Conservatism correction for the market-to-book ratio and Tobin’s q

Abstract

We decompose the market-to-book ratio into two additive components: a conservatism correction factor and a future-to-book ratio. The conservatism correction factor exceeds the benchmark value of one whenever the accounting for past transactions has been subject to an (unconditional) conservatism bias. The observed history of a firm’s past investments allows us to calculate the magnitude of its conservatism correction factor, resulting in an average value that is about two-thirds of the overall market-to-book ratio. We demonstrate that our measure of Tobin’s q, obtained as the market-to-book ratio divided by the conservatism correction factor, has greater explanatory power in predicting future investments than the market-to-book ratio by itself. Our model analysis derives a number of structural properties of the conservatism correction factor, including its sensitivity to growth in past investments, the percentage of investments in intangibles, and the firm’s cost of capital. We provide empirical support for these hypothesized structural properties.

Appendices

Appendix 1: Description of variables

MV T	= Market value of equity at end of fiscal year T (CSHO * PRCC_ F)
MB un T	= Market value of equity at time T divided by book value of equity for fiscal year T
FA T	= Net financial assets at end of fiscal year t, measured as total assets minus net plant and intangibles minus liabilities (AT − PPENT − INTAN − LT)
MB T	= $\frac{\displaystyle{MV_T - FA_T}}{\displaystyle{OA^{o}_T}} = \frac{\displaystyle{MV_T - FA_T}}{\displaystyle{BV^{o}_T - FA_T}}$. Adjusted market-to-book ratio
OA o t	= Net plant + intangibles, at end of fiscal year T, (PPENT + INTAN)
Total Investment	= Advertising expense plus R & D expense plus Capital expenditures for period T − 1 to T (XAD + XRD + CAPXV)
OIADP	= Operating income after depreciation, amortization
Expenses	= SALE-OIADP
α t	= Conservatism in fiscal year t, measured as (XAD + XRD)/(XAD + XRD + CAPXV), where XAD is advertising expense; XRD is Research and Development expense; and CAPXV is capital expenditures
α a T	= Growth weighted average of directly expensed investments
	$\frac{\displaystyle{\alpha_1 (1 + \lambda_1) + \cdots + \alpha_{T} \prod\nolimits^{T}_{i=1}(1 + \lambda_i)}}{\displaystyle{ (1 + \lambda_1) + \cdots+\prod\nolimits^{T}_{i=1} (1 + \lambda_i)}}$
T	= Useful life of plant and intangibles, (average gross\ plant+intangibles)/ (depreciation+amortization of intangibles) measured as (average PPEGT + INTAN)/dep
λ t	= $\frac{\displaystyle{XAD_{t}+XRD_{t}+CAPXV_{t}}}{\displaystyle{XAD_{t-1}+XRD_{t-1}+CAPXV_{t-1}}} -1$
λ a T	= Geometric mean of growth over T periods
λ a3	= Geometric mean of growth rates (λ T , λ T−1, λ T−2)
r e	= Equity cost of capital for firm i and year T, estimated with coefficients from the Fama and French (1992) two-factor model:
	$R_i - R_f = \delta_0 + \delta_1 (R_m - R_f) + \delta_2 (SMB) + \epsilon$
	using CRSP monthly returns from five preceding years and Ken French’s data on market and size factors (SMB)
τ T	= Statutory income tax rate in year T
r d	= Cost of Debt: $(1 - \tau_T) \cdot$ Interest Expense divided by the average of beginning and ending balance of interest-bearing debt
r	= Weighted average cost of capital: $\frac{\displaystyle{BV^{o}_T}}{\displaystyle{AT^{o}_T}}\cdot r_e + \frac{\displaystyle{AT^{o}_T - BV^{o}_T}}{\displaystyle{AT^{o}_T}}\cdot r_d \cdot (1-\tau_T)$
$\Upgamma^n$	$\equiv \gamma + \gamma^2 + \dots + \gamma^n$
N T	= $\Upgamma^1 + \Upgamma^2 (1 + \lambda_2) + \dots + \Upgamma^T \prod\limits^{T}_{i=2} (1 + \lambda_i)$
D T	= $(1 - \alpha_1) \left( 1 - \frac{\displaystyle{T-1}}{\displaystyle{T}}\right) + (1-\alpha_2) \left( 1 - \frac{\displaystyle{T-2}}{\displaystyle{T}}\right) (1 + \lambda_2) + \dots$
	$(1 - \alpha_{T-1}) \prod\limits^{T-1}_{i=2} (1 + \lambda_i) \cdot \left( 1 - \frac{\displaystyle{1}}{\displaystyle{T}}\right) + (1 - \alpha_{T}) \prod\limits^{T}_{i=2} (1 + \lambda_i)$
$\Upgamma^T$	= $\frac{\displaystyle{1}}{\displaystyle{1+r}}$ + $\frac{\displaystyle{1}}{\displaystyle{{(1+r)}^2}} +\cdots +$ $\frac{\displaystyle{1}}{\displaystyle{{(1+r)}^T}}$
CC T	= $\frac{\displaystyle{N_T}}{\displaystyle{D_T}} \cdot \frac{\displaystyle{1}}{\displaystyle{\Upgamma^T}}$ with T, r, γ, and λ as defined above
CC λ T	= Same as CC T except that λ t  = λ a T for all t
FB T	= MB T  − CC T
$\widehat{FB}_T$	= Estimated Future-to-Book Value, defined in equation (23)
$\widehat{CC}_T$	= $MB_T- \widehat{FB}_T$
EconCost T	= Expenses T  − dep T + $\frac{\displaystyle{1}}{\displaystyle{\Updelta_T}}(dep_T + r\cdot OA_{T-1})$
$\Updelta_T$	= $\Upgamma^T\cdot \frac{\displaystyle{u_0 + u_1 (1 + \lambda_1) + \cdots + u_{T-1} \prod\nolimits^{T-1}_{i=1} (1 + \lambda_i)+ \alpha_T \cdot \prod\nolimits^{T}_{i=1} (1 + \lambda_i)}}{\displaystyle{1 + (1 + \lambda_1) + \cdots + \prod\nolimits^{T-1}_{i=1} (1 + \lambda_i)}}$
u t	= $(1-\alpha_t)\left[\frac{\displaystyle{1}}{\displaystyle{T}} + r \cdot\left(1- \frac{\displaystyle{T-1-t}}{\displaystyle{T}}\right) \right]$ for 0 ≤ t ≤ T − 1
$\mathcal{I}\{x\}$	= x if x ≥ 0 and $\mathcal{I}\{x\}=0$ if x ≤ 0
$\Upgamma_{\lambda}^5$	= Capitalization factor, given by $\sum_{i=1}^{5} \left(\frac{\displaystyle{1+\lambda^a_3}}{\displaystyle{1+r}}\right)^i$

