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I want to know if my thinking is correct. Look at the following game.

Simultaneous game

As the game has only one subgame (i.e., the game itself) then the Nash Equilibria will coincide with the subgame perfect equilibria.

In this case, we have two Nash equilibria: {U, u} and {D, d}. By my statement before, the subgame perfect equilibria will be {U, u} and {D, d} too.

Am I right?

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    $\begingroup$ You are correct! $\endgroup$ – afreelunch Nov 6 '18 at 15:11
  • $\begingroup$ The presentation suggests that it's a sequential game. If so, then "u" is not a strategy, so {U,u} is not a Nash equilibrium. $\endgroup$ – Acccumulation Nov 6 '18 at 20:22
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As afreelunch only answered in the comment section, I'll answer it here.

The SPNE (Subgame Perfect Nash Equilibrium) is a refinement of the NE (Nash Equilibrium). So, let's call $S$ the set of all SPNE and $N$ the set of all NE in a game. Then:

$$S \subseteq N.$$

If the sequential game has only one subgame (see the subgame definition here), the game itself, then we will have.

$$S = N.$$

And my reasoning was right.

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