A biotech company discovers an early drug candidate, e.g. a molecular-glue degrader hitting a disease-relevant protein.

Bringing it to market takes many costly development steps and the outcome is uncertain. The biotech won't go it alone: it will sell or license the asset to a pharmaceutical company. The choice is when to sell:

• sell the raw candidate, undeveloped, or
• invest to develop it first, and sell a developed asset.

Possible reasons why the timing of sale matters:

• Financial reasons: the biotech is cash-constrained and sells early.
• Cost advantage: pharma can develop the asset more cheaply.
• Risk allocation: the party better able to bear risk should develop.

We shut all three down: no development-cost advantage, risk-neutral parties, no financial frictions.

What determines the timing of sale, and is it efficient?

A framework with a buyer and a seller, where:

• the seller owns a project with uncertain value that requires costly development;
• neither party has a development-cost advantage;
• the seller chooses whether to develop before sale and the price;
• the buyer can explore alternative projects, before and after purchasing.

We then:

• obtain a tractable formulation of the seller's problem and characterize the equilibrium;
• conduct welfare analysis;
• study comparative statics in development cost and the set of alternatives.

• Seller owns one box $(F,c)$.
• Buyer owns a random collection of boxes $\{(F_i,c_i)\}_{i=1}^N$.
  • $\mathcal{B}$ is the set of collections; $\mu\in\Delta(\mathcal{B})$ its distribution.
• For simplicity, all boxes have support in $[0,\bar v]$.

• $t=0$, Seller: chooses to open ($d=0$) or leave closed ($d=1$), and sets a price $P$.
• $t\ge 1$, Buyer: observes its collection $B$ and sequentially decides to:
  • purchase the seller's box (add it to the collection),
  • open one of its closed boxes,
  • stop and adopt an opened box, or stop with nothing.

$O$ = boxes opened by the buyer, $i$ = box adopted, $t$ = transfer.

Reservation value $r$ uniquely solves

• Search: open the closed box with the highest reservation value.
• Stopping: stop once the highest realized value exceeds all remaining reservation values.
• Adoption: adopt the open box with the highest realized value.

The reservation value is well-defined for every box: each box $(F_i,c_i)$ has its own $r_i$, computed from its distribution and cost via the equation above.

With a price $P$, the buyer treats the seller's box as a box with reservation $r(P)$ solving

The right-hand side grows with $P$, so the reservation value $r(P)$ is decreasing in $P$.

Now $v$ is the realized value of box 0; the buyer compares $v-P$ against the alternatives.

Suppose the buyer has no other boxes, $\mathcal{B}=\{\emptyset\}$, so the seller's box is the only option.

• Closed: price $E[v]-c$, sells with probability 1 $\Rightarrow$ revenue $E[v]-c$.
• Open, symmetric: set price $=v$ for each realization $\Rightarrow$ expected revenue $E[v]-c$. Indifferent.
• Open, asymmetric: a single price $P$ gives $P\,[1-F(P)]-c\le E[v]-c$. Closed strictly better.

Without alternatives, conditioning the price on $v$ is exactly what an open box cannot do, so the asymmetry strictly hurts.

The buyer will not buy the closed box iff there is an alternative box $i$ with $r_i>r(P)$ and $v_i>r(P)$.

• Fixing $B$, let $z_i=\min\{r_i,v_i\}$ and $z_B=\max_i\{z_i\}$.
• Let $\displaystyle Q(r)=\int \Pr(z_B<r)\,d\mu$.
• A closed box sells at price $P$ iff $z_i<r(P)$ for all $i$, with probability $Q(r(P))$.
• An open box of value $v$ sells iff $z_i<v-P$ for all $i$, with probability $Q(v-P)$.

Compare the value of selling open vs. closed:

Proof sketch. The threshold structure follows from a single-crossing property plus two boundary checks:

• Single crossing: $V_0$ is steeper in $c$ than $V_1$, so the two values cross at most once.
• Boundary check $\bar c\ge 0$: open is (weakly) better at $c=0$.
• Boundary check $\bar c\le E[v]$: closed is (weakly) better at $c=E[v]$.

Change of variables. Optimize over the reservation $r$ instead of the price $P$. Since $$P=\int_r^{\bar v}(v-r)\,dF(v)-c$$,

Single crossing. Comparing the slopes in $c$:

• $V_0'(c)=-1$: an open-box seller fully absorbs cost increases.
• By the envelope theorem, $V_1'(c)=-Q(r^*(c))\ge -1$ $\Rightarrow$ $V_1$ is flatter.
• $V_0$ steeper than $V_1$ $\Rightarrow$ the two values cross at most once: single crossing.

It suffices to show $V_0(0)\ge V_1(0)$. Let $r^*$ be the optimal reservation for the closed box at $c=0$,

The high price $2\bar v$ never sells, so the box sells only when $v>r^*$. Then

The open seller can always replicate the closed allocation, so $V_0(0)\ge V_1(0)$, hence $\bar c\ge 0$.

At $c=E[v]$ the seller cannot make a positive profit by opening the box:

By charging a price of $0$, a closed-box seller can always guarantee a zero profit:

So $V_1(E[v])\ge V_0(E[v])$: closed is (weakly) better at $c=E[v]$, hence $\bar c\le E[v]$.

The seller can no longer tailor the price to $v$:

• $\tilde V_0$ is parallel to $V_0$ but at a lower level.
• The closed-box value $V_1$ is unchanged.

Being unable to condition the price on $v$ shrinks the returns to opening: the open region shrinks ($\underline c\le\bar c$), and possibly $\underline c<0$.

We could have $\underline c<0$. When instead is $\underline c>0$? Set $c=0$ and ask when opening is strictly optimal:

Open: price $x$, sells w.p. $Q(v-x)$.  ·  Closed: reservation $x$, price $\int (v-x)\,dF(v)$, sells w.p. $Q(x)$.

The two sides differ only in where $Q$ is evaluated: at the buyer's net value $v-x$ (open) versus at the reservation $x$ (closed).

Convenient because $z_i=\min\{r_i,v_i\}\in\{0,r_i\}$: if such a box does not fail, an optimally-searching buyer adopts it at once.

An open box of value $v$ at price $P$ is purchased with

• probability $0$ if $v<P$;
• probability $1$ if $v-P>r$;
• probability $q$ if $v-P\in[0,r)$.

So the open-box seller's problem is

Hence, by the envelope theorem,

$\tilde V_0$ rises with slope in $[0,E[v]]$, so it crosses the closed envelope from below then above, giving the interval $(q_L,q_H)$.

Suppose the buyer has a single box, with zero cost and uniform value in $[0,1]$. Then

With $Q(x)=x$, the two problems coincide at $c=0$:

• for an open box, $x$ is the price;
• for a closed box, $x$ is the reservation value.

The maximizer $x^*=\arg\max_{x}\int_x^1 x\,(v-x)\,dF(v)$ satisfies the FOC

Claim: $x^*\ge p^\star$. From the FOC,

so $x^*$ lies weakly above the pivot. The optimal closed-box price sets $r(P)=x^*$, which requires $P<p^\star$: the closed price is lower.

Two sellers, each with a box; $v_1,v_2\sim F$ and $c_1,c_2$ on $[0,1]$. Sellers simultaneously choose open/closed and a price; the buyer searches sequentially.

Efficient benchmark: open the cheapest box first (if its cost $<\tfrac12$); open the second only if $\displaystyle\int_{v_1}^{\bar v}(v-v_1)\,dF(v)\ge c_2.$

• Sequential search & Pandora's rule: Weitzman; Doval (2018) on buying without inspecting; Chen (2021) and Pan (2023) on search deterrence.
• Advertising & disclosure: Anderson & Renault (2006), Wang & Sarvary (2017), Dogan & Hu (2022): firms reveal product value to shape search.
  • In contrast, here the buyer invests in its own outside option, not the seller's product.
• Contingent pricing: Armstrong & Zhou (2016), Taylor (2018): exploding offers and search-dependent prices.