Appendix 2: Proofs

Proof of Proposition 1 We can set α to 0 without loss of generality for this proof. From (8), CC _T equals (after dividing through by the common term $I_0\cdot (1+\lambda_{1})$):

$$ CC_T = \frac{BV_{T}({\bf I}_T, {\bf d}^*)}{BV_{T}({\bf I}_T, {\bf d}^o)} = \frac{bv^{*}_{T-1}+ bv^{*}_{T-2} \cdot(1+\lambda_{2})+\cdots+ bv^{*}_{0}\cdot\prod^{T}_{i=2}(1+\lambda_{i})}{bv^{o}_{T-1}+ bv^{o}_{T-2} \cdot(1+\lambda_{2})+\cdots+ bv^{o}_{0}\cdot\prod^{T}_{i=2}(1+\lambda_{i})}. $$

(26)

This ratio is decreasing in λ _t if and only if the sequence $\frac{bv^*_t}{bv^{o}_t}$ is an increasing function of t. ³¹ Given the replacement cost accounting rule, d ^*, we know from (6) and (11) that z ^* _t  = x _t  = 0 for 1 ≤ t ≤ L − 1 and $z^{*}_{t}= c \cdot x_{t}$, for L ≤ t ≤ T. It follows that for all $t, z^{*}_{t}=c\cdot x_t$. We recall the following identity linking book values to future “residual income charges.”

