Ok this one requires only a bit more mathematics then the last two. It is a classic example, introducing a concept called "asymmetric information".

There are 2 players, P1 and P2.

P1 owns a business. The business is worth V to P1.

P2 is considering buying the business. P2 values the business at 1.5V (Maybe P2 is a shrewder entrepenuer and thus can get more out of the business or P1 is subject to a 33% tax on earnings from which P2 is exempt).

Here's where asymmetric information comes in. Only P1 knows V. P2 only knows that V is uniformly distributed on the interval [0,1].

So P1 knows V.
P2 knows V~U(0,1).

Now P2 makes P1 a take-it-or-leave-it offer for the business. If the offer is greater than or equal to V, P1 will accept. If the offer is less than V, P1 will refuse.

Originally posted by telerion Ok this one requires only a bit more mathematics then the last two. It is a classic example, introducing a concept called "asymmetric information".

There are 2 players, P1 and P2.

P1 owns a business. The business is worth V to P1.

P2 is considering buying the business. P2 values the business at 1.5V (Maybe P2 is a shrewder entrepenuer and thu ...[text shortened]... will accept. If the offer is less than V, P1 will refuse.

What offer should P2 make to P1?

could you please explain your notation?

what does V~U(0,1) mean. keep in mind that i read your entire post and still couldn't figure it out.

Originally posted by Acolyte Does P2 know that P1 knows V? Does P1 know that, and so on ad infinitum?

Uhm . . . yes. Yes I think we must go with that. I hesitate only because last time I let common knowledge out of her cage too early. This time though we must assume it from the start.

Suppose P1 offers to sell at price x. Then P2 knows that x is less than or equal to V, as P1 wouldn't sell at a loss. Therefore V must lie somewhere between 0 and x. Since this is a one-off transaction, P2 has no reason to believe that P1 is not trying to oversell the business, so from P2's p-o-v V~U[0,x] (or U[0,1] if x > 1). The business therefore has a value to P2 of 3x/4 or 3/4, whichever is less, so he declines the offer.

You thinking along the correct lines. This is where you go astray just a bit.

Originally posted by Acolyte Suppose P1 offers to sell at price x.

But notice from the OP that P2 is the player making the offer to P1. Amend your reasoning with this in mind, and I'm sure you will have it in a snap. Good work by the way.

Originally posted by telerion You thinking along the correct lines. This is where you go astray just a bit.

Originally posted by Acolyte [b]Suppose P1 offers to sell at price x.

But notice from the OP that P2 is the player making the offer to P1. Amend your reasoning with this in mind, and I'm sure you will have it in a snap. Good work by the way. [/b]

ðŸ˜³ Oops, misread the question.

Suppose P2 offers to buy at x. Then P1 will only sell if V < x, so on average if P1 does sell, P2 will get something worth 3/4 or 3x/4 to him, whichever is lower, for a net profit of -x/4 or worse. If P1 doesn't sell then of course P2 breaks even, so overall for x > 0 P2 has an expected loss. Therefore P2 should offer nothing for the business.

I should say just a bit about the puzzle. Since P2 values the business more than P1, trade would occur under complete information (ceteris peribus and assuming P2 can afford V). Since it is a take-it-or-leave-it offer, the bid would equal V. If they could negotiate, then the accepted bid would lie somewhere between V and 1.5*V. In this case, the uncertainty built into the model prevents trade.

Since V is uniformly distributed in the interval of (0,1), and the business worth 1.5V to P2. P2 must make a take-it-or-leave-it offer which is in the interval of (0.6,1).