Dynamic beauty contests: Learning from the winners to win?

Abstract

I build a dynamic game consisting of a continuum of players to investigate the effects of previous winners’ actions on the spreading of subsequent players’ actions. In each stage, besides the private signal, each player also observes actions taken by the winners of all previous stages as public signals. A unique equilibrium of the game is found and characterized. I then define the variances of three forms of gap: the gap between the average play and the underlying fundamental value, the gap between a generic player’s action and the average play, and the gap between a generic player’s action and the winner’s play. By checking their dynamics in the equilibrium, it is shown that the accumulation of private signals always reduces the first variance and the accumulation of the public signals always reduces the second variance. However, the effects of accumulated public (private) signals on the first (second) variance are rather ambiguous. Based on a theoretical finding that expresses the third variance as a weighted sum of the other two, I conduct an empirical study on the Miss Korea pageant during 1994–2013. I find a descending trend in the variance of the gap between the average face and the underlying “true beauty” face over these years. Moreover, this process is accompanied by ascending trends in the other two variances, indicating that contestants’ faces have been converging to the “true beauty” overall but diverging from each other over the two decades, which is consistent with our theoretical results.

Appendices

Appendix

Proofs

The proofs below use extensively the tools of Bayesian inference and the conjugate distribution property of normal distributions. An uninitiated reader can check sections 10.1 and 10.2 in Vives (2008) for detailed introductions.

1.1 Proof of Lemma 1.

To prove it, let us assume there exists an equilibrium such that for some i and t we have $a_{it}\ne (1-\gamma )E_{it}(\theta ) +\gamma \,E_{it}\left( \int _0^1 a_{jt} \,\mathrm {d}j\right)$. Let $a'_{it}$ be defined as in Eq. (6), and substitute $a_{it}$ with $a'_{it}$ while keeping player i’s other stage strategies as well as all other players’ strategies the same as in the initial equilibrium. Because Eq. (6) is the first order condition for maximization of the expected stage payoff, player i’s stage-t payoff increases after the substitution. This substitution has effects on player i’s expected payoffs for future stages if and only if she turns out to be the most successful player of stage t, which is a zero-probability event according to the continuum setting of the player set. As a result, $a_{it}$ cannot be part of an equilibrium, contradiction. So in any equilibrium, the equality must be satisfied for each player in each stage. $\square$

1.2 Proof of Lemma 2.

To see that it is optimal for player i to adopt $a_{i1}=\frac{\beta _1 y_1+(1-\gamma )\alpha _1 x_{i1}}{\beta _1+(1-\gamma )\alpha _1}$ given others are doing so, notice that the conditional distribution of $\theta$ on $h_{i1}$ is

$$\begin{aligned} \theta |h_{i1} \sim \mathcal {N}\left( \frac{\beta _1 y_1+\alpha _1 x_{i1}}{\beta _1+\alpha _1},\frac{1}{\beta _1+\alpha _1}\right) , \end{aligned}$$

(29)

given all players are taking this stage strategy, we have

$$\begin{aligned} E_{i1}\left( \int _0^1 a_{j1}\, \mathrm {d} j \right)= & {} E_{i1}\left( \int _0^1 \frac{\beta _1 y_1+(1-\gamma )\alpha _1 x_{j1}}{\beta _1+(1-\gamma )\alpha _1}\, \mathrm {d} j \right) \nonumber \\= & {} E_{i1}\left( \frac{\beta _1 y_1 +(1-\gamma )\alpha _1 \theta }{\beta _1+(1-\gamma )\alpha _1} +\frac{(1-\gamma )\alpha _1}{\beta _1+(1-\gamma )\alpha _1} \int _0^1 \varepsilon _{j1}\, \mathrm {d} j \right) \nonumber \\= & {} \frac{\beta _1 y_1 +(1-\gamma )\alpha _1 E_{i1}(\theta )}{\beta _1+(1-\gamma )\alpha _1}. \end{aligned}$$

(30)

So the RHS of Eq. (6) could be expressed as

$$\begin{aligned} (1-\gamma )E_{it}(\theta ) +\gamma \,E_{it}\left( \int _0^1 a_{jt} \,\mathrm {d}j\right)= & {} \frac{(1-\gamma )\left[ \beta _1+(1-\gamma )\alpha _1+\gamma \alpha _1 \right] E_{i1}(\theta )+\gamma \beta _1 y_1}{\beta _1+(1-\gamma )\alpha _1} \nonumber \\= & {} \frac{\beta _1 y_1+(1-\gamma )\alpha _1 x_{i1}}{\beta _1+(1-\gamma )\alpha _1}, \end{aligned}$$

(31)

which is exactly the LHS of the equation under the stage strategy. So the first-order condition, Eq. (6), is fulfilled.