$$ bv_t({\bf d}) = \sum_{i=t+1}^{T} z_{i}({\bf d}) \cdot\gamma^{i-t}. $$

Denoting bv ^o _t ≡ bv _t (d ^o), we have:

$$ \frac{bv^*_t}{bv^o_t } = \frac{\displaystyle \sum\nolimits_{i=t+1}^{T} z_{i}({\bf d}^*) \cdot \gamma^{i-t}}{\displaystyle \sum\nolimits_{i=t+1}^{T} z_{i}({\bf d}^o) \cdot\gamma^{i-t}} = \frac{\displaystyle c \cdot \sum\nolimits_{i=t+1}^{T} x_{i} \cdot\gamma^{i}}{\displaystyle \sum\nolimits_{i=t+1}^{T} z_{i}({\bf d}^o) \cdot\gamma^{i}}. $$

(27)

Analogously, we have:

$$ \frac{bv^*_{t-1}}{bv^o_{t-1}} = \frac{\displaystyle c \cdot \sum\nolimits_{i=t}^{T} x_{i} \cdot\gamma^{i}}{\displaystyle \sum\nolimits_{i=t}^{T} z_{i}({\bf d}^o) \cdot\gamma^{i}} = \frac{\displaystyle c \cdot \left[x_t \cdot \gamma^t + \sum\nolimits_{i=t+1}^{T} x_{i} \cdot\gamma^{i}\right]}{\displaystyle z_{t}({\bf d}^o) \cdot \gamma^t + \sum\nolimits_{i=t+1}^{T} z_{i}({\bf d}^o) \cdot\gamma^{i}}. $$

(28)

To establish that (27) ≥ (28), we note that, for 1 ≤ t ≤ L − 1, the inequality follows immediately since x _t  = 0 and z _t (d ^o) ≥ 0 (see Definition 2). For t ≥ L, the result holds if and only if

$$ \frac{z_{t}({\bf d}^o)}{x_t} \ge \frac{\displaystyle \sum\nolimits_{i=t+1}^{T} z_{i}({\bf d}^o)\cdot\gamma^{i}}{\displaystyle \sum\nolimits_{i=t+1}^{T} x_{i} \cdot\gamma^{i}}. $$

(29)

We demonstrate (29) by induction. For t = T − 1, (29) requires:

$$ \frac{z_{T-1}({\bf d}^o)}{x_{T-1}} \ge \frac{z_{T}({\bf d^o})}{x_{T}}, $$

which is true as $\frac{z_t({\bf d}^o)}{x_t}$ decreases in t. Now suppose that (29) holds for t = k. Then, for t=k−1,

$$ \frac{z_{k-1}({\bf d}^o)}{x_{k-1}} \ge \frac{z_k({\bf d}^o)}{x_k} = \frac{z_k({\bf d}^o) \cdot \gamma^k}{x_k \cdot \gamma^k} \ge \frac{\displaystyle z_k({\bf d}^o) \cdot \gamma^k+\sum\nolimits_{i=k+1}^{T} z_{i}({\bf d}^o) \cdot\gamma^{i}}{\displaystyle x_k \cdot \gamma^k+\sum\nolimits_{i=k+1}^{T} x_{i} \cdot\gamma^{i}} = \frac{\displaystyle \sum\nolimits_{i=k}^{T} z_{i}({\bf d}^o) \cdot\gamma^{i}}{\displaystyle \sum\nolimits_{i=k}^{T} x_{i} \cdot\gamma^{i}}, $$

where the second inequality arises from the induction hypothesis. We have thus shown that (29) holds. To conclude, note that we have stated the proof in terms of weak inequalities. However, if either z _t (d ^o) > 0 for some t ≤ L − 1 or $\frac{z_t({\bf d^o})}{x_t}$ strictly decreases in t for t ≥ L, it follows that $\frac{bv^*_t}{bv^o_t}$ strictly increases for some subset of values of t and therefore that CC _T is monotone decreasing in each λ _t .

