# Tag Info

1

It's not done in two dimensions because it's easier to draw on paper. It's done in two dimensions because if you can grasp the concepts at work in that minimal $n=2$ variable setting, you can generalize the concepts from there to all $n>2$ variable settings. So, for pedagogical reasons of simplicity of teaching, instructors keep it to $n=2$. I don't have ...

2

Does this exist? Of course the 3D representation exist. Below you can see 3D representation of Cobb-Douglas maximization problem with two factors and budget constrain, which is very similar to the problem solved in the video you linked (indeed you would probably not be able to find any qualitative difference in graph for that problem in the video even ...

1

The probabilities are obtained using Bayes updating. Let $f_i = L$ be the event that firm $i$ is low and let $f_i = H$ be the event that firm $i$ is a high type. Assume that firm 1 knows she herself is a high type then:  \begin{align*} \Pr(f_2 = H|f_1 = H) &= \frac{\Pr(f_2 = H \text{ and } f_1 = H)}{\Pr(f_1 = H)},\\ &= \frac{1/3}{1/3 + 1/6},\\ &...

1

Consider a type space $\mathcal T$ of size $T$, an action space $\mathcal A$ of size $A$, and the corresponding set of pure strategies $\mathcal S$ of size $S$. By definition, a pure strategy is a mapping from $\mathcal T$ to $\mathcal A$, implying that $\mathcal S=\mathcal A^\mathcal T$. Therefore, $S=A^T$.

3

Consider a type space $\mathcal T=\{1,2,\dots,T\}$ and an action space $\mathcal A=\{1,\dots,A\}$. With $A=T=2$, you correctly found that there are $S=4$ different pure strategies mapping $\mathcal T \to \mathcal A$. Now fix $A$ and add one more type, so $T=3, A=2$. Type 1 and 2 still have the same number of different actions (4), and the strategy can be ...

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