<p>This seems like a basic issue about probability calculus/theory.</p>

<p>The intuition behind the uniform distribution over $[a,b]$ is that all outcomes between and $a$ and $b$ are equally likely. Because of this, one can get an intuitive grasp about the probability that the outcome $x$ falls in a subinterval $[a_1,b_1] \subset [a,b]$. As the ratio of their length is
$$
\frac{b_1 - a_1}{b - a},
$$
the subinterval $[a_1,b_1]$ contains $(b_1 - a_1)/(b-a)$ share of all the points in $[a,b]$. Each point is an equally likely outcome, therefore
$$
P\left(x \in [a_1,b_1]\right) = \frac{b_1 - a_1}{b - a}.
$$
A problem arises if the subinterval is in fact a single point, $a_1$. What is 
$P\left(x \in \left\{a_1\right\}\right)$?
<br> By the above intuitive reasoning
$$
P\left(x \in [a_1,a_1]\right) = \frac{a_1 - a_1}{b - a} = 0.
$$
A more rigorous approach:
<br> Let us denote this probability $P\left(x \in \left\{a_1\right\}\right)$ by $\alpha$. As the distribution is uniform, all points are equally likely outcomes and hence for any $a_i \in [a,b]$ we have
$$
P\left(x \in \left\{a_i\right\}\right) = \alpha.
$$
There are infinitely many (disjoint) points in $[a,b]$. Let $A$ be an arbitrary set of infinitely many such points, that is
$$
A = \bigcup_i^{\infty} a_i.
$$
Then as probability is <a href="https://en.wikipedia.org/wiki/Sigma_additivity" rel="nofollow noreferrer">sigma additive</a> over disjoint sets
$$
P\left(x \in A \right) = P\left(x \in \bigcup_i^{\infty} a_i \right)
= \sum_i^{\infty} P\left(x \in \left\{a_i\right\}\right)
= \sum_i^{\infty} \alpha.
$$
However as $A \subseteq [a,b]$, $P\left(x \in A \right) \leq 1$. But for any $\alpha &gt; 0$ the sum $\sum_i^{\infty} \alpha$ would be infinite. Therefore $\alpha$ cannot be positive, it is zero.</p>

CC BY-SA 3.0
2017-11-08T14:45:33.120Z
2017-11-09T09:45:56.070Z
2017-11-09T09:45:56.070Z
Giskard
<p>Consider a 2 player all-pay auction, bidding for $1.</p>

<p>Each submits a bid that is a real number, thus $S_i=[0,\infty)$. The player with a higher bid wins $1, but both players must pay the submitted bid.</p>

<p>Player $i$'s payoff function is:</p>

<p>$v_i(s_i,s_{j})=$</p>

<p>$-s_i$ if $s_i&lt;s_{j}$</p>

<p>($\frac{1}{2}-s_i$) if $s_i=s_{j}$</p>

<p>($1-s_i$) if $s_i&gt;s_j$.</p>

<p>Suppose player $j$ plays a mixed strategy in which she is uniformly choosing a bid between 0 and 1.</p>

<p>I am only interested in the expected payoff from player $i$'s bidding less than 1, because if $s_i$>1, he would win, but also incur a negative payoff.</p>

<p>My question:</p>

<p>Why is Pr$(s_i=s_j)$=0 for any $s_i\in[0,1]$ if player $j$'s mixed strategy is U[0,1]?</p>

<p>My confusion is: the PDF of player $j$ is 1 for any $s_j\in(0,1)$. So if I only consider $s_i&lt;1$ and Pr$(s_i=s_j)$, how do I justify it becomes $0$?</p>

<game-theory><auctions>
2 Player All-Pay Auction
Frank Swanton
# 2 Player All-Pay Auction

(Asked by: username_0 on 2017-11-08)

Consider a 2 player all-pay auction, bidding for $1.

Each submits a bid that is a real number, thus $S\_i=[0,\infty)$. The player with a higher bid wins $1, but both players must pay the submitted bid.

Player $i$'s payoff function is:

$v\_i(s\_i,s\_{j})=$

$-s\_i$ if $s\_i

($\frac{1}{2}-s\_i$) if $s\_i=s\_{j}$

($1-s\_i$) if $s\_i>s\_j$.

Suppose player $j$ plays a mixed strategy in which she is uniformly choosing a bid between 0 and 1.

I am only interested in the expected payoff from player $i$'s bidding less than 1, because if $s\_i$>1, he would win, but also incur a negative payoff.

My question:

Why is Pr$(s\_i=s\_j)$=0 for any $s\_i\in[0,1]$ if player $j$'s mixed strategy is U[0,1]?

My confusion is: the PDF of player $j$ is 1 for any $s\_j\in(0,1)$. So if I only consider $s\_i<1$ and Pr$(s\_i=s\_j)$, how do I justify it becomes $0$?

**Tags:** game-theory, auctions

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## 1 answers

### Answer by username_1 on 2017-11-08. Score: 6 | Accepted answer

This seems like a basic issue about probability calculus/theory.

The intuition behind the uniform distribution over $[a,b]$ is that all outcomes between and $a$ and $b$ are equally likely. Because of this, one can get an intuitive grasp about the probability that the outcome $x$ falls in a subinterval $[a\_1,b\_1] \subset [a,b]$. As the ratio of their length is
$$
\frac{b\_1 - a\_1}{b - a},
$$
the subinterval $[a\_1,b\_1]$ contains $(b\_1 - a\_1)/(b-a)$ share of all the points in $[a,b]$. Each point is an equally likely outcome, therefore
$$
P\left(x \in [a\_1,b\_1]\right) = \frac{b\_1 - a\_1}{b - a}.
$$
A problem arises if the subinterval is in fact a single point, $a\_1$. What is
$P\left(x \in \left\{a\_1\right\}\right)$?
  
 By the above intuitive reasoning
$$
P\left(x \in [a\_1,a\_1]\right) = \frac{a\_1 - a\_1}{b - a} = 0.
$$
A more rigorous approach:
  
 Let us denote this probability $P\left(x \in \left\{a\_1\right\}\right)$ by $\alpha$. As the distribution is uniform, all points are equally likely outcomes and hence for any $a\_i \in [a,b]$ we have
$$
P\left(x \in \left\{a\_i\right\}\right) = \alpha.
$$
There are infinitely many (disjoint) points in $[a,b]$. Let $A$ be an arbitrary set of infinitely many such points, that is
$$
A = \bigcup\_i^{\infty} a\_i.
$$
Then as probability is [sigma additive](https://en.wikipedia.org/wiki/Sigma_additivity) over disjoint sets
$$
P\left(x \in A \right) = P\left(x \in \bigcup\_i^{\infty} a\_i \right)
= \sum\_i^{\infty} P\left(x \in \left\{a\_i\right\}\right)
= \sum\_i^{\infty} \alpha.
$$
However as $A \subseteq [a,b]$, $P\left(x \in A \right) \leq 1$. But for any $\alpha > 0$ the sum $\sum\_i^{\infty} \alpha$ would be infinite. Therefore $\alpha$ cannot be positive, it is zero.


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