Proof of Proposition 2

Consider CC _T as represented in equation (26). The denominator, BV _T , is determined by the depreciation scheme under consideration and is independent of the cost of capital, r. So it is sufficient to show that the numerator, BV ^* _T , increases in r, or, equivalently, that it decreases in γ. We use the following formulation of BV _T (I _T , d ^*):

$$ BV_{T}({\bf I}_T, {\bf d}^*) = bv^{*}_{T-1}\cdot I_1 + bv^{*}_{T-2} \cdot I_2 +\cdots+ bv^{*}_{0}\cdot I_T. $$

As in the proof of Proposition 1, we set $bv^*_t = c \cdot \sum_{i=t+1}^{T} x_{i} \cdot\gamma^{i-t}$. BV _T (I _T , d ^*) therefore equals:

$$ c \cdot \left[I_1\cdot \sum_{i=T}^{T} x_i \cdot \gamma^{i-(T-1)} + I_2 \cdot \sum_{i=T-1}^{T} x_i \cdot \gamma^{i-(T-2)} + \ldots+ I_T\cdot \sum_{i=1}^{T} x_i \cdot \gamma^{i}\right]. $$

As $c=\frac{1}{\sum_{i=L}^{T} x_{i} \cdot\gamma^{i}}$, we need to show that the following expression decreases in γ:

$$ \frac{\displaystyle I_1\cdot x_{T} \cdot \gamma + I_2\cdot \sum\nolimits_{i=T-1}^{T} x_i \cdot \gamma^{i-(T-2)} + \ldots+ I_T\cdot \sum\nolimits_{i=1}^{T} x_i \cdot \gamma^{i}}{\sum\nolimits_{i=L}^{T} x_{i} \cdot\gamma^{i}}. $$

(30)

We do so one term at a time. Ignoring the positive constant I _t , an arbitrary term in (30) is of the form:

$$ \frac{\displaystyle{\sum_{i=k}^{T} x_i \cdot \gamma^{i-k+1}}}{\displaystyle{\sum_{i=L}^{T} x_i \cdot \gamma^{i}}}, \quad k\in\{1,2,\ldots, T\}. $$

(31)

Consider k ≤ L. As $x_1=\cdots=x_{L-1}=0$, (31) is equivalent to:

$$ \frac{\displaystyle{\sum_{i=L}^{T} x_i \cdot \gamma^{i-k+1}}}{\displaystyle{\sum_{i=L}^{T} x_i \cdot \gamma^{i}}} = \gamma^{-(k-1)}, $$

(32)

which is decreasing in γ as k ≥ 1.

For k > L, (31) decreases in γ if and only if

$$ \left(\sum_{i=L}^{T} x_i \cdot \gamma^i\right)\cdot \left[\sum_{i=k}^{T} x_i \cdot (i-k+1)\cdot \gamma^{i-k}\right]\leq \left[\sum_{i=k}^{T} x_i \cdot \gamma^{i-k+1}\right] \cdot \left(\sum_{i=L}^{T} x_i\cdot i\cdot \gamma^{i-1}\right),\quad \hbox {or} \frac{\displaystyle{\sum_{i=L}^{T} x_i \cdot \gamma^{i-1}}}{\displaystyle{\sum_{i=L}^{T} x_i\cdot i\cdot \gamma^{i-1}}} \leq \frac{\displaystyle{ \sum\nolimits_{i=k}^{T} x_i \cdot \gamma^{i-k}}}{\displaystyle{\sum\nolimits_{i=k}^{T} x_i \cdot (i-k+1)\cdot \gamma^{i-k}}}. $$

(33)

