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\begin{document}

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\begin{center}
{\normal \scshape Humphreys, Sanchez de la Sierra, and van der Windt, 2012. \\ Fishing. Online Appendix B }
\end{center}
\setcounter{figure}{1}



\section*{Proofs of claims for the model discussed in Section 5.3}
\subsection*{Assumptions}
Let the researcher's objective function be given by:

\begin{equation}
U(n |\overline{u},\underline{u}, \overline{\beta}) =\underline{u} + (\overline{u} - \underline{u}) \int_{\overline{\beta}-a}^{\overline{\beta}+a}{\pi(\beta,n)f(\beta)d\beta} - c(n)
\end{equation}

where $c$ is a strictly convex cost function, $\pi$ is the power of the test, and $f$ denotes prior beliefs about $\beta$ which we assume to be a uniform distribution over $[\overline{\beta}-a, \overline{\beta}+a]$ for $0<\overline{\beta}<a$. We examine a simple case where treatment and control groups are of equal size and potential outcomes in both groups are distributed normally with unit variance; in this case the power function is given by:  
\begin{equation}
\pi(\beta,n)=1-\Phi\left(t-\beta \sqrt{\frac{n}{4}}\right)
\end{equation}
where $\Phi()$ is the standard normal cdf and $t$ is the $t$ statistic employed for this test. Note that we assume that one sided tests are employed and that the 95\% confidence level is adopted, implying that $t$ takes a value of approximately $1.65$.

Note that: 
\begin{equation} \frac{\partial\pi}{\partial n}=\phi\left(t-\beta \sqrt{\frac{n}{4}}\right)\frac{\beta}{4\sqrt{n}} \end{equation}

And so power is increasing in $n$ if $\beta$ is positive and decreasing if $\beta$ is negative. 

However a useful result is that for positive $\beta$:  
\begin{equation}
\frac{\partial\pi(\beta,n)}{\partial n}+\frac{\partial\pi(-\beta,n)}{\partial n}=\left(\phi\left(t-\beta \sqrt{\frac{n}{4}}\right)- \phi\left(t+\beta \sqrt{\frac{n}{4}}\right)\right)\frac{\beta}{4\sqrt{n}}>0
\end{equation}


The inequality follows from the fact that $t-\beta \sqrt{\frac{n}{4}}$ is closer to 0 than $t+\beta \sqrt{\frac{n}{4}}$. It follows from our assumptions on the distribution of $\beta$ that the expected marginal effect of $n$ on power is positive.
\begin{equation} \int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial \pi(\beta,n)}{\partial n}f(\beta)d\beta} > 0 \end{equation} \label{ME}


\subsection*{Concavity of the objective function}
We first establish that this objective function is strictly concave. Taking the second derivative of the objective function we have:

\begin{equation}
\frac{\partial^{2} U}{\partial n^{2}} =(\overline{u} - \underline{u}) \int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial^{2} \pi(\beta,n)}{\partial n^{2}}f(\beta)d\beta} - c''(n)
\end{equation}

From convexity $c''(n)>0$ and hence we focus on $\int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial^{2} \pi(\beta,n)}{\partial n^{2}}f(\beta)d\beta} $. 

\textbf{Claim A}: $\int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial^{2} \pi(\beta,n)}{\partial n^{2}}f(\beta)d\beta} \leq 0$ 

Claim A is difficult to establish because under our assumptions on $t$ the power function is concave for $\beta>0$ and convex for $\beta<0$. Our strategy is instead to show that for any $\beta>0$  ${\frac{\partial^{2} \pi(\beta,n)}{\partial n^{2}} + \frac{\partial^{2} \pi(-\beta,n)}{\partial n^{2}} \leq 0$.\footnote{Note that we prove this for our assumption on $t$ but note that the result does not necessarily  obtain for values outside of this range. As an illustration set $\beta=.5, n=4, t=2$. Then  $\pi''(\beta) = -0.00025$ and  $\pi''(-\beta) = 0.00031$.}


\textbf{Proof.} The second derivative of the power function is given by:
 
\begin{equation}
\pi'' = -\frac{\beta}{(4n)^{\frac{3}{2}}}\left[ \beta^2\frac{n}{4} -t\beta\frac{\sqrt{n}}{2} + 1 \right]\phi\left(t-\beta\sqrt{\frac{n}{4}}\right)
\end{equation}

