Structural properties of the price-to-earnings and price-to-book ratios

Abstract

We examine the structural properties of a firm’s price-to-earnings (P/E) and price-to-book (P/B) ratios and the relation between these two ratios. A benchmark result is obtained under the hypothesis that firms use replacement cost accounting to value their operating assets, so that the P/B ratio coincides with Tobin’s q. The firm’s P/E ratio can then be expressed as a convex combination of the P/E ratios suggested respectively by the permanent earnings model and the Gordon growth model, with the relative weight to be placed on these two endpoints determined entirely by Tobin’s q. Under current financial reporting rules, the accounting for operating assets is likely to be more conservative than replacement cost accounting. Our findings characterize how the magnitude and behavior of the P/E and P/B ratios are jointly shaped by several key variables, including both past and anticipated future growth, economic profitability, and accounting conservatism

Appendix

Proof of Proposition 1

The residual earnings valuation model provides the identity:

$$\begin{aligned} P_{T}=BV_{T}+\sum _{t=1}^{\infty }RI_{T+t}\cdot \gamma ^{t}, \end{aligned}$$

for any accounting rules, provided the condition of comprehensive income measurement is met; see Preinreich (1935) and Feltham and Ohlson (1995, 1996). Since income is measured comprehensively in our model, we can apply the residual income formula to replacement cost accounting in particular. Thus:

$$\begin{aligned} P_{T}=BV_{T}^{*}+\sum _{t=1}^{\infty }RI_{T+t}^{*}\cdot \gamma ^{t}=BV_{T}^{*}+\sum _{t=1}^{\infty }\pi _{T+t}^{o}\cdot \gamma ^{t}. \end{aligned}$$

As argued in the main text, the proportionate growth assumption in (A1) implies that $\pi _{t+1}^{o}=(1+\mu _{t+1})\cdot \pi _{t}^{o}$. We then obtain:

$$\begin{aligned} P_{T}=BV_{T}^{*}+\frac{RI_{T+1}^{*}}{r-s(\varvec{\mu })}=BV_{T}^{*}+\frac{\pi _{T+1}^{o}}{r-s(\varvec{\mu })}, \end{aligned}$$

(30)

where

$$\begin{aligned} \frac{1}{r-s(\varvec{\mu })}\equiv \gamma +\gamma ^{2}\cdot \left( 1+\mu _{T+2}\right) +\gamma ^{3}\cdot \left( 1+\mu _{T+2}\right) \cdot \left( 1+\mu _{T+3}\right) +\cdots \end{aligned}$$

Dividing both sides in (30) by $P_{T}$ yields

$$\begin{aligned} 1=\frac{1}{r-s(\varvec{\mu })}\frac{E_{T+1}^{*}}{P_{T}} -\frac{s(\varvec{\mu })/q_{T}}{r-s(\varvec{\mu })}, \end{aligned}$$

which, in turn, implies

$$\begin{aligned} \frac{P_{T}}{E_{T+1}^{*}}=\frac{1}{r-s(\varvec{\mu })\cdot \frac{q_{T}-1}{q_{T}}}. \end{aligned}$$

$\square$

Some of the proofs below will rely on the following auxiliary lemma.

Lemma

For any numbers $a_{1},\ldots,a_{n}$, positive numbers $b_{1},\ldots,b_{n}$, and growth rates $\xi _{2},\ldots,\xi _{n}\ge -1$, the function

$$\begin{aligned} f\left( \xi _{1},\ldots,\xi _{n}\right) =\frac{a_{n}+\left( 1+\xi _{2}\right) a_{n-1}+\cdots +\left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{n}\right) a_{1}}{b_{n}+\left( 1+\xi _{2}\right) b_{n-1}+\cdots +\left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{n}\right) b_{1}} \end{aligned}$$

is everywhere increasing (decreasing) in each $\xi _{i}$ for $2\le i\le n$, if the sequence $\frac{a_{i}}{b_{i}}$ is decreasing (increasing) in i.