We then adopt the method used in Morris and Shin (2002) to show that there are no other choices for $a_{i1}$. Define operator $\bar{E}_1(\cdot )=\int _0^1 E_{j1}(\cdot )\, \mathrm {d} j$ as well as

$$\begin{aligned} \bar{E}_1^n(\theta )=\underbrace{\bar{E}_1(\bar{E}_1(\cdots (\bar{E}_1}_n (\theta )\cdots ))). \end{aligned}$$

(32)

The initial first order condition (6) can thus be rewritten as

$$\begin{aligned} a_{i1}= & {} (1-\gamma )E_{i1}(\theta ) +\gamma \,E_{i1}\left( \int _0^1 a_{j1} \,\mathrm {d}j\right) \nonumber \\= & {} (1-\gamma )E_{i1}(\theta ) +\gamma \,E_{i1}\left[ \int _0^1 (1-\gamma )E_{j1}(\theta )\,\mathrm {d}j+ \int _0^1 \gamma E_{j1}\left( \int _0^1 a_{\ell 1} \,\mathrm {d}{\ell }\right) \,\mathrm {d}j\right] \nonumber \\= & {} (1-\gamma ) \sum _{n=0}^{\infty } \gamma ^{n} E_{i1} \left( \bar{E}_1^n(\theta ) \right) , \end{aligned}$$

(33)

where the second equality adopts the fact that the first order condition (6) must hold for all players in the symmetric equilibrium, and the third equality uses the definitions above as well as the iteration rule.

Define $\mu _1=\alpha _1/(\beta _1+\alpha _1)$, based on (29) and (32), it is straightforward to have $\bar{E}^n_1(\theta )=(1-\mu _1^n)y_1+\mu _1^n \theta$ and $E_{i1}(\bar{E}^n_1(\theta ))=(1-\mu _1^{n+1})y_1+\mu _1^{n+1} x_{i1}$. Also notice that $\mu _1<1$ always holds, so the RHS of (33) is bounded and is rewritten as

$$\begin{aligned} a_{i1}= & {} (1-\gamma ) \sum _{n=0}^{\infty } \gamma ^{n} E_{i1} \left( \bar{E}_1^n(\theta ) \right) \nonumber \\= & {} (1-\gamma ) \sum _{n=0}^{\infty }\left[ (\gamma ^n -\gamma ^n \mu _1^{n+1})y_1 +\gamma ^n \mu _1^{n+1} x_{i1} \right] \nonumber \\= & {} (1-\gamma ) \left[ \left( \frac{1}{1-\gamma }-\frac{\mu _1}{1-\gamma \mu _1} \right) y_1 +\frac{\mu _1}{1-\gamma \mu _1} x_{i1} \right] \nonumber \\= & {} \frac{\beta _1 y_1 +(1-\gamma )\alpha _1 x_{i1}}{\beta _1+(1-\gamma )\alpha _1}. \end{aligned}$$

(34)

This is the unique solution that satisfies condition (6), and it also makes sure every player is maximizing her stage-1 payoff given others’ strategies, so the first half of the statement is now proved. For the second half, notice that for individuals $i\ne j$ we have $x_{i1}-x_{j1}=\theta +\varepsilon _{i1}-\theta -\varepsilon _{j1}=\varepsilon _{i1}-\varepsilon _{j1}$, so in the equilibrium play

$$\begin{aligned} a_{i1}-\theta= & {} \frac{\beta _1(y_1-\theta ) +(1-\gamma )\alpha _1 \varepsilon _{i1}}{\beta _1+(1-\gamma )\alpha _1}\end{aligned}$$

(35)

$$\begin{aligned} a_{i1}-a_{j1}= & {} \frac{(1-\gamma )\alpha _1 (\varepsilon _{i1}-\varepsilon _{j1})}{\beta _1+(1-\gamma )\alpha _1}. \end{aligned}$$

(36)