And so we seek to establish for $\beta>0$:
\begin{eqnarray*}
\pi''(\beta) + \pi''(-\beta)& \leq 0  \\ 
& \leftrightarrow  \\ 
-\frac{\beta}{(4n)^{\frac{3}{2}}} \left[ \beta^2\frac{n}{4} -t\beta\frac{\sqrt{n}}{2} + 1 \right]\phi\left(t-\beta\sqrt{\frac{n}{4}}\right) +    
\frac{\beta}{(4n)^{\frac{3}{2}}} \left[ \beta^2\frac{n}{4} +t\beta\frac{\sqrt{n}}{2} + 1 \right]\phi\left(t+\beta\sqrt{\frac{n}{4}}\right) &\leq 0\\ 
&\leftrightarrow  \\ 
-\left[ \beta^2\frac{n}{4} -t\beta\frac{\sqrt{n}}{2} + 1 \right]e^{\beta t \sqrt{n}} 
+ \left[ \beta^2\frac{n}{4} +t\beta\frac{\sqrt{n}}{2} + 1 \right]& \leq 0 
\end{eqnarray*}


Defining $\kappa = \beta\sqrt{n}$, we then seek to show:


\begin{equation}
G \equiv  -  e^{t\kappa}\left({1+ \frac{\kappa^2}{4}  - \frac{\kappa t}{2}  }\right) + \left(1+ \frac{\kappa^2}{4}  + \frac{\kappa t}{2} \right) 
 \leq 0 \end{equation}



We now show that for $t \leq \sqrt{3}$ (which includes the range of interest for this analysis) $G$ is decreasing in $\kappa$ and takes on its maximum value at $\kappa =0 $; at this maximum value the condition holds with equality. 

We have:
\begin{eqnarray*}
\frac{\partial G}{ {\partial \kappa}}  &=& \frac{\kappa + t}{2}-e^{\kappa t}\left(\frac{\kappa-t} {2}\right)-te^{\kappa t}\left({1+ \frac{\kappa^2}{4}  - \frac{\kappa t}{2}  }\right) \\
&=& \frac{1}{2}\left( \kappa + t -e^{\kappa t}\left({\frac{t\kappa^2}{2}  - {\kappa t^2} + {\kappa+t} }\right) \right) 
\end{eqnarray*}

Note that $\frac{\partial G}{ {\partial \kappa}}  = 0 $ for $\kappa = 0$.

We now show that  $\frac{\partial G}{ {\partial \kappa}}  \leq 0 $ for all $\kappa$ by showing that $\frac{\partial ^2G}{ {\partial \kappa ^2}}  \leq 0$ 

We have:
\begin{eqnarray*}
\frac{\partial ^2 G}{ {\partial \kappa ^2}} &=& \frac{1}{2}\left(1 - e^{\kappa t}\left( 1+ \frac{(\kappa t)^2}{2} + \kappa t(2- t^2) \right) \right) \\
&=& \frac{1}{2}\left(1 - e^{\kappa t}\left( 1+ \psi \kappa t^*+ \frac{(\kappa t)^2}{2}  \right)\right)
\end{eqnarray*}
where $\psi=\equi 2-t^2$ takes a value in $(-1,2)$ for $t \in [0, \sqrt{3}]$.


Now from the expansion of $e^x$ we have:
\begin{equation}  e^{\kappa t} \geq 1 + \kappa t + \frac{(\kappa t)^2}{2} \end{equation}
 
Note that $\left( 1+ \psi \kappa t^*+ \frac{(\kappa t)^2}{2}  \right)$ is positive (to check this note that the smallest value this expression can take in this range is for $t=\sqrt{3}$ and at this value the minimum is achieved when $\kappa = \frac{1}{\sqrt{3}}$ which yields a value of $\frac{1}{2}$ for the expression).  We then have:
\begin{equation}
 e^{\kappa t}(1 +\psi \kappa t + \frac{(\kappa t)^2}{2}) \geq (1 + \kappa t + \frac{(\kappa t)^2}{2}) (1 +\psi \kappa t + \frac{(\kappa t)^2}{2}) = 1 +(1+\psi)(\kappa t^*+(\kappa t)^2 + \frac{(\kappa t)^3}{2}) + \frac{(\kappa t)^4}{4}\geq 1
\end{equation}

