )

The industry method, introduced by Gebhardt et al. (2001), employs a residual income valuation model derived from a 12-year forecast horizon. The following model results:

$$ P_0 = b_0+\sum^{11}_{t=1} \gamma_{GLS}^{t} \cdot (ROE_t-r_{GLS}) \cdot b_{t-1}+\frac{\gamma_{GLS}^{12}}{r_{GLS}} \cdot(ROE_{12}- r_{GLS})\cdot b_{11} $$

where

• ${ROE_t = \frac{eps_t}{b_{t-1}}}$ is the forecasted return on equity for period t.

• eps _t = forecasted earnings per share in year t,

• b _t = book value per share in year t,

• r _GLS = estimated cost of equity capital,

• ${\gamma_{GLS} \equiv \frac{1}{1+ r_{GLS}}}$.

2.1.3 Method 3: PEG Ratio Method (r _PEG )

The PEG method, introduced by Easton (2004), proceeds as follows:

$$ r_{PEG} = \sqrt{\frac{eps_2 - eps_1}{P_0}}. $$

Our method is similar, but we use long-run earnings forecasts (eps ₅ and eps ₄) in place of eps ₂ and eps ₁ (consistent with Botosan and Plumlee 2005). Accordingly, the empirical specification of the equation we employ to estimate r _PEG is given by:

$$ r_{PEG} = \sqrt{\frac{eps_5 - eps_4}{P_0}}. $$