Timing of Nature Drawing Types

Question

Consider a simple static, incomplete-information entry game.
Two players, $i=1,2$. where player $1$ is new entrant, and $2$ is incumbent.
Player 2 can be two types: rational or belligerent, with common prior $(p,1-p)$ over her types.
Action set: $A_1=\{Enter, Out\}$, $A_2=\{Fight, Accomodate\}$

When we turn this extensive form game into normal form, we compute the expected payoff for each pair of strategies $(x,zy)$ where $x$ refers to what player 1 chooses at the information set, and $zy$ refers to what player 2 chooses in case of each type.

My question:
When we make this incomplete information game into a complete-imperfect one, we introduce Nature and let her draw the type for player 2. However, she knows her type when she is called upon to move, so why do we calculate an expected payoff for player 2 when converting the game into an normal-form? Is this because when we collapse the extensive form into the normal-form, player 2's pure strategy set takes into account each type that she can be (rational or belligerent), thus she is also forced to calculate an expected pay off? I mean, for any pair of pure strategy set, you have to deal with two terminal nodes and their payoff, so it makes sense in this perspective, you need some sort of expected payoff calculation. But it seems odd that player 2 who supposedly knows her type also has to form expectation on her payoff.

My guess is the way strategy is defined in incomplete information is that it prescribes each type of a player what she should do if this is the type Nature draws. So, as long as the cardinality of type space is greater 1, any pair of pure strategy will force each player to face more than 1 terminal nodes (eg. $(Enter, (Fight, Fight)$).

It just sounds odd that all along, Harsanyi allows a player's type to be unknown only to other players, but in the end, no matter what, whether you have various types or not, the payoffs must be computed in expectation.

Answer 1

I suppose you have in mind a game that looks like this:

she knows her type when she is called upon to move, so why do we calculate an expected payoff for player 2 when converting the game into an normal-form?

The fact that "she knows her type" and therefore has no need to calculate an expected payoff presupposes that the play has reach a particular information set for player 2. Formally, let $\theta\in\{R,B\}$ be P2's type. Her utility conditional on her type being $\theta$ is \begin{equation} u_2(\sigma_1,\beta_2(\color{red}{\theta})\vert\color{red}{\theta}). \end{equation} where $\beta_2(\theta)$ is P2's behavior strategy at the information set associated with $\theta$ (i.e. the left decision node for P2 when $\theta=R$ and the right decision node for P2 if $\theta=R$). Here, no expected value is calculated since the uncertainty is resolved for P2. In contrast, in the normal form representation, P2's utility is an unconditional one: \begin{equation} u_2(\sigma_1,\sigma_2)=\Pr(R)u_2(\sigma_1,\beta_2(R)\vert R)+\Pr(B)u_2(\sigma_1,\beta_2(B)\vert B) \end{equation} where expectation is to be taken.

The reason is that by converting an extensive form game into the normal form, the temporal aspect of the game is lost and everything is seen from the ex ante perspective.

Answer 2

I think the point is that player 2 has no way of revealing her type to player 1 (and even if she did, she might not want to). So regardless of the fact that she knows her type she is going to have to rely on player 1 calculating the expectation when making layer 1's decision to enter or stay out, making the payoff of player 2 also expected.

You're right of course that the expectation collapses if both types result in the same terminal node, because in that case no expectation is needed.