Variance matters (in stochastic dividend discount models)

Abstract

Stochastic dividend discount models (Hurley and Johnson in Financ Anal J 50–54. http://www.jstor.org/stable/4479761, 1994, J Portf Manag 27–31. doi:10.3905/jpm.1998.409658, 1998; Yao in J Portf Manag 99–103. doi:10.3905/jpm.1997.409618, 1997) present expressions for the expected value of stock prices when future dividends, periodically received by shareholders as a reward for their risky investment, evolve through time in a Markovian setting by the means of a discretely distributed random rate of growth. Such result extends and makes more flexible the classical textbook formula for stock prices known as Gordon model. This paper introduces a closed-form expression for the variance of random stock prices, determines how their variance is affected by the variance of the dividend rate of growth, establishes that, in this framework, the dividend process is non-stationary, and perform a simple econometric analysis applying real market data.

Appendices

Appendix 1

Let conditions (C1), (C2) and (C3) hold. The inner sums in (6) return

$$\begin{aligned}&\prod _{t=1}^n\left( 1+g_t\right) ^{2s_t} \sum _{z_1=s_1}^{s_1+p-j} \ldots \sum _{z_n=s_n}^{s_n+p-j} \mathbf {1}_{\left\{ B\right\} } \left( {\begin{array}{c}p-j\\ \left( z_1-s_1\right) \ldots \left( z_n-s_n\right) \end{array}}\right) \\&\quad \left( \left( 1+g_1\right) q_1\right) ^{z_1-s_1} \ldots \left( \left( 1+g_n\right) q_n\right) ^{z_n-s_n} = \prod _{t=1}^n\left( 1+g_t\right) ^{2s_t} \left( \sum _{t=1}^n\left( 1+g_t\right) q_t\right) ^{p-j} \\&\quad = \prod _{t=1}^n\left( 1+g_t\right) ^{2s_t} \left( \sum _{t=1}^n q_t + \sum _{t=1}^n g_t q_t\right) ^{p-j} = \left( 1 + E\left[ \tilde{g}\right] \right) ^{p-j} \prod _{t=1}^n\left( 1+g_t\right) ^{2s_t} \end{aligned}$$

where $B=\left\{ z_1-s_1+\cdots +z_n-s_n = p-j\right\}$.

Letting, instead, $C=\left\{ s_1+\cdots +s_n = j\right\}$, the outer sums become

$$\begin{aligned}&E\left[ \tilde{d}_p \cdot \tilde{d}_j \right] = d_0^2 \left( 1 + E\left[ \tilde{g}\right] \right) ^{p-j} \\&\quad \cdot \sum _{s_1=0}^j \ldots \sum _{s_n=0}^j \mathbf {1}_{\left\{ C\right\} } \left( \left( 1+g_1\right) ^{2s_1} q_1^{s_1} \right) \ldots \left( \left( 1+g_n\right) ^{2s_n} q_n^{s_n} \right) \left( {\begin{array}{c}j\\ s_1 \ldots s_n\end{array}}\right) \\&\quad = d_0^2 \left( 1 + E\left[ \tilde{g}\right] \right) ^{p-j} \left( \sum _{t=1}^n \left( 1+g_t\right) ^2 q_t \right) ^j = d_0^2 \left( 1 + E\left[ \tilde{g}\right] \right) ^{p-j} E^j \left[ \left( 1 + \tilde{g} \right) ^2 \right] \end{aligned}$$

Appendix 2

Recalling that, when $-1<y<1, \sum _{u=1}^{+\infty } y^u = \frac{y}{1-y}$, the variance of $\tilde{P}_{0}$, if $E[\tilde{g}] < k$, is obtained as follows

$$\begin{aligned} Var\left[ \tilde{P}_{0}\right]&= \sum _{j=1}^{+\infty }\sum _{p=1}^{+\infty }\dfrac{\text {Cov}\left( \tilde{d}_{j};\tilde{d}_{p}\right) }{\left( 1+k\right) ^{j+p}} \\&= d_{0}^{2}\sum _{j=1}^{+\infty }\sum _{p=1}^{+\infty }\dfrac{\left( 1+E\left[ \tilde{g}\right] \right) ^{p-j}\left[ E^{j}\left[ \left( 1+\tilde{g}\right) ^{2}\right] -\left( 1+E\left[ \tilde{g}\right] \right) ^{2j}\right] }{\left( 1+k\right) ^{j+p}} \\&= d_{0}^{2}\sum _{j=1}^{+\infty }\dfrac{E^{j}\left[ \left( 1+\tilde{g}\right) ^{2}\right] -\left( 1+E\left[ \tilde{g}\right] \right) ^{2j}}{\left[ \left( 1+E\left[ \tilde{g}\right] \right) \left( 1+k\right) \right] ^{j}}\sum _{p=1}^{+\infty }\left( \dfrac{1+E\left[ \tilde{g}\right] }{1+k}\right) ^{p} \\&= d_{0}^{2}\dfrac{1+E\left[ \tilde{g}\right] }{k-E\left[ \tilde{g}\right] }\left[ \sum _{j=1}^{+\infty }\left[ \dfrac{1+2E\left[ \tilde{g}\right] +E\left[ \tilde{g}^{2}\right] }{\left( 1+E\left[ \tilde{g}\right] \right) \left( 1+k\right) }\right] ^{j}-\sum _{j=1}^{+\infty }\left( \dfrac{ 1+E\left[ \tilde{g}\right] }{1+k}\right) ^{j}\right] \\&= d_{0}^{2}\dfrac{1+E\left[ \tilde{g}\right] }{k-E\left[ \tilde{g}\right] }\left( \dfrac{1+2E\left[ \tilde{g}\right] +E\left[ \tilde{g}^{2}\right] }{k-E\left[ \tilde{g}\right] +kE\left[ \tilde{g}\right] -E\left[ \tilde{g}^{2}\right] }-\dfrac{1+E\left[ \tilde{g}\right] }{k-E\left[ \tilde{g}\right] }\right) . \end{aligned}$$