With regard to the left-hand side of (33), note that:

$$ \frac{\displaystyle \sum\nolimits_{i=L}^{T} x_i \cdot \gamma^{i-1}}{\displaystyle \sum\nolimits_{i=L}^{T} x_i\cdot i\cdot \gamma^{i-1}} < \frac{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i \cdot \gamma^{i-1}}{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i\cdot i\cdot \gamma^{i-1}}. $$

since each additional term in the former has a numerator-to-denominator ratio of less than 1/(L + T − k). So it is sufficient to demonstrate that

$$ \begin{aligned} \frac{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i \cdot \gamma^{i-1}}{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i \cdot i\cdot \gamma^{i-1}} \leq& \frac{\displaystyle \sum\nolimits_{i=k}^{T} x_i \cdot \gamma^{i-k}}{\displaystyle \sum\nolimits_{i=k}^{T} x_i \cdot (i-k+1) \cdot \gamma^{i-k}}\\ \Leftrightarrow \quad \frac{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i \cdot \gamma^{i-L}}{\displaystyle \sum\nolimits_{i=k}^{T} x_i \cdot \gamma^{i-k}} \leq& \frac{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i \cdot i \cdot \gamma^{i-L}}{\displaystyle \sum\nolimits_{i=k}^{T} x_i \cdot (i-k+1)\cdot \gamma^{i-k}}. \end{aligned} $$

(34)

Since x _i /x _i+1 increases in i, we know that $\frac{x_L}{x_k}\leq \frac{x_{L+1}}{x_{k+1}} \cdots\leq \frac{x_{L+T-k}}{x_T}.$ The left-hand side of (34) places equal weight on these ratios, which implies that

$$ \frac{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i \cdot \gamma^{i-L}}{\displaystyle \sum\nolimits_{i=k}^{T} x_i \cdot \gamma^{i-k}} < \frac{\displaystyle \sum\nolimits_{i=L}^{L+T-k} x_i \cdot (i-L+1) \cdot \gamma^{i-L}}{\displaystyle \sum\nolimits_{i=k}^{T} x_i \cdot (i-k+1)\cdot \gamma^{i-k}}, $$

(35)

since the expression on the right-hand side of (35) places increasingly higher weights on the higher ratios. Finally, L ≥ 1 implies that the right-hand side of (34) exceeds the right-hand side of (35). Hence, the inequality in (34) holds, and we have shown that BV _T (I _T , d ^*) (and hence CC _T ) increases in r.

Proof of Proposition 3

For L = 1, we have $x_t=1-\beta\cdot (t-1)$. The capital charges in (16) simplify to:

$$ z_t = \frac{1-\alpha}{T} \cdot \left[1+r \cdot (T-t+1)\right]. $$

Using these expressions, as well as the definition of c, we can rewrite CC _T in (15) as:

$$ CC_T = \frac{T}{\displaystyle \sum\nolimits_{i=1}^{T} [1-\beta\cdot (i-1)] \cdot \gamma^i}\cdot \frac{\displaystyle \sum\nolimits_{i=1}^{T} [1-\beta\cdot (i-1)] \cdot (\gamma^i-\mu^i)}{\displaystyle \sum\nolimits_{i=1}^{T} [1+r \cdot (T-i+1)] \cdot (\gamma^i-\mu^i)}\cdot \frac{1}{1-\alpha}. $$

(36)

Expanding this expression, it can then be shown that the limit values of the CC _T function are as follows:

$$ \begin{aligned} &\lim_{\lambda \rightarrow -1} CC_T(\cdot) \thinspace= \left(\frac{1}{1-\alpha}\right)\cdot \frac{T \cdot r^2 \cdot (1+r)^{T-1} \cdot [1-\beta \cdot(T-1)]}{(r-\beta)\cdot [(1+r)^T-1]+\beta \cdot r \cdot T} \\ &\lim_{\lambda \rightarrow 0} CC_T(\cdot) \enspace = \left(\frac{1}{1-\alpha}\right) \cdot \left[\frac{T \cdot [2+\beta \cdot (1-T)] \cdot r^2 \cdot (1+r)^{T}}{(r-\beta) \cdot [(1+r)^T-1]+\beta \cdot r \cdot T}-2\right] \cdot \frac{1}{r \cdot [T+1]}\\ &\lim_{\lambda \rightarrow \infty} CC_T(\cdot) \thinspace= \frac{1}{1-\alpha}. \end{aligned} $$

(37)

The limit results for the β = 0 case follow directly from these expressions.

