<ol>
<li><p>Roughly speaking, <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> are affiliated because they have the common component <span class="math-container">$T$</span>. That is, if <span class="math-container">$X_1$</span> is large, <span class="math-container">$X_2$</span> tends to be large as well, because a large <span class="math-container">$X_1$</span> makes a large <span class="math-container">$T$</span> likelier than without this information.
The variables are affiliated if for the joint density <span class="math-container">$f_{X_1,X_2}$</span> 
<span class="math-container">$$f_{X_1,X_2} (x',y)  f_{X_1,X_2} (x,y') \geq f_{X_1,X_2} (x,y) f_{X_1,X_2} (x',y') \quad \forall x'\geq y, y'\geq y,$$</span><br>
which here holds with equality. That is, they are weakly affiliated. See below.</p></li>
<li><p>In Krishna's book <span class="math-container">$G$</span> (resp. <span class="math-container">$g$</span>) is usually the CDF (resp. density) of the highest value among the other bidders, <span class="math-container">$Y_1$</span>. Here, there is only one other bidder such that <span class="math-container">$Y_1=X_2$</span> and <span class="math-container">$g(y|x) = f_{X_2|X_1=x}(y |x)$</span>. </p></li>
</ol>

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

<p><a href="https://i.sstatic.net/CE0xY.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/CE0xY.png" alt="enter image description here"></a>
 So, how to get there?</p>

<ol start="3">
<li><span class="math-container">$S_1,S_2, T$</span> are i.i.d. draws from the uniform distribution.
Let us first consider the joint CDF of <span class="math-container">$X_1, X_2$</span> conditional on <span class="math-container">$T=t$</span>. That is easy, because they are independent now! Once we have this conditional CDF, we use that
<span class="math-container">$$F_{X_1,X_2} (x,y) = \int_0^1 F_{X_1,X_2|T=t}(x,y|t) f_T (t) dt,$$</span>
where <span class="math-container">$f_T(t)= 1$</span> if <span class="math-container">$t\in [0,1]$</span> and <span class="math-container">$f_T(t)= 0$</span> otherwise.
Then, we take derivatives with respect to <span class="math-container">$x$</span> and <span class="math-container">$y$</span> to find <span class="math-container">$f_{X_1,X_2}(x,y)$</span>. This should look like what Krishna has in his picture or alternatively, like the function in the appendix of <a href="https://academic.oup.com/restud/article-abstract/65/2/185/1580729" rel="nofollow noreferrer">Avery (ReStud 1998)</a>, 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 
<span class="math-container">$$F_{X_1,X_2|T=t}(x,y|t) = \mbox{Pr}(X_1&lt;x,X_2&lt;y|T=t) = \mbox{Pr}(S_1&lt;x-t,S_2&lt;y-t)$$</span> 
and then pay attention to the support of the uniform distribution. The CDF is <span class="math-container">$F_{S_1}(s)=s$</span> for <span class="math-container">$s\in [0,1]$</span> and zero for smaller <span class="math-container">$s$</span> and one for larger <span class="math-container">$s$</span>.</li>
</ol>

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

<ol start="4">
<li>The conditional density of is given by the equation
<span class="math-container">$$f_{X_1,X_2} (x,y) = f_{X_1|X_2=y} (x|y) f_{X_2} (y),$$</span>
where
<span class="math-container">$$f_{X_2}(y) = \begin{cases} y &amp;\mbox{if } y \in [0,1] \\
2-y &amp;\mbox{if } y \in [1,2],
\end{cases} $$</span>
which follows from the fact that <span class="math-container">$X_2 =S_2+T$</span>, and <span class="math-container">$S_2$</span> and <span class="math-container">$T$</span> are independently uniformly distributed. You get there with <a href="https://math.stackexchange.com/questions/357672/density-of-sum-of-two-uniform-random-variables-0-1">convolution</a>. What you get should look like this plot from the Avery paper<a href="https://i.sstatic.net/GAPpm.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GAPpm.png" alt="enter image description here"></a></li>
</ol>

<p>Then, you integrate to find <span class="math-container">$G(y|x)$</span> and go back to my reply 2.</p>

CC BY-SA 4.0
2019-03-28T16:42:19.523Z
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.
Bayesian
CC BY-SA 4.0
2019-03-28T18:42:40.860Z
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.
superhulk
CC BY-SA 4.0
2019-03-28T16:23:37.760Z
2019-03-28T16:44:12.540Z
2019-03-28T16:44:12.540Z
Bayesian
<p>This problem is an example given in Vijay Krishna's <em>Auction Theory</em> (2nd Edition, Chapter-6, Example 6.2). The problem is as follows:</p>

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

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

<p>I have four questions:</p>

<blockquote>
  <ol>
  <li>How are <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span> affiliated?</li>
  <li>How is <span class="math-container">$\frac{g(t|t)}{G(t|t)}$</span> calculated in this example?</li>
  <li>How do I find the joint density of <span class="math-container">$X_1$</span> and <span class="math-container">$X_2$</span>?</li>
  <li>How do I find the conditional density of <span class="math-container">$X_2$</span> given that <span class="math-container">$X_1=x$</span>?</li>
  </ol>
</blockquote>

CC BY-SA 4.0
2019-03-23T16:57:30.393Z
Can anybody help with this?
superhulk
CC BY-SA 4.0
2019-03-23T17:24:34.593Z
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.
Bayesian
CC BY-SA 4.0
2019-03-24T01:33:29.277Z
@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 $X_1,X_2$.
superhulk
CC BY-SA 4.0
2019-03-24T04:21:17.410Z
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.
superhulk
CC BY-SA 4.0
2019-03-25T16:04:52.187Z
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.
Bayesian
CC BY-SA 4.0
2019-03-25T17:11:42.013Z
Yes, the main problem that I'm facing is how to find the joint density. I'll be grateful if you could write an answer for the same.
superhulk
CC BY-SA 4.0
2019-03-28T15:32:12.913Z
I will reformulate your question and then answer it.
Let me know if what I wrote sounds reasonable to you!
Bayesian
<self-study><auctions>
Auctions with affiliation
superhulk
# Auctions with affiliation

(Asked by: username_0 on 2019-03-22)

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 $X\_1=x$?

**Tags:** self-study, auctions

> **Comments:**
> **username_0**: Can anybody help with this?  
> **username_1**: 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.  
> **username_0**: @username_1 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 $X_1,X_2$.  
> **username_0**: 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.  
> **username_1**: 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.  
> **username_0**: Yes, the main problem that I'm facing is how to find the joint density. I'll be grateful if you could write an answer for the same.  
> **username_1**: I will reformulate your question and then answer it. Let me know if what I wrote sounds reasonable to you!  

---

## 1 answers

### Answer by username_1 on 2019-03-28. Score: 4 | Accepted answer

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](https://i.sstatic.net/CE0xY.png)](https://i.sstatic.net/CE0xY.png)
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)](https://academic.oup.com/restud/article-abstract/65/2/185/1580729), 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
   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,
$$f\_{X\_1,X\_2}(x,y) = \frac{\partial^2}{\partial x \partial y} \left( \int\_0^x (x-t)(y-t) dt + \int^1\_x 0 dt \right) = x.$$
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](https://math.stackexchange.com/questions/357672/density-of-sum-of-two-uniform-random-variables-0-1). What you get should look like this plot from the Avery paper[![enter image description here](https://i.sstatic.net/GAPpm.png)](https://i.sstatic.net/GAPpm.png)

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

> **Comments:**
> **username_1**: 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.  
> **username_0**: 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.  

---

