6. Manager's Myopic Action under Efficient Market Assumption

This chapter studies Stein's (1989) signal-jamming problem: under an efficient-market assumption, why does a manager with stock-based compensation do things that benefit the firm in the short run but hurt it in the long run? The key is the probability of losing control of the firm (being taken over) and the resulting incentive to manipulate earnings to prop up the price before the takeover. (Signal jamming: an agent deliberately sends untrue signals to the market to maximize its own benefit.)

We study the steady state of a discrete-time model.

• In each period \(t\), the firm generates natural (pre-manipulation) earnings

$$e_t^n=z_t+v_t,\qquad v_t\sim\mathcal{N}\!\left(0,\sigma_v^2\right)$$

where

$$z_t=z_{t-1}+u_t,\qquad u_t\sim\mathcal{N}\!\left(0,\sigma_u^2\right)$$

is a random walk capturing the permanent shock in period \(t\), and \(v_t\) captures the transitory shock in period \(t\).

• The reported earnings are denoted \(e_t\) such that

$$e_t=e_t^n+b_t-C(b_{t-1})$$

where \(b_t\) is the artificially manipulated extra earnings and \(C(b_{t-1})\) is the cost of last period's manipulation, paid in this period.

• Assume \(C'(\cdot)>0\), \(C''(\cdot)>0\) and \(C'(0)=1+r\), where \(r\) is the firm's discount rate.
• \(C''(\cdot)>0\) can be justified by an increasing marginal borrowing cost or inefficient liquidation of projects, etc.

The following problem pins down the first-best manipulation \(b_t\):

$$\max_{\{b_t\}_{t=0}^{\infty}}\ \sum_{t=1}^{\infty}\frac{1}{(1+r)^t}\left(e_t^n+b_t-C(b_{t-1})\right)$$

The f.o.c. w.r.t. \(b_t\) for all \(t\) is

$$\frac{1}{(1+r)^{t+1}}C'(b_t)=\frac{1}{(1+r)^t}\ \Rightarrow\ C'(b_t)=1+r$$

However, we know that \(C'(b_t)=1+r\) only when \(b_t=0\), so the first-best solution is \(b_t=0\) for all \(t\).

Intuition: from the firm's (the long-run owner's) point of view, manipulating earnings merely borrows future value into today, and since the cost is convex, the optimum is no manipulation at all.

Now we focus on the steady state of this earnings-control problem, characterized by

$$e_t=e_t^n+\bar b-C(\bar b)$$

The random walk of natural earnings implies that in steady state the expectation of any future period is the same:

$$\mathbb{E}[e_{t+1}\mid\mathcal{F}_t]=\mathbb{E}[e_{t+j}\mid\mathcal{F}_t],\qquad j=1,2,\dots \tag{6.1}$$

where \(\mathcal{F}_t\) is the period-\(t\) information. So the price today (post today's earnings distribution) in steady state is

$$P_t=\mathbb{E}\!\left[\sum_{j=1}^{\infty}\frac{e_{t+j}}{(1+r)^j}\,\Big|\,\mathcal{F}_t\right]=\sum_{j=1}^{\infty}\frac{1}{(1+r)^j}\,\mathbb{E}[e_{t+1}\mid\mathcal{F}_t]=\frac{1}{r}\,\mathbb{E}[e_{t+1}\mid\mathcal{F}_t]$$

(The second equality uses (6.1) to replace every \(\mathbb{E}[e_{t+j}\mid\mathcal{F}_t]\) by \(\mathbb{E}[e_{t+1}\mid\mathcal{F}_t]\); the geometric series \(\sum_{j\ge1}(1+r)^{-j}=1/r\).)

The market uses a Kalman filter to extract the permanent state \(z_t\) from the observed earnings \(e_t\) (see the derivation below), ending with \(\mathbb{E}[e_{t+1}\mid\mathcal{F}_t]\) as a weighted average of the past expectation and current earnings, so the market price is

$$P_t=\frac{1}{r}\Big[(1-\alpha)\,\mathbb{E}[e_t\mid\mathcal{F}_{t-1}]+\alpha\,e_t\Big],\qquad \alpha=\frac{\sigma_u^2}{\sigma_u^2+\sigma_v^2}$$

where \(\alpha\) is the market's pricing weight on current earnings \(e_t\): the larger the permanent-shock variance \(\sigma_u^2\), the more earnings reflect permanent information, and the more strongly the market reacts (larger \(\alpha\)). It is precisely this \(\alpha>0\) that gives the manager leverage to prop up the price through \(b_t\).

Suppose there is a probability \(\pi\in(0,1)\) of being taken over by outsiders in every period. The manager solves the following dynamic problem in every period \(t\):

$$V_t(b_{t-1})=\max_{b_t}\ e_t+\pi P_t+(1-\pi)\frac{V_{t+1}(b_t)}{1+r}$$

(Reading: the manager gets the private benefit of current earnings \(e_t\); with probability \(\pi\) he is taken over, in which case his value is anchored to the current price \(P_t\); with probability \(1-\pi\) he stays, entering next period's continuation value \(V_{t+1}\).) The f.o.c. is

$$\frac{de_t}{db_t}+\pi\frac{dP_t}{db_t}+(1-\pi)\frac{1}{1+r}\frac{dV_{t+1}(b_t)}{db_t}=0 \tag{6.2}$$

Note that we have

$$\frac{de_t}{db_t}=1,\qquad \frac{dV_{t+1}(b_t)}{db_t}=-C'(b_t),\qquad \frac{dP_t}{db_t}=\frac{\alpha}{r}$$

(\(b_t\) enters \(P_t=\frac1r[\cdots+\alpha e_t]\) through \(e_t\) with weight \(\alpha\), so \(dP_t/db_t=\alpha/r\); \(b_t\) enters \(V_{t+1}\) only through next period's cost \(-C(b_t)\).) So (6.2) becomes

$$1+\pi\frac{\alpha}{r}-(1-\pi)\frac{1}{1+r}C'(b_t)=0\ \Rightarrow\ C'(b_t)=\frac{(1+r)(r+\alpha\pi)}{r(1-\pi)}$$

Since we are thinking about the steady-state problem, we can drop the time subscript: the manager always chooses to manipulate by the amount \(\bar b\) such that

$$C'(\bar b)=\frac{(1+r)(r+\alpha\pi)}{r(1-\pi)} \tag{6.3}$$