In the equilibrium, player i’s stage-1 payoff becomes

$$\begin{aligned} U_{i1}= & {} -(1-r)(a_{i1}-\theta )^2 -rL_{i1} \nonumber \\= & {} -\frac{(1-\gamma )\left[ \beta _1(y_1-\theta ) +(1-\gamma )\alpha _1 \varepsilon _{i1} \right] ^2 +\gamma \left[ (1-\gamma )\alpha _1 \right] ^2 \int _0^1 (\varepsilon _{i1}-\varepsilon _{j1})^2 \, \mathrm {d} j}{\left( \beta _1+(1-\gamma )\alpha _1 \right) ^2}.\nonumber \\ \end{aligned}$$

(37)

Now suppose player i turns out to be the most successful player of stage 1, Assumption 1 makes sure that $\varepsilon _{i1}$ is a maximizer of the RHS of (37) in $\mathbb {R}$. Further denote by $\varepsilon ^*_1$ the value of private noise of the first stage’s most successful player, it thus solves the first order condition

$$\begin{aligned} \beta _1(y_1-\theta )+(1-\gamma )\alpha _1 \varepsilon ^*_1+\gamma \alpha _1 \int _0^1(\varepsilon ^*_1-\varepsilon _{j1})\,\mathrm {d}j=0, \end{aligned}$$

(38)

which in turn gives us

$$\begin{aligned} \varepsilon ^*_1=\frac{\beta _1}{\alpha _1} \left( \theta -y_1\right) . \end{aligned}$$

(39)

So the most successful player in stage 1 receives private signal $x^*_1=\theta +\varepsilon ^*_1=\theta +\frac{\beta _1}{\alpha _1}(\theta -y_1)$. The lemma is proved. $\square$

1.3 Proof of Proposition 1.

We prove the proposition by induction. Lemma 2 has already assured the validity of the statement for the first stage. Now for $t\ge 2$, suppose the statement is true for stage $t-1$, it suffices to show the statement is also true for stage t. Suppose that in stage $t-1$ we have

$$\begin{aligned} \theta |h_{i,t-1} \sim \mathcal {N}\left( \frac{\tilde{\beta }_{t-1} \tilde{y}_{t-1}+\tilde{\alpha }_{t-1} \tilde{x}_{i,t-1}}{\tilde{\beta }_{t-1}+\tilde{\alpha }_{t-1}},\frac{1}{\tilde{\beta }_{t-1}+\tilde{\alpha }_{t-1}}\right) . \end{aligned}$$

(40)

Then in stage t, there are two signals added into player i’s history:

$$\begin{aligned} x_{it}= & {} \theta +\varepsilon _{it},\end{aligned}$$

(41)

$$\begin{aligned} y_{t}= & {} a^{*}_{t-1}+\xi _t, \end{aligned}$$

(42)

where $a^{*}_{t-1}$ is the action taken by the most successful player in stage $t-1$. Based on the validity of the statement in stage $t-1$, we can adopt the same logic of the proof for $x^*_1=\theta +\frac{\beta _1}{\alpha _1}(\theta -y_1)$ in Lemma 2, and get

$$\begin{aligned} \tilde{x}^{*}_{t-1}=\theta +\frac{\tilde{\beta }_{t-1}}{\tilde{\alpha }_{t-1}}(\theta -\tilde{y}_{t-1}). \end{aligned}$$

(43)

Inserting the stage strategy $a_{i,t-1}$ and (43) into (42) yields

$$\begin{aligned} y_t= & {} \frac{\tilde{\beta }_{t-1} \tilde{y}_{t-1} +(1-\gamma )\tilde{\alpha }_{t-1}\tilde{x}_{t-1}^{*} }{\tilde{\beta }_{t-1} +(1-\gamma )\tilde{\alpha }_{t-1} } +\xi _t \nonumber \\= & {} \frac{\tilde{\beta }_{t-1} \tilde{y}_{t-1} + (1-\gamma )\tilde{\alpha }_{t-1} \theta +(1-\gamma )\tilde{\beta }_{t-1}(\theta -\tilde{y}_{t-1})}{\tilde{\beta }_{t-1} +(1-\gamma )\tilde{\alpha }_{t-1}}+\xi _t. \end{aligned}$$

(44)