Proof of Lemma

The claim obviously holds for $n=2$. For $n>2$, the function $f\left( \xi _{1},\ldots,\xi _{n}\right)$ can be written as:

$$\begin{aligned} f\left( \xi _{1},\ldots,\xi _{n}\right) =\frac{A_{2}+\left( 1+\xi _{i}\right) A_{1}}{B_{2}+\left( 1+\xi _{i}\right) B_{1}}, \end{aligned}$$

(31)

where

$$\begin{aligned} A_{2}&=a_{n}+\left( 1+\xi _{2}\right) a_{n-1}+\cdots +\left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{i-1}\right) a_{n-i+2},\\ B_{2}&=b_{n}+\left( 1+\xi _{2}\right) b_{n-1}+\cdots +\left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{i-1}\right) b_{n-i+2},\\ A_{1}&=\frac{1}{\left( 1+\xi _{i}\right) }\left\{ \left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{i}\right) a_{n-i+1}+\cdots +\left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{n}\right) a_{1}\right\} ,\\ B_{1}&=\frac{1}{\left( 1+\xi _{i}\right) }\left\{ \left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{i}\right) b_{n-i+1}+\cdots +\left( 1+\xi _{2}\right) \ldots\left( 1+\xi _{n}\right) b_{1}\right\} . \end{aligned}$$

The fact that $\frac{a_{i}}{b_{i}}$ is decreasing in i implies that

$$\begin{aligned} \frac{A_{2}}{B_{2}}\le \frac{a_{n-i+2}}{b_{n-i+2}}\le \frac{a_{n-i+1}}{b_{n-i+1}}\le \frac{A_{1}}{B_{1}}. \end{aligned}$$

(32)

The representation in (31) and the inequality above reduce the problem for a general n to the special case of $n=2$. $\square$

Proof of Proposition 2

From Proposition 1, we know that:

$$\begin{aligned} \frac{P_{T}}{E_{T+1}^{*}}=\frac{1}{r-s(\varvec{\mu })\cdot \frac{q_{T}-1}{q_{T}}}. \end{aligned}$$

It thus suffices to show that Tobin’s q is decreasing in $\lambda _{t}$. Tobin’s q can be rewritten as

$$\begin{aligned} q_{T}= & {} \frac{P_{T}}{BV_{T}^{*}}=\frac{E_{T+1}^{*}-s(\varvec{\mu })\cdot BV_{T}^{*}}{\left( r-s(\varvec{\mu })\right) \cdot BV_{T}^{*}}\\= & {} \frac{1}{\left( r-s(\varvec{\mu })\right) }\cdot \frac{\left( x_{1}p^{o}-d_{1}^{*}-s(\varvec{\mu })\cdot bv_{0}^{*}\right) I_{T}+\cdots +\left( x_{T}p^{o}-d_{T}^{*}-s(\varvec{\mu })\cdot bv_{T-1}^{*}\right) I_{1}}{bv_{0}^{*}I_{T}+\cdots +bv_{T-1}^{*}I_{1}}. \end{aligned}$$

If we show that

$$\begin{aligned} \frac{x_{i+1}p^{o}-d_{i+1}^{*}-s(\varvec{\mu })\cdot bv_{i}^{*}}{bv_{i}^{*}} \end{aligned}$$

increases in i, then the monotonicity of Tobin’s q will follow from Lemma A by setting $n=T$, $a_{i}=x_{i}p^{o}-d_{i}^{*}-s(\varvec{\mu })\cdot bv_{i-1}^{*}$, $b_{i}=bv_{i-1}^{*}$, and $\xi _{i}=\lambda _{i}$.