We next prove the claim regarding the bounds on the ratios of the CC _T variables. Note that the term 1/(1 − α) enters in a multiplicative fashion in each of the CC _T expressions in (37) and, as such, can be ignored. Also, when T = 2, direct computations on (37) reveal that the ratio in question always equals $\frac{2}{3}$. We therefore restrict attention to values of T > 2. We first show the upper bound result that

$$ \frac{CC_T(\lambda=-1)-CC_T(\lambda=0)}{CC_T(\lambda=-1)-CC_T(\lambda=\infty)} \leq \frac{T}{T+1}. $$

To do so, we will demonstrate the equivalent result that

$$ \frac{CC_T(\lambda=0)-CC_T(\lambda=\infty)}{CC_T(\lambda=-1)-CC_T(\lambda=\infty)}\geq \frac{1}{T+1}. $$

(38)

Using the limits in (37), (38) reduces to the following inequality:

$$ \begin{aligned} &(T+1)\cdot \left[\frac{1}{r(1+T)}\cdot \frac{T r^2 (1+r)^T(2+\beta-\beta T)}{\beta r T +(r-\beta)[(1+r)^T-1]}-\frac{2}{r(1+T)}-1\right] \\ &\quad \geq \frac{T r^2 (1+r)^{T-1}[1+\beta-\beta T]}{\beta r T+(r-\beta)[(1+r)^T-1]}-1. \end{aligned} $$

$$ \begin{aligned} &\Leftrightarrow (T+1)T r^2 (1+r)^T (2+\beta-\beta T)-(T+1)\left[2 + r (1+T)\right] \cdot \left[\beta r T+ (r-\beta) \cdot \left[(1+r)^T -1\right]\right]\\ &\quad \geq T r^2 (1+r)^{T-1}[1+\beta-\beta T]r(1+T)-r(1+T)\big[\beta r T+(r-\beta)[(1+r)^T-1]\big]\\ &\Leftrightarrow T r^2 (1+r)^{T-1}(1+\beta-\beta T)+T r^2 (1+r)^T-\big[\beta r T+(r-\beta)[(1+r)^T-1]\big](2+rT) \geq 0\\ & \Leftrightarrow T r^2(1+r)^{T-1}(2+\beta-\beta T+r)-(2+rT)\big[\beta r T+(r-\beta)[(1+r)^T-1]\big] \geq 0. \end{aligned} $$

(39)

But (39) is a linear function of β. At $\beta=\frac{r}{1+rT}$, (39) equals

$$ \frac{T r^2 (1+r)^T(2+rT)}{(1+rT)}-\frac{T r^2 (1+r)^T (2+rT)}{(1+rT)}=0. $$

At β = 0, the expression in (39) reduces to (after dividing through by r):

$$ T\cdot r \cdot (1+r)^{T-1}(2+r)-(2+rT)[-1+(1+r)^T] \geq 0. $$

Letting s = (1 + r) ≥ 1, this inequality holds if and only if

$$ T\cdot (s-1)\cdot s^{T-1}\cdot (s+1)-2(s^T-1)-(s^T-1)\cdot T \cdot (s-1) \geq 0, $$

or

$$ T\cdot (s-1)(s^{T-1}+1)-2(s^T-1) \geq 0. $$

But this function and its first derivative equal 0 at s = 1, while the second derivative is

$$ (s-1)\cdot T\cdot (T-1)s^{T-3}\cdot (T-2) \geq 0, $$

for all s ≥ 1. So the function is convex and positive everywhere. Thus (39) ≥ 0 for all $\beta \in [0, \frac{r}{1+rT}]$. We have therefore shown that (38) holds.