Rearranging (44) gives us

$$\begin{aligned} \frac{\left( \tilde{\beta }_{t-1} +(1-\gamma )\tilde{\alpha }_{t-1}\right) y_t -\gamma \tilde{\beta }_{t-1} \tilde{y}_{t-1} }{(1-\gamma )(\tilde{\alpha }_{t-1} +\tilde{\beta }_{t-1})} =\theta +\frac{\tilde{\beta }_{t-1}+(1-\gamma )\tilde{\alpha }_{t-1}}{(1-\gamma )(\tilde{\alpha }_{t-1} +\tilde{\beta }_{t-1})} \xi _t. \end{aligned}$$

(45)

Denote the LHS of (45) by $\Upsilon _t$ and clearly we have

$$\begin{aligned} \frac{\tilde{\beta }_{t-1}+(1-\gamma )\tilde{\alpha }_{t-1}}{(1-\gamma )(\tilde{\alpha }_{t-1} +\tilde{\beta }_{t-1})} \xi _t \sim \mathcal {N}\left( 0, \frac{1}{\Omega _t \beta _t} \right) , \end{aligned}$$

(46)

where $\Omega _t=\left[ \frac{(1-\gamma )(\tilde{\alpha }_{t-1}+\tilde{\beta }_{t-1})}{\tilde{\beta }_{t-1}+(1-\gamma )\tilde{\alpha }_{t-1} } \right] ^2$. Player i can thus update her belief on $\theta$ using (41) and (45). The two noise terms in these are independent and both normally distributed, the conjugacy property of normal prior and posterior distributions gives us

$$\begin{aligned} \theta |h_{it} \sim \mathcal {N}\left( \frac{\tilde{\beta }_{t-1} \tilde{y}_{t-1}+\tilde{\alpha }_{t-1} \tilde{x}_{i,t-1}+\alpha _t x_{it} + \Omega _t \beta _t \Upsilon _t}{\tilde{\beta }_{t-1}+\tilde{\alpha }_{t-1}+\alpha _t +\Omega _t \beta _t }\,,\,\frac{1}{\tilde{\beta }_{t-1}+\tilde{\alpha }_{t-1} +\alpha _t +\Omega _t \beta _t }\right) .\nonumber \\ \end{aligned}$$

(47)

The proposition also defines $\tilde{\alpha }_t=\tilde{\alpha }_{t-1}+\alpha _t$, $\tilde{x}_{it}=(\tilde{\alpha }_{t-1} \tilde{x}_{i,t-1} +\alpha _t x_{it} ) /\tilde{\alpha }_t$, $\tilde{\beta }_t=\tilde{\beta }_{t-1} +\Omega _t \beta _t$, and $\tilde{y}_t=(\tilde{\beta }_{t-1} \tilde{y}_{t-1} + \Omega _t \beta _t \Upsilon _t )/\tilde{\beta }_t$. We thus rewrite (47) as

$$\begin{aligned} \theta |h_{it} \sim \mathcal {N}\left( \frac{\tilde{\beta }_{t} \tilde{y}_{t}+\tilde{\alpha }_{t} \tilde{x}_{it}}{\tilde{\beta }_{t}+\tilde{\alpha }_{t}},\frac{1}{\tilde{\beta }_{t}+\tilde{\alpha }_{t}}\right) . \end{aligned}$$

(48)

We then impose here again the logic in the proof of Lemma 2 by analogously defining $\mu _t=\tilde{\alpha }_t/(\tilde{\beta }_t+\tilde{\alpha }_t)$ as well as getting $\bar{E}^n_t(\theta )=(1-\mu _t^n)\tilde{y}_t+\mu _t^n \theta$ and $E_{it}(\bar{E}^n_t(\theta ))=(1-\mu _t^{n+1})\tilde{y}_t+\mu _t^{n+1} \tilde{x}_{it}$. As a result, we must have in an equilibrium

$$\begin{aligned} a_{it}=\frac{\tilde{\beta }_t \tilde{y}_t +(1-\gamma )\tilde{\alpha }_t \tilde{x}_{it}}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t}. \end{aligned}$$

(49)

So in the equilibrium, given the statement of the proposition as well as property (48) are both true for stage $t-1$, it must also be the case for stage t. And the validity of both the statement and the property for stage 1 are shown in Lemma 2, uniqueness is depicted by (49) and the proof is thus completed by induction. $\square$