Observe that

$$\begin{aligned}&\frac{x_{i+1}p^{o}-d_{i+1}^{*}-s(\varvec{\mu })\cdot bv_{i}^{*}}{bv_{i}^{*}}=\frac{x_{i+1}p^{o}-d_{i+1}^{*}-r\cdot bv_{i}^{*}+\left( r-s(\varvec{\mu })\right) bv_{i}^{*}}{bv_{i}^{*}}\\&\quad =\frac{p^{o}-c}{\left( bv_{i}^{*}/x_{i+1}\right) }+r-s(\varvec{\mu }). \end{aligned}$$

It remains to show that $bv_{i}^{*}/x_{i+1}$ is decreasing in i. To that end, we note that

$$\begin{aligned} \frac{bv_{i-1}^{*}}{bv_{i}^{*}}=\frac{\gamma x_{i}+\cdots +\gamma ^{T-i+1}x_{T}}{\gamma x_{i+1}+\cdots +\gamma ^{T-i+1}x_{T+1}}, \end{aligned}$$

where $x_{T+1}=0$.

We can apply Lemma A to the sequences defined by the following equations:

$$\begin{aligned}&\left( a_{1},\ldots,a_{n}\right) =\left( x_{T},\ldots,x_{i}\right) ,\\&\left( b_{1},\ldots,b_{n}\right) =\left( x_{T+1},\ldots,x_{i+1}\right) ,\\&\left( 1+\xi _{2},\ldots,1+\xi _{n}\right) =\left( \gamma ,\ldots,\gamma \right) ,\\&\left( 1+\xi '_{2},\ldots,1+\xi '_{n}\right) =\left( 0,\ldots,0\right) . \end{aligned}$$

Since the productivity pattern satisfies the condition:

$$\begin{aligned} \frac{x_{t}-x_{t+1}}{x_{t}}\ge \frac{x_{t-1}-x_{t}}{x_{t-1}}, \end{aligned}$$

it follows that:

$$\begin{aligned} \frac{a_{t-1}}{b_{t-1}}\ge \frac{a_{t}}{b_{t}}. \end{aligned}$$

Therefore the function f from Lemma A will be increasing in each $\xi _{i}$, and its value at $\left( \xi _{2},\ldots,\xi _{n}\right)$ will be greater than its value at $\left( \xi '_{2},\ldots,\xi '_{n}\right)$. Hence,

$$\begin{aligned} \frac{bv_{i-1}^{*}}{bv_{i}^{*}}\ge \frac{\gamma \cdot x_{i}+0\cdot x_{i+1}+\cdots +0\cdot x_{T}}{\gamma \cdot x_{i+1}+0\cdot x_{i+2}+\cdots +0\cdot x_{T+1}}=\frac{x_{i}}{x_{i+1}}, \end{aligned}$$

and it follows that the sequence $\frac{bv_{i-1}^{*}}{x_{i}}$ is decreasing in i. $\square$

Proof of Observation 2

Assume that $\varvec{d}$ is more conservative than $\varvec{d}^{\prime }$. Since

$$\begin{aligned} \sum d_{\tau }=1, \end{aligned}$$

we can rewrite $d_{1}$ as:

$$\begin{aligned} d_{1}=\frac{1}{1+\frac{d_{2}}{d_{1}}+\cdots +\frac{d_{2}}{d_{1}}\frac{d_{3}}{d_{2}}\ldots\frac{d_{T}}{d_{T-1}}}. \end{aligned}$$

(33)

Since $\varvec{d}$ is more conservative than $\varvec{d}^{\prime }$,

$$\begin{aligned} \frac{d_{\tau +1}}{d_{\tau }}\le \frac{d_{\tau +1}^{^{\prime }}}{d_{\tau }^{^{\prime }}}. \end{aligned}$$

Equation (33) then implies that $d_{1}\ge d_{1}^{^{\prime }}$. Note that if $d_{\tau }\le d_{\tau }^{^{\prime }}$ for some $\tau$, then $d_{\tau +1}\le d_{\tau }\frac{d_{\tau +1}^{^{\prime }}}{d_{\tau }^{^{\prime }}}\le d_{\tau +1}^{^{\prime }}$. Applying the same argument iteratively, one can verify that $d_{\tau }\le d_{\tau }^{^{\prime }}$ implies $d_{\tau +i}\le d_{\tau +i}^{\prime }$ for any i.

