<p>$p_i (t_j) $ is the PDF of $i$'s belief over $j$'s types.</p>

<p>Recall that if $t_j$ is uniformly distributed on $[0,x]$, then the probability that $t_j \in [a,b]$, where $0 \leq a \leq b \leq x$ is given by</p>

<p>$$ \Pr ( a \leq t_j \leq b) = \int_a^b p_i (t_j) d t_j $$</p>

<p>In particular,</p>

<p>$$ \Pr (t_j = a) = 0 $$</p>

<p>for any $a$.</p>

CC BY-SA 3.0
2016-12-19T11:28:07.203Z
2016-12-19T11:28:07.203Z
Theoretical Economist
<p>My book defines player $i$s 'belief' about player $j$s type as $p_i(t_j)$.</p>

<p>It then goes on to say that if $t_j$ is presumed to be uniformly distributed on $[0,x]$, then $p_i(t_j) = 1/x$ for all $t_j$.</p>

<p>That makes no sense to me. What do they mean by $1/x$? So if $x = 1$, then $p_i(t_j = 2) = 1$? Even though $2 \notin T_j$??</p>

CC BY-SA 3.0
2016-12-20T23:51:52.497Z
$1/x$ is a density not a probability.  And if $x=1$ then $0 \le t_j \le 1$ so cannot be $2$ (or is presumed not to be)
Henry
<game-theory>
What's a "belief" in Bayesian games?
Jian
# What's a "belief" in Bayesian games?

(Asked by: username_0 on 2016-12-19)

My book defines player $i$s 'belief' about player $j$s type as $p\_i(t\_j)$.

It then goes on to say that if $t\_j$ is presumed to be uniformly distributed on $[0,x]$, then $p\_i(t\_j) = 1/x$ for all $t\_j$.

That makes no sense to me. What do they mean by $1/x$? So if $x = 1$, then $p\_i(t\_j = 2) = 1$? Even though $2 \notin T\_j$??

**Tags:** game-theory

> **Comments:**
> **username_1**: $1/x$ is a density not a probability.  And if $x=1$ then $0 \le t_j \le 1$ so cannot be $2$ (or is presumed not to be)  

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## 1 answers

### Answer by username_2 on 2016-12-19. Score: 5

$p\_i (t\_j) $ is the PDF of $i$'s belief over $j$'s types.

Recall that if $t\_j$ is uniformly distributed on $[0,x]$, then the probability that $t\_j \in [a,b]$, where $0 \leq a \leq b \leq x$ is given by

$$ \Pr ( a \leq t\_j \leq b) = \int\_a^b p\_i (t\_j) d t\_j $$

In particular,

$$ \Pr (t\_j = a) = 0 $$

for any $a$.


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