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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 $2 \notin T_j$??

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    $\begingroup$ $1/x$ 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) $\endgroup$ – Henry Dec 20 '16 at 23:51
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$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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