Let $\delta \left( i\right) =bv_{i}-bv_{i}^{^{\prime }}$. We have the following observations:

• 1. $\delta \left( 1\right) =bv_{1}-bv_{1}^{^{\prime }}\le 0.$

• 2. ${\text {If }}\,\delta ( \tau ) -\delta ( \tau +1) \le 0\,{\text{ for}}\,{\text{ some }}\,\tau,\,{\text { then }}\,\delta ( \tau +i) -\delta \left( \tau +i+1\right) \le 0\,{\text { for}}\, {\text{any }}\,i\ge 0.$

• 3. $\delta \left( T\right) =0.$

The function $\delta$ is negative at one, and once it becomes increasing, it continues to increase up to the end of the useful life. Therefore $\delta \left( i\right)$ can only cross zero once and this happens at $i=T$. The three observations hence imply that $\delta \left( i\right) \le 0$ and $bv_{i}\le bv_{i}^{^{\prime }}$ for all i. $\square$

Proof of Observation 3

Assume that

$$\begin{aligned} d_{1}'=\left( 1-\eta \right) d_{1}+\eta , \end{aligned}$$

and

$$\begin{aligned} d'_{\tau }=\left( 1-\eta \right) d_{\tau } \end{aligned}$$

for $\tau >1$ and some $\eta >0$.

Note that

$$\begin{aligned} \frac{d_{\tau }}{d_{\tau +1}}=\frac{d'_{\tau }}{d'_{\tau +1}} \end{aligned}$$

for $\tau >1$. For $\tau =1$, we have

$$\begin{aligned} \frac{d'_{1}}{d'_{2}}=\frac{\left( 1-\eta \right) d_{1}+\eta }{\left( 1-\eta \right) d_{2}}>\frac{d_{1}}{d_{2}}. \end{aligned}$$

Therefore $\varvec{d}'$ is more conservative than $\varvec{d}$. $\square$

Proof of Proposition 3

Let $D_{T+1}$ and $D'_{T+1}$ denote the aggregate depreciation expenses under rules $\varvec{d}$ and $\varvec{d}'$, respectively, where $\varvec{d}$ is more conservative than $\varvec{d}'.$ We will show that $D_{T+1}\ge D_{T+1}^{\prime }$ ($D_{T+1}\le D_{T+1}^{\prime }$) if investments $I_{1},\ldots,I_{T}$ are monotonically increasing (decreasing). From this it will follow that if $\varvec{d}$ is more conservative than replacement cost accounting and the firm is growing, then

$$\begin{aligned} PE_{T}\left( \varvec{\lambda },\varvec{d}\right) =\frac{P{}_{T}\left( \varvec{\lambda }\right) }{E_{T+1}\left( \varvec{\lambda },\varvec{d}\right) }\ge \frac{P{}_{T}\left( \varvec{\lambda }\right) }{E_{T+1}^{*}}=\frac{1}{r-s\left( \varvec{\mu }\right) \cdot \frac{q_{T}\left( \varvec{\lambda }\right) -1}{q_{T}\left( \varvec{\lambda }\right) }}. \end{aligned}$$

Observe that

$$\begin{aligned} D_{T+1}= & {} I_{1}\cdot d_{T}+I_{2}\cdot d_{T-1}+\cdots +I_{T}\cdot d_{1}\\= & {} I_{1}+\left( I_{2}-I_{1}\right) \cdot \left( 1-bv_{T-1}\right) +\cdots +\left( I_{T}-I_{T-1}\right) \cdot \left( 1-bv_{1}\right) . \end{aligned}$$