And so: 
\begin{equation}
\frac{\partial ^2 G}{ {\partial \kappa ^2}}  \leq 0 \end{equation}


Thus the highest value $G$ can take is for $\kappa=0$ in which case $G=0$ and we are done. \blacksquare



\subsection*{Comparative statics}

The first order condition defines the optimum implicitly:
\begin{equation}
\frac{\partial U(n^* | \overline{u},\underline{u}, \overline{\beta})}{\partial n} = (\overline{u} - \underline{u}) \int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial \pi(\beta,n^*)}{\partial n}f(\beta)d\beta} - c'(n^*)=0
\end{equation}

For comparative statics, we use the implicit function theorem around $n^*$. Define $F(n, \overline{u},\underline{u}, \overline{\beta})= (\overline{u} - \underline{u}) \int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial \pi(\beta,n)}{\partial n}f(\beta)d\beta} - c'(n)$. We have:
\begin{equation}
\frac{\partial n^\ast}{\partial \underline{u} } = -\frac{\frac{\partial F(n^\ast,\overline{u},\underline{u}, \overline{\beta})}{\partial \underline{u}}}{\frac{\partial F(n^\ast,\overline{u},\underline{u}, \overline{\beta})}{\partial n}}
\end{equation}

Both the denominator and numerator are negative. For the numerator recall that we have from Equation (3) that $\int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial \pi(\beta,n)}{\partial n}f(\beta)d\beta}>0$.  The denominator is negative from the strict concavity of the objective function, as established above. This yields: $\frac{\partial n^\ast}{\partial \underline{u} } <0$. Similar reasoning follows for $\frac{\partial n^*}{\partial \overline{\beta}}<0$.


\subsection*{Comparative statics of the mean prior threshold}
Assume now that data gathering costs are such that in some cases the costs of implementing the optimal design are so great that a researcher would do better forgoing a study entirely. In our model the expected value of any study is an increasing function of $\overline{\beta}$ and so it is helpful to focus on problems for which prior distributions with expectation greater than some value $\overline{\beta}^*$ yield positive expected value while distributions with expectation below $\overline{\beta}^*$ yield negative expected value. 

In particular, let $V(\overline{u}, \underline{u}, a, \overline{\beta})$ denote the value of $U$ when the optimal value of $n$ is chosen and consider ranges in which $\overline{\beta}^\ast(\overline{u}, \underline{u}, a)$ is defined implicitly by the condition: 
\begin{equation}V(\overline{u}, \underline{u}, a, \overline{\beta}^*)=0 \end{equation}

A result from the Envelope theorem is: 
\begin{equation}
\frac{\partial V(\overline{u},\underline{u},\overline{\beta})}{\partial \overline{\beta}}=\frac{\partial U(n^*,\overline{u},\underline{u},\overline{\beta})}{\partial \overline{\beta}}+ \frac{\partial U(n^*,\overline{u},\underline{u},\overline{\beta})}{\partial n }\times \frac{\partial n^*}{\partial \overline{\beta}}=\frac{\partial U(n^*,\overline{u},\underline{u},\overline{\beta})}{\partial \overline{\beta}}
\end{equation}

Applying differentiation under the integral sign, note that: 
\begin{equation}
\frac{\partial U(n^*,\overline{u},\underline{u},\overline{\beta})}{\partial \overline{\beta}}=(\overline{u} - \underline{u})(\pi(\overline{\beta}+a,n)-\pi(\overline{\beta}-a,n))+ \int_{\overline{\beta}-a}^{\overline{\beta}+a}{\frac{\partial \pi(\beta,n)}{\partial \overline{\beta}}f(\beta)d\beta}
\end{equation}

The first quantity is positive since power is increasing in effect sizes. The second quantity is 0 since $\pi$ does not depend on $\overline{\beta}$ except through the limits of integration. 

We then have:
\begin{equation}
\frac{\partial \overline{\beta}^\ast}{\partial \underline{u}}=-\frac{\frac{\partial V}{\partial \underline{u}}}{\frac{\partial V}{\partial \overline{\beta} }}
\end{equation}

From the envelope theorem the numerator is clearly positive and the denominator has already been shown to be positive. This establishes that:
$\frac{\partial \overline{\beta}^\ast}{\partial \underline{u}}<0$.

\end{document}