We next demonstrate the lower bound inequality:

$$ \frac{CC_T(\lambda=-1)-CC_T(\lambda=0)}{CC_T(\lambda=-1)-CC_T(\lambda=\infty)}\geq \frac{2}{3}. $$

Expanding these expressions using (37), we seek to show that

$$ \begin{aligned} &\frac{Tr^2(1+r)^{T-1}(1+\beta-\beta T)}{(r-\beta)[(1+r)^T-1]+\beta r T}- \frac{6 T r^2(1+r)^T(2+\beta-\beta T)}{2r(1+T)\big[\beta r T+(r-\beta)[(1+r)^T-1]\big]}+\frac{6}{r(1+T)}+2 \geq 0,\\ &\Leftrightarrow T r^3 (1+r)^{T-1}(1+T)(1+\beta-\beta T)-3T r^2(1+r)^T(2+\beta-\beta T)+6\big[\beta r T+(r-\beta)[(1+r)^T-1]\big] +2r(1+T)\big[\beta r T+(r-\beta)[(1+r)^T-1]\big] \geq 0,\\ &\Leftrightarrow T r^2 (1+r)^{T-1}(1+\beta-\beta T)[r+rT-3(1+r)]-3T r^2 (1+r)^T+2[3+r+rT]\big[\beta r T+(r-\beta)[(1+r)^T-1]\big]\geq 0. \end{aligned} $$

(40)

Again, this is a linear function of β, so it suffices to show that (40) holds at its end points. At $\beta=\frac{r}{1+rT}$, the expression reduces to

$$ \begin{aligned} &T r^2 (1+r)^{T-1} \frac{(1+r)}{(1+rT)}[r(1+T)-3(1+r)]-3Tr^2(1+r)^T+2[3+r+rT]\cdot \frac{r^2(1+r)^T T}{(1+rT)} \\ =&\frac{T r^2 (1+r)^T}{(1+rT)}[r+rT-3-3r+6+2r+2rT]-3Tr^2(1+r)^T \\ =&\frac{T r^2 (1+r)^T}{(1+rT)}[3(1+rT)]-3Tr^2(1+r)^T=0. \end{aligned} $$

At β = 0, we need to show that

$$ \begin{aligned} & Tr^3(1+r)^{T-1}(1+T)-6Tr^2(1+r)^T+6r((1+r)^T-1)+2r^2(1+T) ((1+r)^T-1)\geq 0\\ & \Leftrightarrow T(1+T)s^{T-1}(s^2-2s+1)-6T(s-1)s^T+6(s^T-1)+2(1+T)(s-1)(s^T-1)\geq 0\\ & \Leftrightarrow T(1+T)s^{T+1}-2T(1+T)s^T+T(1+T)s^{T-1}-6Ts^{T+1}+6T s^T+6s^T-6 +2(1+T)s^{T+1}-2(1+T)s^T-2s(1+T)+2(1+T)\geq 0\\ & \Leftrightarrow (T-2)(T-1)s^{T+1}-2(T-2)(T+1)s^T+T(1+T)s^{T-1}-2(1+T)s+2T-4\geq 0. \end{aligned} $$

Again, this equals 0 at s = 1. In addition, its derivative is

$$ \begin{aligned} (T-2)(T-1)(T+1)s^T-2(T-2)(T+1)Ts^{T-1}+T(1+T)(T-1)s^{T-2}-2(1+T)\\ \propto (T-2)(T-1)s^T-2(T-2)Ts^{T-1}+T(T-1)s^{T-2}-2. \end{aligned} $$

This equals 0 at s = 1. Its derivative in turn is

$$ \begin{aligned} &T(T-2)(T-1)s^{T-1}-2T(T-1)(T-2)s^{T-2}+T(T-1)(T-2)s^{T-3} \\ =&T(T-1)(T-2)s^{T-3}(s-1)^2 > 0, \end{aligned} $$

for all T > 2 and all s > 1. We have thus established that (40) is strictly positive for values of β between 0 and $\frac{r}{1+rT}$. We conclude that for any level of decay in that range, the ratio bounds of $\frac{2}{3}$ and $\frac{T}{T+1}$ hold.