Therefore,

$$\begin{aligned} D_{T+1}-D_{T+1}^{\prime }=\left( I_{2}-I_{1}\right) \cdot \left( bv'_{T-1}-bv_{T-1}\right) +\cdots +\left( I_{T}-I_{T-1}\right) \cdot \left( bv'_{1}-bv_{1}\right) \ge 0, \end{aligned}$$

where the last inequality holds by Observation 2 if investments are increasing and $\varvec{d}$ is more conservative than $\varvec{d}'$. $\square$

Proof of Observation 4

In the proof of Proposition 2, we have verified that $\frac{bv_{i-1}^{*}}{x_{i}}$ is decreasing in i. Recall that

$$\begin{aligned} \frac{d_{i}^{*}+r\cdot bv_{i-1}^{*}}{x_{i}}=c. \end{aligned}$$

Since $\frac{bv_{i-1}^{*}}{x_{i}}$ is decreasing in i, the sequence $\frac{d_{i}^{*}}{x_{i}}$ must increase in i. Therefore,

$$\begin{aligned} \frac{d_{i}^{*}}{d_{i+1}^{*}}\le \frac{x_{i}}{x_{i+1}}=\frac{d_{i}^{p}}{d_{i+1}^{p}}. \end{aligned}$$

$\square$

Proof of Proposition 4

We expand the P/E ratio as:

$$\begin{aligned}&\frac{P_{T}}{E_{T+1}}=\frac{1}{\left( r-s\left( \varvec{\mu }\right) \right) }\cdot \frac{E_{T+1}^{*}-s\left( \varvec{\mu }\right) \cdot BV_{T}^{*}}{E_{T+1}}\\&\quad =\frac{1}{\left( r-s\left( \varvec{\mu }\right) \right) }\cdot \frac{I_{T}\left( x_{1}p^{o}-d_{1}^{*}-s\left( \varvec{\mu }\right) \cdot bv_{0}^{*}\right) +\cdots +I_{1}\left( x_{T}p^{o}-d_{T}^{*}-s\left( \varvec{\mu }\right) \cdot bv_{T-1}^{*}\right) }{I_{T}\left( x_{1}p^{o}-d_{1}\right) +\cdots +I_{1}\left( x_{T}p^{o}-d_{T}\right) }. \end{aligned}$$

To apply Lemma A, we need to show that

$$\begin{aligned} \frac{x_{i}p^{o}-d_{i}^{*}-s\left( \varvec{\mu }\right) \cdot bv_{i-1}^{*}}{x_{i}p^{o}-d_{i}} \end{aligned}$$

is decreasing in i. Note that

$$\begin{aligned}&\frac{x_{i}p^{o}-d_{i}^{*}-s\left( \varvec{\mu }\right) \cdot bv_{i-1}^{*}}{x_{i}p^{o}-d_{i}}=\frac{x_{i}p^{o}-d_{i}^{*}-r\cdot bv_{i-1}^{*}+\left( r-s\left( \varvec{\mu }\right) \right) bv_{i-1}^{*}}{x_{i}\left( p^{o}-\frac{d_{i}}{x_{i}}\right) }\nonumber \\&\quad =\frac{p^{o}-c+\left( r-s\left( \varvec{\mu }\right) \right) \frac{bv_{i-1}^{*}}{x_{i}}}{p^{o}-\frac{d_{i}}{x_{i}}}. \end{aligned}$$

(34)

Assumption (1) implies that $\frac{bv_{i-1}^{*}}{x_{i}}$ is decreasing in i (see the proof of Proposition 2), and since $r>s\left( \varvec{\mu }\right)$, the numerator is decreasing in i. If $\varvec{d}$ corresponds to proportional depreciation, then the denominator does not depend on i. If $\varvec{d}$ is more conservative than the proportional depreciation rule, then $\frac{d_{i}}{x_{i}}$ is decreasing in i, and the denominator is increasing in i. Therefore the ratio (34) is decreasing in i.