1.4 Proof of Corollary 1.

For the first half of the statement, based on the inductive logic used in the proof above, the steps we showed (43) also prove $\tilde{x}^*_t=\theta +\frac{\tilde{\beta }_t}{\tilde{\alpha }_t}(\theta -\tilde{y}_t)$. For the second half, notice that

$$\begin{aligned} a_{it}-\theta =\frac{\tilde{\beta }_t (\tilde{y}_t-\theta ) +(1-\gamma )\tilde{\alpha }_t (\tilde{x}_{it}-\theta )}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t}. \end{aligned}$$

(50)

We rewrite the two noise terms in the equation above as

$$\begin{aligned} \tilde{\alpha }_t (\tilde{x}_{it}-\theta )= & {} \tilde{\alpha }_{t-1}\tilde{x}_{i,t-1} +\alpha _t x_{it} -(\tilde{\alpha }_{t-1}+\alpha _t ) \theta \nonumber \\= & {} \sum _{n=1}^{t} \alpha _n (x_{in}-\theta ) \nonumber \\= & {} \sum _{n=1}^{t} \alpha _n \varepsilon _{in}\,, \end{aligned}$$

(51)

$$\begin{aligned} \tilde{\beta }_t (\tilde{y}_t -\theta )= & {} \tilde{\beta }_t \tilde{y}_{t-1} +\Omega _t \beta _t \Upsilon _t -(\tilde{\beta }_t+\Omega _t \beta _t)\theta \nonumber \\= & {} \beta _1(y_1-\theta )+\sum _{n=2}^t \Omega _n \beta _n (\Upsilon _n-\theta ) \nonumber \\= & {} \beta _1(y_1-\theta ) +\sum _{n=2}^t \sqrt{\Omega _n} \beta _n \xi _n\,. \end{aligned}$$

(52)

The last equality of (52) results from (45). And we already know $\varepsilon _{in} \sim \mathcal {N}(0,{1}/{\alpha _{n}})$, $\sqrt{\Omega _n} \xi _n \sim \mathcal {N}(0,{\Omega _n}/{\beta _{n}})$ for $n\ge 2$, and $(y_1-\theta ) \sim \mathcal {N}(0,{1}/{\beta _{1}})$. Because of the independence between noises, we thus have

$$\begin{aligned} \textit{Var}\left[ \tilde{\alpha }_t (\tilde{x}_{it}-\theta )\right]= & {} \sum _{n=1}^{t} \alpha _n^2 \cdot \frac{1}{{\alpha }_n} = \tilde{\alpha }_t \,, \end{aligned}$$

(53)

$$\begin{aligned} \textit{Var}\left[ \tilde{\beta }_t (\tilde{y}_t -\theta )\right]= & {} \beta _1 +\sum _{n=2}^t \beta _n^2 \cdot \frac{\Omega _n}{\beta _n} = \tilde{\beta }_t. \end{aligned}$$

(54)

All the properties above together suggest $(a_{it}-\theta )$ is normally distributed with mean 0 and variance $\frac{\tilde{\beta }_t + (1-\gamma )^2 \tilde{\alpha }_t}{\left( \tilde{\beta }_t + (1-\gamma )\tilde{\alpha }_t \right) ^2}$. The whole statement is proved. $\square$

1.5 Proof of Proposition 2.

According to the definition of $d_{1t}$, we have

$$\begin{aligned} d_{1t}= & {} \frac{\tilde{\beta }_t (\tilde{y}_t-\theta ) +(1-\gamma )\int _0^1\tilde{\alpha }_t (\tilde{x}_{it}-\theta )\,\mathrm {d} i}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t} \nonumber \\= & {} \frac{\tilde{\beta }_t (\tilde{y}_t-\theta ) +(1-\gamma )\int _0^1\sum _{n=1}^t \alpha _n \varepsilon _{in} \,\mathrm {d} i}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t} \nonumber \\= & {} \frac{\tilde{\beta }_t (\tilde{y}_t-\theta ) +(1-\gamma )\sum _{n=1}^t \left[ \int _0^1\alpha _n \varepsilon _{in} \,\mathrm {d} i\right] }{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t} \nonumber \\= & {} \frac{\tilde{\beta }_t (\tilde{y}_t-\theta )}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t}. \end{aligned}$$

(55)

The second equality above results from (51) and the last equality is because $\int _0^1 \alpha _n \varepsilon _{in} \, \mathrm {d}i=0$ for each n. Property (54) thus leads to

$$\begin{aligned} d_{1t} \sim \mathcal {N} \left( 0 \,,\, \frac{\tilde{\beta }_t}{\left( \tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t \right) ^2}\right) , \end{aligned}$$

