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 $X_1=x$?
  • $\begingroup$ Can anybody help with this? $\endgroup$
    – superhulk
    Commented Mar 23, 2019 at 16:57
  • $\begingroup$ What do you mean by " how are $X_1$ and $X_2$ affiliated"? Affiliation is a form of positive correlation. There is a definition in the book. Moreover, Krishna usually uses $G$ as the cdf of the highest order statistic and $g$ as its density. $\endgroup$
    – Bayesian
    Commented Mar 23, 2019 at 17:24
  • $\begingroup$ @Bayesian Let me rephrase my query. Say there are two random variables(rvs) $X_1,X_2$. We say that the rvs are affiliated, if a higher value of one rv may lead to statistically higher value of the other. Mathematically, we say that the joint density is log-supermodular. My question is, how did the author deduce that $X_1,X_2$ are affiliated, given that there is no information about their joint density. Moreover, in the next step he has calculated $\frac{g(t|t)}{G(t|t)}$, which I think cannot be calculated without the joint density of $X_1,X_2$. $\endgroup$
    – superhulk
    Commented Mar 24, 2019 at 1:33
  • $\begingroup$ Also, the joint density of $X_1$ and $Y_1$ given in Figure 6.2 of the book, which I'm not able to figure out. $\endgroup$
    – superhulk
    Commented Mar 24, 2019 at 4:21
  • $\begingroup$ I can back out the joint density and conditional denisty from the fact that each of the three components ($S_1,S_2,T$) is uniformly distributed. I will put that into an answer later. Let me know if that is what you mean. $\endgroup$
    – Bayesian
    Commented Mar 25, 2019 at 16:04

1 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:

enter image description here So, how to get there?

  1. $S_1,S_2, T$ 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.

  1. 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 paperenter image description here

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

  • $\begingroup$ Disclaimer: I am not a probability theorist - far from it. This is how I approached this example from scratch. I put in some work, because I was curious - I actually taught this example once and I took Krishna's density for granted without knowing how to get there. I would be happy to make this answer more detailed so I know where to come back to if I have to teach this again and forget how to do it. $\endgroup$
    – Bayesian
    Commented Mar 28, 2019 at 16:42
  • 1
    $\begingroup$ Thanks. The expression $F_{X_1,X_2|T=t}(x,y|t) = \mbox{Pr}(X_1<x,X_2<y|T=t) = \mbox{Pr}(S_1<x-t,S_2<y-t)$ was something I was looking for. From here it is relatively easy to work out the solution. $\endgroup$
    – superhulk
    Commented Mar 28, 2019 at 18:42

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