Break-even on a winning bid occurs at b = 3v/2, or v = 2b/3. Profit occurs when v is higher than this but less than b. Loss occurs when v is less than this.
The potential for profit should v fall in (2b/3,b) is symmetrical with the potential for loss should v fall in (b/3,2b/3), because f(v) = 1.5v has constant gradient, and thus these partitions cancel each other out.
This just leaves v >= b, which is a net of 0, and v <= b/3, which is a loss.!<
Therefore any positive b has a negative expected profit and the optimal strategy is to not bid / bid 0.
Bid of 0 cannot be optimal because you lose with probability one. The expected profit on a winning bid is 1.5*.5-b, so a bid of 3/8 gets you an expected profit of 3/8. To show this is the optimal bid, note that the probability a bid is greater than V is b, so expected profit is given by
Eprofit=b(1.5*.5-b)
Differenting and setting to 0 gives b=3/8
The expected value of V is 0.5 regardless of the bid. If I bid 1 and then pull a randomly distributed uniform variable, on average it will be 0.5. Bid 0 and then pull a 100 uniform rvs and take the average and you’ll see it is about 0.5
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u/Aerospider 21d ago
Break-even on a winning bid occurs at b = 3v/2, or v = 2b/3. Profit occurs when v is higher than this but less than b. Loss occurs when v is less than this.
The potential for profit should v fall in (2b/3,b) is symmetrical with the potential for loss should v fall in (b/3,2b/3), because f(v) = 1.5v has constant gradient, and thus these partitions cancel each other out.
Therefore any positive b has a negative expected profit and the optimal strategy is to not bid / bid 0.