Summing up the two fractions into brackets results a fraction whose numerator can be written as $Var\left[ \tilde{g}\right] \left( 1+k\right)$ so that

$$\begin{aligned} Var\left[ \tilde{P}_{0}\right]&= \dfrac{d_{0}^{2}Var\left[ \tilde{g}\right] \left( 1+E\left[ \tilde{g}\right] \right) \left( 1+k\right) }{\left( k-E \left[ \tilde{g}\right] +kE\left[ \tilde{g}\right] -E\left[ \tilde{g}^{2} \right] \right) \left( k-E\left[ \tilde{g}\right] \right) ^{2}} \nonumber \\&= \dfrac{E\left[ \tilde{P}_{0}\right] Var\left[ \tilde{g}\right] d_{0}\left( 1+k\right) }{\left( k-E\left[ \tilde{g}\right] +k E\left[ \tilde{g}\right] - E\left[ \tilde{g}^{2}\right] \right) \left( k-E\left[ \tilde{g}\right] \right) }. \end{aligned}$$

(8)

Consider the first parenthesis in the denominator of the expression above. By summing and subtracting $E^2[\tilde{g}]$ one gets

$$\begin{aligned} k + k E\left[ \tilde{g}\right] - E\left[ \tilde{g}\right] - E^2\left[ \tilde{g}\right] - \left( E\left[ \tilde{g}^{2}\right] - E^2\left[ \tilde{g}\right] \right)&= \left( 1 + E\left[ \tilde{g}\right] \right) \left( k-E\left[ \tilde{g}\right] \right) \\&\quad - Var\left[ \tilde{g}\right] . \end{aligned}$$

This means that

$$\begin{aligned} Var\left[ \tilde{P}_{0}\right] = \dfrac{E\left[ \tilde{P}_{0}\right] Var\left[ \tilde{g}\right] d_{0}\left( 1+k\right) }{\left( \left( 1+ E\left[ \tilde{g}\right] \right) \left( k-E\left[ \tilde{g}\right] \right) - Var\left[ \tilde{g}\right] \right) \left( k-E\left[ \tilde{g}\right] \right) }. \end{aligned}$$

Finally, recalling once again (3), the variance can be expressed as

$$\begin{aligned} Var\left[ \tilde{P}_{0}\right] = \dfrac{E^2\left[ \tilde{P}_{0}\right] Var\left[ \tilde{g}\right] \left( 1+k\right) }{\left( \left( 1 + E\left[ \tilde{g}\right] \right) \left( k-E\left[ \tilde{g}\right] \right) - Var\left[ \tilde{g}\right] \right) \left( 1 + E\left[ \tilde{g}\right] \right) }. \end{aligned}$$

Appendix 3

From (8), for the convergence and positiveness of $Var\left[ \tilde{P}_{0}\right]$ both

$$\begin{aligned} k-E\left[ \tilde{g}\right] +k E\left[ \tilde{g}\right] - E\left[ \tilde{g}^{2}\right] \end{aligned}$$

and $k-E\left[ \tilde{g}\right]$ must be strictly positive. This is the case if

$$\begin{aligned} k>\max \left( \dfrac{E\left[ \tilde{g}\right] +E\left[ \tilde{g}^{2}\right] }{1+E\left[ \tilde{g}\right] };E\left[ \tilde{g}\right] \right) . \end{aligned}$$

Inequality

$$\begin{aligned} \dfrac{E\left[ \tilde{g}\right] +E\left[ \tilde{g}^{2}\right] }{1+E\left[ \tilde{g}\right] }>E\left[ \tilde{g}\right] \end{aligned}$$

is always true as, being equivalent to $E\left[ \tilde{g}\right] +E\left[ \tilde{g}^{2}\right] >E\left[ \tilde{g}\right] +E^{2}\left[ \tilde{g}\right]$, it boils down to $E\left[ \tilde{g}^{2}\right] -E^{2}[\tilde{g}] =Var\left[ \tilde{g}\right] > 0.$

This implies that $Var\left[ \tilde{P}_{0}\right]$ returns positive and finite values as long as

$$\begin{aligned} k>\dfrac{E\left[ \tilde{g}\right] +E\left[ \tilde{g}^{2}\right] }{1+E\left[ \tilde{g}\right] }. \end{aligned}$$