# Second Price Auction - adjusting PDF for reservation price

The situation:

There is a second price auction with 2 players. Consider a second-price auction with 2 players. Their valuations of the object at auction are and are independently and identically distributed with pdf $f$ and cdf $F$. over $[0,\hat v]$. Assume $f$ is continuous and positive over $[0,\hat v]$.

Now here comes the question: a reservation bid $r$ is now implemented - the winner pays the second of the highest bids including the reservation price, or if both bid lower no-one wins. I want to find the pdf that both bids are above $r$ and above some $x$, and add this to an equation calculating expected revenue for the auctioneer.

I have already found the pdf for both bids being above some value of $x$: $2f(x)(1-F(x))$. The pdf for both being above $r$ is $(1-F(r))^2$.

I've had a look at an answer for the problem, and it suggests that the combined pdf is $\frac{2f(x)(1-F(x))}{(1-F(r))^2}$. Could someone explain to me how this is so?

Then, when calculating expected revenue for the auctioneer, we have for the case where both bids are above $r$: $(1-F(r))^2\int_r^\hat v{\frac{2f(x)(1-F(x))}{(1-F(r))^2}}dx$. I'm also quite confused why we multiply by $(1-F(r))^2$.

To find expected revenue of the seller in a second price auction with reserved price consisting of two bidders who bid their valuations in equilibrium, we do the following :

Given that valuations are i.i.d with pdf $f$ and CDF $F$ over $[0, \hat{v}]$, the revenue that the seller gets for different realizations of players' valuations are as indicated in the graph below: Therefore, expected revenue of the seller is : \begin{eqnarray*} &&\int_r^\hat{v}\int_r^{v_2} v_1 f(v_1)f(v_2)dv_1dv_2 + \int_r^\hat{v}\int_{v_2}^\hat{v} v_2 f(v_1)f(v_2)dv_1dv_2 \\ &&+\int_0^r\int_r^\hat{v} r f(v_1)f(v_2)dv_1dv_2 + \int_r^\hat{v}\int_0^r r f(v_1)f(v_2)dv_1dv_2 \\ &=& \int_r^\hat{v}v_1 (1-F(v_1))f(v_1)dv_1 + \int_r^\hat{v} v_2 (1-F(v_2))f(v_2)dv_2 + 2rF(r)(1-F(r)) \\ &=& 2\int_r^\hat{v} v_2 (1-F(v_2))f(v_2)dv_2 + 2rF(r)(1-F(r))\end{eqnarray*}

I think that the first part of your question must ask for a conditional probability. In other words, for $v_1$ and $v_2$ representing valuations of player 1 and player 2, we should have been asked for the derivation (namely, the density) of the following probability:

$$Pr( v_1>x \land v_2>x \;|\; r<v_1 \land r<v_2)$$ Thus it is equal to: $$\frac{d\left( \frac{(1-F(x))^2}{(1-F(r))^2} \right)}{dx} = \frac{2f(x)(1-F(x))}{(1-F(r))^2}$$

From your words, however, it is deduced that what it asks for is the derivation (so the density) of the following probability:

$$Pr(v_1>max(r,x) \;\land\; v_2>max(x,r))$$ But, assuming wlog x>r, this is just equal to $\frac{d\left( (1-F(x))^2 \right)}{dx} = 2f(x)(1-F(x))$, so there is no place for $r$ since if $v_1$ and $v_2$ are grater than $x$, and $x$ is greater than $r$, then $v_i$>$x$ implies $v_i$>$r$ for $i=1,2$. So no need to bother with $r$ in this case. But I don't think that your question asks for this.

For the last part of your question, I'm again suspicious whether there is no requirement for $v_1$ and $v_2$ to be higher than $x$. If the only constraint for them is to be higher than the reserve price $r$, then Amit above has given a very well answer except that $v_1$ or $v_2$ can't get any value less than $r$ in equilibrium, so in his calculations the integrals $\int_0^r\int_r^\hat{v} r f(v_1)f(v_2)dv_1dv_2$ and $\int_r^\hat{v}\int_0^r r f(v_1)f(v_2)dv_1dv_2$ are useless. When we remove these two integrals, we get the answer $2\int_r^\hat{v} v_2 (1-F(v_2))f(v_2)dv_2$ (you can replace variable $v_2$ by $x$ here to get an identical result with your answer).