To show that the price-to-book ratio is decreasing in each $\lambda _{t}$, we recall that the firm’s market value is given by:

$$\begin{aligned} P_{T}=BV_{T}^{*}+\frac{\left( p^{o}-c\right) K_{T+1}^{o}}{r-s \left( \varvec{\mu }\right) }. \end{aligned}$$

$PB_{T}$ is therefore equal to:

$$\begin{aligned} PB_{T}=\frac{BV_{T}^{*}}{BV_{T}} +\frac{\left( p^{o}-c\right) }{r-s\left( \varvec{\mu }\right) }\cdot \frac{K_{T +1}^{o}}{BV_{T}}. \end{aligned}$$

(35)

It remains to show that both $\frac{BV_{T}^{*}}{BV_{T}}$ and $\frac{K_{T+1}^{o}}{BV_{T}}$ decrease in each $\lambda _{t}.$ ⁴⁴ Note that

$$\begin{aligned} \frac{BV_{T}^{*}}{BV_{T}}=\frac{bv_{0}^{*}I_{T}+\cdots +bv_{T-1}^{*}I_{1}}{bv_{0}I_{T} +\cdots +bv_{T-1}I_{1}}. \end{aligned}$$

By Lemma A, to show that $BV_{T}^{*}/BV_{T}$ is declining in $\lambda _{t}$, it suffices to check that $\frac{bv_{i}^{*}}{bv_{i}}$ is increasing in i. Recall that

$$\begin{aligned} bv_{i}^{*}=\gamma cx_{i+1}+\cdots +\gamma ^{T-i}cx_{T} \end{aligned}$$

and

$$\begin{aligned} bv_{i}=d_{i+1}+\cdots +d_{T}. \end{aligned}$$

Using the assumption that $\frac{x_{\tau }}{x_{\tau +1}}$ increases in $\tau$ and Lemma A, we obtain:⁴⁵

$$\begin{aligned} \frac{bv_{i}^{*}}{bv_{i+1}^{*}}=\frac{\gamma cx_{i+1}+\cdots +\gamma ^{T-i}cx_{T}}{\gamma cx_{i+2}+\cdots +\gamma ^{T-i-1}cx_{T}}\le \frac{x_{i+1}+\cdots +x_{T}}{x_{i+2}+\cdots +x_{T}}. \end{aligned}$$

If we now show that

$$\begin{aligned} \frac{bv_{i}}{bv_{i+1}}\ge \frac{x_{i+1}+\cdots +x_{T}}{x_{i+2}+\cdots +x_{T}}, \end{aligned}$$

(36)

it will follow that $\frac{bv_{i}^{*}}{bv_{i}}$ increases in i.

Inequality (36) is equivalent to:

$$\begin{aligned} \frac{d_{i+1}+\cdots +d_{T}}{x_{i+1}+\cdots +x_{T}}\ge \frac{d_{i+2}+\cdots +d_{T}}{x_{i+2}+\cdots +x_{T}}. \end{aligned}$$

The inequality above holds by Lemma A, since $\frac{d_{\tau }}{x_{\tau }}$ decreases in $\tau$. This concludes the proof that $\frac{BV_{T}^{*}}{BV_{T}}$ declines in each $\lambda _{t}$.

To verify that $\frac{K_{T+1}^{o}}{BV_{T}}$ also decreases in $\lambda _{t}$, we will check that $\frac{K_{T+1}^{o}}{BV_{T}^{*}}$ declines in $\lambda _{t}$. Note that

$$\begin{aligned} \frac{K_{T+1}^{o}}{BV_{T}^{*}}=\frac{x_{1}I_{T}+\cdots +x_{T}I_{1}}{bv_{0}^{*}I_{T}+\cdots +bv_{T-1}^{*}I_{1}}. \end{aligned}$$

(37)