(56)

the first part is proved. It is then straightforward to see $\partial V_{1t}/ \partial \tilde{\alpha }_{t}<0$. To check the last property, we have

$$\begin{aligned} \frac{\partial V_{1t}}{\partial \tilde{\beta }_{t}}= & {} \frac{\left( \tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t \right) ^2-2\tilde{\beta }_t (\tilde{\beta }_t+(1-\gamma ) \tilde{\alpha }_t)}{\left( \tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t \right) ^4} \nonumber \\= & {} \frac{(1-\gamma )^2 \tilde{\alpha }^2_t -\tilde{\beta }_t^2}{\left( \tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t \right) ^4}, \end{aligned}$$

(57)

so $\partial V_{1t}/ \partial \tilde{\beta }_{t}<0$ holds if and only if $\frac{\tilde{\beta }_t}{\tilde{\alpha }_t}>1-\gamma$. $\square$

1.6 Proof of Proposition 3.

The definition of $d_{2t}$ gives us

$$\begin{aligned} d_{2t}= & {} \frac{\tilde{\beta }_t \tilde{y}_t +(1-\gamma )\tilde{\alpha }_t \tilde{x}_{it}}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t} -\frac{\tilde{\beta }_t \tilde{y}_t +(1-\gamma ) \int _0^1 \tilde{\alpha }_t \tilde{x}_{jt} \, \mathrm {d} j}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t} \nonumber \\= & {} \frac{(1-\gamma )\tilde{\alpha }_t (\tilde{x}_{it}-\theta )}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t}, \end{aligned}$$

(58)

combining with (53) we have

$$\begin{aligned} d_{2t} \sim \mathcal {N} \left( 0 \, , \, \frac{(1-\gamma )^2 \tilde{\alpha }_t}{\left( \tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t \right) ^2} \right) . \end{aligned}$$

(59)

The first half is proved. For the second part of the statement, it is straightforward to see it is true by simply checking the signs of $\partial V_{2t}/ \partial \tilde{\beta }_{t}$ and $\partial V_{2t}/ \partial \tilde{\alpha }_{t}$. $\square$

1.7 Proof of Proposition 4.

Based on the previous two propositions we have

$$\begin{aligned} V_{1t}+\frac{V_{2t}}{1-\gamma }=\frac{1}{\tilde{\beta }_t+(1-\gamma )\tilde{\alpha }_t}. \end{aligned}$$

(60)

Obviously, the RHS of the equation above is always decreasing from t to $t'=t+1$ for any t. So as long as $V_{1t}<V_{1t'}$, we must have $V_{2t}>V_{2t'}$; and given $V_{2t}<V_{2t'}$, we must have $V_{1t}>V_{1t'}$. $\square$

1.8 Proof of Proposition 5.

In the equilibrium we have

$$\begin{aligned} d_{3t}= & {} \frac{(1-\gamma )\tilde{\alpha }_t (x_{it}-x^{*}_t)}{\tilde{\beta }_t +(1-\gamma ) \tilde{\alpha }_t} \nonumber \\= & {} \frac{(1-\gamma )\left[ \tilde{\alpha }_t (x_{it}-\theta )+\tilde{\beta }_t (\tilde{y}_t -\theta ) \right] }{\tilde{\beta }_t +(1-\gamma ) \tilde{\alpha }_t}. \end{aligned}$$

(61)

$d_{3t}$ is thus normally distributed with mean zero and variance

$$\begin{aligned} V_{3t}=\frac{(1-\gamma )^2(\tilde{\alpha }_t +\tilde{\beta }_t)}{\left( \tilde{\beta }_t +(1-\gamma ) \tilde{\alpha }_t\right) ^2} = V_{2t}+(1-\gamma )^2 V_{1t}. \end{aligned}$$

(62)

This completes the proof. $\square$

Data and codes

All datasets and codes used in the empirical study could be downloaded from the following link: https://www.dropbox.com/sh/kkepe449sd0lxsk/AACwSK8SwTNCQW8RaSfS4hsua?dl=0

To run the programs, you need to download the whole folder, have Matlab program with version later than R2012a installed on the computer, and follow the instructions in file “Read_Me”.