A Help Applying Search Rule Weitzman (1979)

[mathlover]

I was wondering if you could help me make sure I have things correct. Weitzman (1979) (Optimal search for the best alternative) considers a decision maker that is facing n boxes, each box has potential reward $x_i $with probability distribution $f(x_i)$ (iid). It costs $c_i$ to open the box and learn its contents. There is a discount factor $\beta_i$ for each box. The paper shows that each box has a reservation price $z_i$ that satisfies:

\begin{equation}
c_i = \beta_i \int_{z_i}^\infty (x_i-z_j)\textrm{d}f(x_i)-(1-\beta_i)z_i
\end{equation}Then, the paper gives a specific example where $\beta_i=1$, each box has reward $R_i$ with probability $p_i$ and reward of 0 with probability $1-p_i$. It is given that in this case:
\begin{equation}
z_i=(p_i*R_i-c_i)/p_i
\end{equation}

Now, I'm trying to find out what the $z_i$ would be if $0<\beta_i<1.$ I believe that it is:
\begin{equation}
z_i=(\beta_i*p_i*R_i-c_i)/(\beta_i*p_i+(1-\beta_i))
\end{equation}

The reason I believe this is the solution is due to the following steps i took:
\begin{equation}
\int_{z_i}^\infty (x_i-z_j)\textrm{d}f(x_i)=p_i*(R_i-z_i)
\end{equation}
(I got this from the solution the paper provided for the case where $\beta_i=1$.)

Then, we have
\begin{equation}
c_i = \beta_i*p_i*(R_i-z_i)-(1-\beta_i)z_i
\end{equation}

And,
\begin{equation}
c_i = \beta_i*p_i*R_i-\beta_i*p_i*z_i-(1-\beta_i)z_i
\end{equation}

Thus,
\begin{equation}
c_i = \beta_i*p_i*R_i-z_i*[\beta_i*p_i+(1-\beta_i)]
\end{equation}

Giving us,
\begin{equation}
z_i=(\beta_i*p_i*R_i-c_i)/(\beta_i*p_i+(1-\beta_i))
\end{equation}

However, I had previously calculated it as
\begin{equation}
z_i=(\beta_i*p_i*R_i-c_i)/(\beta_i*p_i)
\end{equation}
(I don't know how I had calculated this) and now I am doubting myself as to which one is the correct solution. Can you help me here?

Thank you.

 

[tnich]

Your newer solution looks correct to me. If you are unsure, why not substitute it for $z_i$ in the original equation and see if it works?