We have shown earlier that $\frac{bv_{\tau -1}^{*}}{x_{\tau }}$ is decreasing in $\tau$, and therefore $\frac{x_{\tau }}{bv_{\tau -1}^{*}}$ is increasing in $\tau$. Then, by Lemma A, the right-hand side of (37) decreases in each $\lambda _{t}$. $\square$

Proof of Observation 5

$\frac{\partial PE_{T}}{\partial p^{o}}$ has the same sign as $\frac{\partial \ln (PE_{T})}{\partial p^{o}}$.

$$\begin{aligned}&\frac{\partial \ln (PE_{T})}{\partial p^{o}}=\frac{\partial \ln P_{T}}{\partial p^{o}}-\frac{\partial \ln (E_{T+1})}{\partial p^{o}}=\frac{1}{P_{T}}\left( \frac{\partial P_{T}}{\partial p^{o}}\right) -\frac{1}{E_{T+1}}\left( \frac{\partial E_{T+1}}{\partial p^{o}}\right) ,\\&\frac{\partial P_{T}}{\partial p^{o}}=\frac{K_{T+1}^{o}}{r-s\left( \varvec{\mu }\right) },\\&\frac{\partial E_{T+1}}{\partial p^{o}}=K_{T+1}^{o}. \end{aligned}$$

Therefore $\frac{\partial PE_{T}}{\partial p^{o}}$ has the same sign as

$$\begin{aligned} \frac{K_{T+1}^{o}}{\left( r-s\left( \varvec{\mu }\right) \right) P_{T}} -\frac{K_{T+1}^{o}}{E_{T+1}}=\left( \frac{1}{r-s\left( \varvec{\mu }\right) } -PE_{T}\right) \frac{K_{T+1}^{o}}{P_{T}}. \end{aligned}$$

$\square$

Proof of Proposition 5

We show that if investments have grown at the constant rate $\lambda$, the difference

$$\begin{aligned} E_{T+1}-\lambda \cdot BV_{T} \end{aligned}$$

(38)

is invariant to the accounting rules. To that end, it suffices to demonstrate that

$$\begin{aligned} D_{T+1}+\lambda \cdot BV_{T}=(1+\lambda )^{T}I_{1}, \end{aligned}$$

for any depreciation schedule, $\varvec{d}$. By definition:

$$\begin{aligned} D_{T+1}&= d_{1}\cdot I_{T}+d_{2}\cdot I_{T-1}+\cdots +d_{t}\cdot I_{1}\\&= [d_{1}\cdot (1+\lambda )^{T-1}+d_{2}\cdot (1+\lambda )^{T-2}+\cdots +d_{T}]\cdot I_{1}. \end{aligned}$$

Similarly,

$$\begin{aligned} BV_{T}&= bv_{0}\cdot I_{T}+bv_{1}\cdot I_{t-1}+\cdots +bv_{T-1}\cdot I_{1}\\&= [bv_{0}\cdot (1+\lambda )^{T-1}+bv_{1}\cdot (1+\lambda )^{T-2}+\cdots +bv_{T-1}]\cdot I_{1} \end{aligned}$$

Thus

$$\begin{aligned} D_{T+1}+\lambda \cdot BV_{T}=(1+\lambda )^{T}\cdot \left[ \sum _{t=1}^{T}(d_{t}+\lambda \cdot bv_{t-1})\cdot (1+\lambda )^{-t}\right] I_{1} \end{aligned}$$

(39)

For any depreciation schedule and any $\lambda >{-}1$, the expression in brackets in the right-hand side of (39) is equal to one.⁴⁶ We may evaluate (38) by supposing replacement cost accounting. Proposition 1 then yields $E_{T+1}^{*}=r\cdot P_{T}$ and $BV_{T}^{*}=P_{T}$. Thus,

$$\begin{aligned} E_{T+1}-\lambda \cdot BV_{T}=(r-\lambda )\cdot P_{T}, \end{aligned}$$

(40)

and the claim follows immediately. $\square$