I would like to clarify a doubt regarding the paper Testing the Rationality of Price Forecasts: New Evidence from Panel Data (by MICHAEL P. KEANE AND DAVID E. RUNKLE) that presents an estimator involving survey means of forecasts.
The article works in a context where we have $_tP_{i,t+k}$ as a $k$-step-ahead prediction of $P_t$ made at time $t$ by forecaster $i$, with $i=1,\ldots,N$ forecasters and $t=1,\ldots,T$. In fact, the paper tests the rationality of individual price forecasts in a panel of professional forecasters. Although this is the main subject of the article, I am now interested in deriving the formula for a certain regression coefficient.
The survey mean forecast at time $t$, which I suppose that is implicitly defined by the authors as: $$ _t\overline{P}_{t+k} = \frac{1}{N}\sum_{i=1}^N\, _tP_{i,t+k}. $$ The paper run three separate tests of rationality (Here we will use just two of them). It gives an expression for an certain coeficient estimator as follows (See Equation (5) of the paper): \begin{equation}\label{sm123} \hat{\alpha}_{1m} = \frac{\sum_{t=1}^T P_{t+k}\left(\sum_{i=1}^N {}_tP_{i,t+k}\right)}{\sum_{t=1}^T\left(\sum_{i=1}^N {}_tP_{i,t+k}^2\right)}\tag{I}. \end{equation} It seems that the article does not fully detail how to derive this particular formula.
To better understand my problem, let’s recall a simple linear regression model: \begin{equation} y_t = \alpha_0 + \alpha_1 x_t + \varepsilon_t, \quad t=1,\ldots,T. \end{equation} In matrix form: $$ Y = X\alpha + \varepsilon, $$ where $X$ is a $T \times 2$ matrix with a column of ones and a column of $x_t$ values, and $\alpha = (\alpha_0,\alpha_1)$. The OLS estimator is given by: \begin{equation} \hat{\alpha} = (X'X)^{-1}X'Y. \end{equation} Defining the sample means of $x_t$ and $y_t$: $$ \bar{x} = \frac{1}{T}\sum_{t=1}^T x_t, \quad \bar{y} = \frac{1}{T}\sum_{t=1}^T y_t, $$ we have the well-known OLS formulas: \begin{equation}\label{ols123} \hat{\alpha}_1 = \frac{\sum_{t=1}^T (x_t - \bar{x})(y_t - \bar{y})}{\sum_{t=1}^T (x_t - \bar{x})^2}, \quad \hat{\alpha}_0 = \bar{y} - \hat{\alpha}_1 \bar{x}.\tag{II} \end{equation}
For an individual forecaster, a test of rationality can be performed by running the regression: \begin{equation} P_{t+k}=\alpha_{0}+\alpha_{1}\,\, _tP_{i,t+k}+\varepsilon_{t,k}^{i} \end{equation} I repeat, I am not interested at this moment in defining what it means for a forecaster to be rational, but it is necessary to write the regressions.
If we assume: $$ \frac{1}{T}\sum_{t=1}^T P_{t+k} = 0 \quad \text{and} \quad \frac{1}{T}\sum_{t=1}^T \, _tP_{i,t+k} = 0, $$ we can use the formula (\ref{ols123}) to conclude that: $$ \hat{\alpha}_1 = \frac{\sum_{t=1}^T \, _tP_{i,t+k} P_{t+k}}{\sum_{t=1}^T \, _tP_{i,t+k}^2}. $$ Taking the mean of the estimated coefficients across individuals, the article shows (in its Equation (3)): \begin{equation} \hat{\alpha}_{1i}:=\frac1N\sum_{i=1}^N\frac{\sum_{t=1}^T P_{t+k} \, _tP_{i,t+k} }{\sum_{t=1}^T \, _tP_{i,t+k}^2 }. \end{equation} Now we turn to deriving the estimator $\hat{\alpha}_{1m}$. When replacing $x_t$ with the survey mean forecast $_t\overline{P}_{t+k}$, one might expect an analogous derivation. If I try to start from the regression: $$ P_{t+k} = \alpha_0 + \alpha_1 \, _t\overline{P}_{t+k} + u_{t+k}, $$ assuming $\frac{1}{T}\sum_{t=1}^T \, _t\overline{P}_{t+k}=0$, and apply OLS in a standard way, I would write: \begin{equation}\label{estimator045} \hat{\alpha}_1 = \frac{\sum_{t=1}^T \, _t\overline{P}_{t+k} P_{t+k} }{\sum_{t=1}^T (_t\overline{P}_{t+k})^2}.\tag{III} \end{equation} Now, let’s attempt to expand $_t\overline{P}_{t+k}$. Since $$ _t\overline{P}_{t+k} = \frac{1}{N}\sum_{i=1}^N {}_tP_{i,t+k}, $$ For the numerator, we have $$ \sum_{t=1}^T (_t\overline{P}_{t+k} P_{t+k}) = \sum_{t=1}^T \left(\frac{1}{N}\sum_{i=1}^N {}_tP_{i,t+k}\right)P_{t+k} = \frac{1}{N}\sum_{t=1}^T P_{t+k}\left(\sum_{i=1}^N {}_tP_{i,t+k}\right). $$ For the denominator: $$ \sum_{t=1}^T (_t\overline{P}_{t+k})^2 = \sum_{t=1}^T \left(\frac{1}{N}\sum_{i=1}^N {}_tP_{i,t+k}\right)^2 = \frac{1}{N^2}\sum_{t=1}^T \left(\sum_{i=1}^N {}_tP_{i,t+k}\right)^2. $$ So the estimator (\ref{estimator045}) can be rewritten as: $$ \hat{\alpha}_1 = \frac{\frac{1}{N}\sum_{t=1}^T P_{t+k}\left(\sum_{i=1}^N {}_tP_{i,t+k}\right) }{ \frac{1}{N^2}\sum_{t=1}^T \left(\sum_{i=1}^N {}_tP_{i,t+k}\right)^2 } = N\frac{\sum_{t=1}^T P_{t+k}\left(\sum_{i=1}^N {}_tP_{i,t+k}\right) }{ \sum_{t=1}^T \left(\sum_{i=1}^N {}_tP_{i,t+k}\right)^2 } . $$ And this seems to be very different to $\hat{\alpha}_{1m}$ given by equation (\ref{sm123}) (Equation (5) from the paper)
How can we proceed to arrive at the exact formula for $\hat{\alpha}_{1m}$ presented in the article?
Thank you in advance for any help or insight.
