Auctions with affiliation

Question

This problem is an example given in Vijay Krishna's Auction Theory (2nd Edition, Chapter-6, Example 6.2). The problem is as follows:

Suppose $S_1,S_2$, and $T $ are uniformly and independently distributed on $[0,1]$. There are two bidders.Bidder 1 receives the signal $X1=S1+T$,and bidder 2 receives the signal $X_2=S_2+T$. The object has a common value for both the bidders, $V=(X_1+X_2)/2$.

Now, we are required to find out the bidding strategy for a first price auction. The equilibrium bidding function is given as $\beta(x)=\int_0^xv(y,y) \,dL(y|x)$. $L(y|x)$ is further equal to $\exp(-\int_y^x\,\frac{g(t|t)}{G(t|t)}\,dt)$.

I have four questions:

1. How are $X_1$ and $X_2$ affiliated?
2. How is $\frac{g(t|t)}{G(t|t)}$ calculated in this example?
3. How do I find the joint density of $X_1$ and $X_2$?
4. How do I find the conditional density of $X_2$ given that $X_1=x$?

Answer 1

1. Roughly speaking, $X_1$ and $X_2$ are affiliated because they have the common component $T$. That is, if $X_1$ is large, $X_2$ tends to be large as well, because a large $X_1$ makes a large $T$ likelier than without this information. The variables are affiliated if for the joint density $f_{X_1,X_2}$ $$f_{X_1,X_2} (x',y) f_{X_1,X_2} (x,y') \geq f_{X_1,X_2} (x,y) f_{X_1,X_2} (x',y') \quad \forall x'\geq y, y'\geq y,$$
   which here holds with equality. That is, they are weakly affiliated. See below.

2. In Krishna's book $G$ (resp. $g$) is usually the CDF (resp. density) of the highest value among the other bidders, $Y_1$. Here, there is only one other bidder such that $Y_1=X_2$ and $g(y|x) = f_{X_2|X_1=x}(y |x)$.

Now, my answers only help you if you know how to find the joint density. Here is what we want to find:

So, how to get there?

3. $S_1,S_2, T$ are i.i.d. draws from the uniform distribution. Let us first consider the joint CDF of $X_1, X_2$ conditional on $T=t$. That is easy, because they are independent now! Once we have this conditional CDF, we use that $$F_{X_1,X_2} (x,y) = \int_0^1 F_{X_1,X_2|T=t}(x,y|t) f_T (t) dt,$$ where $f_T(t)= 1$ if $t\in [0,1]$ and $f_T(t)= 0$ otherwise. Then, we take derivatives with respect to $x$ and $y$ to find $f_{X_1,X_2}(x,y)$. This should look like what Krishna has in his picture or alternatively, like the function in the appendix of Avery (ReStud 1998), where the example is from (I guess). These calculations are cumbersome work with a bunch of case distinctions and thus it is easy to mess up, but you start with $$F_{X_1,X_2|T=t}(x,y|t) = \mbox{Pr}(X_1<x,X_2<y|T=t) = \mbox{Pr}(S_1<x-t,S_2<y-t)$$ and then pay attention to the support of the uniform distribution. The CDF is $F_{S_1}(s)=s$ for $s\in [0,1]$ and zero for smaller $s$ and one for larger $s$.

Let us take an exemplary region, $y>x>1$: $$F_{X_1,X_2}(x,y) = \int_0^{x-1} 1 dt + \int_{x-1}^{y-1}(x-t)dt+ \int_{y-1}^1 (x-t)(y-t)dt,$$ which gives us something of which Mathematica says it has a derivateive $\partial x \partial y$ equal to $2-y$. Now, we can verify one of the regions of Krishna's graphic. One more for fun, if $x<y<1$, $$f_{X_1,X_2}(x,y) = \frac{\partial^2}{\partial x \partial y} \left( \int_0^x (x-t)(y-t) dt + \int^1_x 0 dt \right) = x.$$ And so on. Now, you can go back to my reply 1.

4. The conditional density of is given by the equation $$f_{X_1,X_2} (x,y) = f_{X_1|X_2=y} (x|y) f_{X_2} (y),$$ where $$f_{X_2}(y) = \begin{cases} y &\mbox{if } y \in [0,1] \\ 2-y &\mbox{if } y \in [1,2], \end{cases} $$ which follows from the fact that $X_2 =S_2+T$, and $S_2$ and $T$ are independently uniformly distributed. You get there with convolution. What you get should look like this plot from the Avery paper

Then, you integrate to find $G(y|x)$ and go back to my reply 2.