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.

\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.

Last edited: