Optimal Tender Price for a Block Trade

Worked Solution

How to Think About It: This is a classic adverse-selection bidding problem. You are quoting a price to buy a block. Bid too low and the seller rejects you; bid too high and you win but overpay. The optimal bid balances the probability of execution against the expected profit per trade. Before any math: your profit is $(\mu - P)$ per accepted trade, and raising $P$ increases acceptance probability $F(P)$ but shrinks the margin. The optimum trades off marginal acceptance probability against marginal profit -- exactly the logic behind any market-making spread.

Quick Estimate: Suppose $\mu = 100$, $\mu_0 = 98$ (seller's location), $\lambda = 1$. Intuitively the optimal bid is somewhere between the seller's midpoint 98 and the true value 100. At $P = 99$, acceptance probability is $F(99) = 1/(1 + e^{-1}) \approx 0.73$, expected profit is $0.73 \times (100 - 99) = 0.73$. At $P = 98$, acceptance is $0.5$, profit is $0.5 \times 2 = 1.0$. At $P = 97$, acceptance is about $0.27$, profit is $0.27 \times 3 = 0.81$. So the optimum is near $P \approx 98$, i.e., close to the seller's location parameter. We will confirm this.

Approach: Write the expected gain, differentiate, and express via the hazard rate. Then specialize to the logistic distribution, whose hazard rate has a particularly clean form.

Formal Solution:

Part 1 -- Expected Gain and FOC

Since $V$ is independent of whether the tender is accepted, the expected gain is:

$$G(P) = F(P) \cdot (\mu - P)$$

Differentiating with respect to $P$:

$$G'(P) = f(P)(\mu - P) - F(P) = 0$$

This gives the FOC:

$$f(P)(\mu - P) = F(P)$$

Part 2 -- Hazard Rate Form

Divide both sides of the FOC by $1 - F(P)$:

$$\frac{f(P)}{1 - F(P)} \cdot (\mu - P) = \frac{F(P)}{1 - F(P)}$$

The left side contains the hazard rate $h(P) = f(P)/(1 - F(P))$, and the right side is the odds ratio. So the FOC becomes:

$$h(P)(\mu - P) = \frac{F(P)}{1 - F(P)}$$

Alternatively, rearranging the original FOC by dividing both sides by $F(P)$:

$$\frac{f(P)}{F(P)} (\mu - P) = 1$$

But the hazard-rate form is more natural. The economic interpretation: at the optimum, the marginal gain from raising $P$ by $dP$ (which increases acceptance probability by $f(P) \, dP$ on a deal worth $\mu - P$) exactly equals the marginal cost (paying $dP$ more on the $F(P)$ fraction of deals that would have been accepted anyway).

Part 3 -- Logistic CDF

For the logistic distribution with $F(P) = \frac{1}{1 + e^{-\lambda(P - \mu_0)}}$, we have:

$$f(P) = \lambda F(P)(1 - F(P))$$

This is the key property of the logistic -- the density factors as the CDF times the survival function times $\lambda$. The hazard rate is:

$$h(P) = \frac{f(P)}{1 - F(P)} = \lambda F(P)$$

Substituting into the FOC $f(P)(\mu - P) = F(P)$:

$$\lambda F(P)(1 - F(P))(\mu - P) = F(P)$$

Since $F(P) > 0$ at any interior optimum, divide by $F(P)$:

$$\lambda(1 - F(P))(\mu - P) = 1$$

Now $1 - F(P) = \frac{e^{-\lambda(P - \mu_0)}}{1 + e^{-\lambda(P - \mu_0)}} = \frac{1}{1 + e^{\lambda(P - \mu_0)}}$. Let $x = \lambda(P - \mu_0)$. Then:

$$\frac{\lambda(\mu - P)}{1 + e^{x}} = 1$$

$$\lambda(\mu - P) = 1 + e^{\lambda(P - \mu_0)}$$

Let $\Delta = \mu - \mu_0$ (the edge -- how much more the block is worth to you than the seller's location). Then $\mu - P = \Delta - (P - \mu_0)$ and writing $u = \lambda(P - \mu_0)$:

$$\lambda \Delta - u = 1 + e^{u}$$

This is a transcendental equation for $u$ (equivalently for $P^{*}$). It does not have a closed form in elementary functions, but can be expressed using the Lambert $W$ function. Rearranging:

$$e^{u} = \lambda \Delta - 1 - u$$

For the special case $\lambda \Delta = 2$, we can verify $u = 0$ works: $e^{0} = 1 = 2 - 1 - 0$. This gives $P^{*} = \mu_0$, meaning you bid at the seller's location when $\mu - \mu_0 = 2/\lambda$.

More generally, the solution satisfies $P^{*} = \mu - \frac{1 + e^{\lambda(P^{*} - \mu_0)}}{\lambda}$, which is easily solved numerically for any given parameters.

Checking our earlier estimate: with $\mu = 100$, $\mu_0 = 98$, $\lambda = 1$, we get $\lambda \Delta = 2$, so $P^{*} = \mu_0 = 98$, confirming our back-of-envelope calculation that the optimum was near 98.

Answer: The expected gain is $G(P) = F(P)(\mu - P)$. The FOC is $f(P)(\mu - P) = F(P)$, or in hazard-rate form: $h(P)(\mu - P) = F(P)/(1 - F(P))$. For the logistic CDF, the optimal bid satisfies $\lambda(1 - F(P^{*}))(\mu - P^{*}) = 1$, which reduces to the transcendental equation $e^{\lambda(P^{*} - \mu_0)} = \lambda(\mu - P^{*}) - 1$.