Here are the results for the simple case of a two-bidder, Vickrey auction, where v_0, v_1 and v_2 are all distributed uniformly on [0,1]. Here v_0 is the common value, v_1 is the private value for bidder B_1, and v_2 is the private value for bidder v_2.
The strategy (x,y) indicates that the bidder purchases x pieces of information on the common value and y pieces of information for her private value. The expected payoff (a,b) indicates that bidder B_1 expects a profit of a, and bidder B_2 expects a profit of b.
B_1: (0,0), B_2: (0,0).
Then neither bidder knows
anything but the item's expected value. Therefore, unless there is collusion,
they both bid 1, and expect zero profit.
Expected payoffs: (0,0)
B_1: (1,0), B_2: (1,0).
Then each bidder knows the common
value exactly, and since neither has any knowledge of the private values, they
each bid v_0 + .5 . Each then wins half the time, and profits by the expected
difference between her private value and .5, which is zero.
Expected payoffs: (0,0).
B_1: (0,1), B_2: (0,1).
Then each one bids .5 plus their known private value,
and since she pays .5 plus the other bidder's private value, profits
by the difference in private values. Since a bidder wins with probability
.5, the conditional expected difference in private values given that you win is 1/3, and the probability that you win is .5, the expected profit is
1/6.
Expected payoffs: (1/6, 1/6).
B_1: (1,1), B_2: (1,1).
Then each bids exactly their
value. Either bidder wins with probability .5 (when they have the higher total
value), and pays the other bidder's total value. Therefore, the bidder profits
by the difference in private values.
Expected payoffs: (1/6, 1/6).
B_1: (0,0), B_2: (1,0).
Then neither knows her private value and only B_2 knows the common value.
In this case B_2 optimally bids v_0+ .5, and B_1 optimally bids 1.5. Since each
pays the other's bid when she wins, B_1 always wins and pays v_0 + .5, for
an expected profit of zero. B_2 never wins, and so also has expected profit
zero.
Expected payoffs: (0,0).
B_1: (0,0), B_2: (0,1).
B_1 optimally
bids 1 and B_2 optimally bids v_2 + .5, giving each an expected profit of
1/8.
Expected payoffs: (1/8, 1/8).
B_1: (0,0), B_2: (1,1).
Here B_1 knows nothing and B_2 knows everything.
Then
B_2 optimally bids v_0 + v_2, her total value. If B_1 is going to pay v_0
+ v_2 for the item, she expects to profit v_1 - v_2 from winning it. Given that
v_0 + v_2 is less than some value c (like the bid of B_1, for example), the
expected value of v_2 is c/3. Then the expected profit of B_1 is (.5-c/3) times
the probability that v_0 + v_2 is less than c. This expected profit has a
maximum value of 1/12 when the bid is c = 1. Then B_2 wins when v_0 + v_2 > 1,
which occurs with probability .5. She profits v_0 + v_2 - 1, which, given
v_0 + v_2 > 1, has expected value 1/3, so her expected profit is 1/6.
Expected payoffs: (0, 1/6).
B_1: (0,1), B_2: (1,0).
In this case, B_2 optimally bids v_0
+ 1/2, because the value v_2 is unknown to both players. Therefore, if B_1 wins
the auction, B_1 must pay v_0 + 1/2 for an item that is worth v_0 + v_1 to her, for a profit of v_1 - 1/2. Since B_1 knows the exact value of v_1, B_1 should
bid only when v_1 > 1/2. Furthermore, since B_1 always makes a positive profit when v_1 > 1/2, B_1 should bid at least 3/2 when v_1 > 1/2, ensuring that her bid will be greater than the bid of v_0 + 0.5 submitted by B_2. Following this optimal strategy, the expected profit for B_1 is 1/8. The question of the expected profit for B_2 is murkier. Clearly B_2 will win when v_1 < 1/2, but it is unclear how much profit B_2 will make in this situation. If B_1 bids zero when B_1 does not want to win (i.e., when v_1 < 1/2), then B_2 expects a profit of 1/2. On the other hand, B_1 could submit a bid as great as 1/2 and would still win the auction with probability zero. This could reduace B_2's expected profit to as little as 1/4. Although B_2's profit is actually indeterminate, we will for now suppose that when faced with strategies that are indifferent with regards to her own profits, B_1 will choose the strategy that minimizes B_2's profits.
Expected payoffs: (1/8, 1/4).
B_1: (0,1), B_2: (1,1).
B_2 optimally bids v_0 + v_2, for the reasons stated above. Then B_1 will
pay v_0 + v_2 whenever she wins the auction, for a profit of (v_0 + v_1) -
(v_0 + v_2) = v_1 - v_2. B_1 knows v_1, so B_1 should bid according to a strategy f(v_1) which is a function of v_1, such that the integral of v_1 - v_2 over
the region f(v_1) > v_0 + v_2 is the supremum over the values correspoding to
all f:[0,1] -> R. It can be shown using basic calculus methods that such
a strategy is f(v_1) = 2v_1, which gives the expected payoffs shown below.
Expected payoffs: (7/48, 7/24).
B_1: (1,0), B_2: (1,1).
Expected payoffs: (1/8, 1/8).
The remaining cases are symmetric to ones presented above. (For reversed strategies, reverse the payoffs)
If we ignore the issue of information cost (or if we let cost of the private value information be less than 1/48), the strategy (1,1) dominates all others.
If we let the information cost be between 1/48 and 1/24, the result is more interesting. Then (1,1) no longer dominates (0,1).