<p>The &quot;trick&quot; is to identify an equilibrium action where you know (or can pin-down) the payoff. You can then impose indifference between this action's known payoff and the payoffs from other actions (which will depend on the mixing probabilities) to determine those equilibrium mixing probabilities.</p>
<p>A nice example can be found in a Varian's &quot;<a href="https://www.jstor.org/stable/1803562?seq=1#page_scan_tab_contents" rel="nofollow noreferrer">A Model of Sales</a>&quot;. Let me describe a simple version of that model before discussing what it teaches us about finding continuous mixed equilibria.</p>
<hr />
<p><strong>Set up</strong></p>
<p>Two firms each set a price, <span class="math-container">$p_i$</span>, for a homogeneous product they product at zero cost. There is a continuum of consumers, each willing to pay up to <span class="math-container">$v$</span> for the product.</p>
<p>A fraction <span class="math-container">$a$</span> of the consumers (call them &quot;shoppers&quot;) will buy from whichever firm sets the lowest price (breaking ties randomly). A fraction <span class="math-container">$\frac{1-a}{2}$</span> of the consumers (call them &quot;1's captives&quot;) always buy from firm <span class="math-container">$1$</span> (as long as <span class="math-container">$p_1&lt;v$</span>). The final  <span class="math-container">$\frac{1-a}{2}$</span> of the consumers (call them &quot;2's captives&quot;) always buy from firm <span class="math-container">$2$</span> (as long as <span class="math-container">$p_2&lt;v$</span>).</p>
<p>You should be able to quickly check that this game has no pure strategy equilibrium (think about why we can't have an equilibrium with <span class="math-container">$0&lt;p_1&lt;p_2$</span>, <span class="math-container">$0&lt;p_1=p_2$</span>, or <span class="math-container">$0=p_1=p_2$</span>).</p>
<p>Thus, we look for a mixed strategy equilibrium where prices are drawn from the distribution <span class="math-container">$F$</span>. Let the support of <span class="math-container">$F$</span> (i.e., the range of prices over which firms mix) be <span class="math-container">$[\underline{p},\overline{p}]$</span>.</p>
<p>No firm will ever choose <span class="math-container">$p_i&gt;v$</span> because it then serves no consumers, but could earn a profit of <span class="math-container">$v(1-a)/2$</span> by serving its captives at a price of <span class="math-container">$v$</span>. Thus, <span class="math-container">$\overline{p}\leq v$</span>.</p>
<p>A firm's profit if it sets price <span class="math-container">$p$</span> is
<span class="math-container">$$\frac{1-a}{2}p+ap[1-F(p)].$$</span>
The first term is the profit from captives (who the firm serves for sure). The second term is the profit from shoppers, who are served only if the rival firm sets a higher prices (which happens with probability <span class="math-container">$1-F(p)$</span>).</p>
<p><strong>Finding the mixed-strategy equilibrium</strong></p>
<p>We know that if the firm is to mix over prices <span class="math-container">$p$</span> and <span class="math-container">$\overline{p}$</span> then it must be indifferent:
<span class="math-container">$$\frac{1-a}{2}p+ap[1-F(p)]=\frac{1-a}{2}\overline{p}+a\overline{p}[1-F(\overline{p})].$$</span>
Notice that if we could eliminate the <span class="math-container">$\overline{p}$</span> terms from the right-hand side then we would have an equation that could be solved for the equilibrium strategy, <span class="math-container">$F$</span>.</p>
<blockquote>
<p>The next part is the key to being able to find the equilibrium, so I'll put it in a box:</p>
<p>Let's think about what happens when a firm sets <span class="math-container">$p_i=\overline{p}$</span>. Because <span class="math-container">$\overline{p}$</span> is the highest price ever charged in equilibrium, we know that the firm never serves the shoppers at that price (i.e., <span class="math-container">$F(\overline{p})=1$</span>). Moreover, since the firm serves only its captives who will buy at any price below <span class="math-container">$v$</span>, it must be the case that <span class="math-container">$\overline{p}=v$</span>.</p>
</blockquote>
<p>We therefore know that:</p>
<ul>
<li><span class="math-container">$\overline{p}$</span> is in the support of <span class="math-container">$F$</span>, meaning firms must be indifferent between <span class="math-container">$\overline{p}$</span> and any other price</li>
<li><span class="math-container">$\overline{p}$</span> results in profit <span class="math-container">$\frac{1-a}{2}\overline{p}$</span></li>
<li><span class="math-container">$\overline{p}=v.$</span></li>
</ul>
<p>Putting these facts together with the indifference condition, we have
<span class="math-container">$$\frac{1-a}{2}p+ap[1-F(p)]=\frac{1-a}{2}v,$$</span>
which implies
<span class="math-container">$$F(p)=1-\frac{\frac{1-a}{2}(v-p)}{ap},$$</span>
which is our mixed strategy equilibrium. The only part of the equilibrium we haven't solved is the value of <span class="math-container">$\underline{p}$</span>. But this is easily found by solving <span class="math-container">$F(\underline{p})=0$</span> for  <span class="math-container">$\underline{p}$</span> now that we know <span class="math-container">$F(\cdot)$</span>.</p>
<hr />
<p><strong>Final remarks</strong></p>
<p>The trick used in the above example is pretty useful more generally.</p>
<p>Unless the equilibrium strategy, <span class="math-container">$F$</span>, has an atom, we know by definition that <span class="math-container">$F(\underline{p})=0$</span> and <span class="math-container">$F(\overline{p})=1$</span>. Thus, if we can use <em>economic reasoning</em> to determine the value of either <span class="math-container">$\overline{p}$</span> or <span class="math-container">$\underline{p}$</span> (as in the box above) then we can eliminate all of the terms that depend on <span class="math-container">$\overline{p}$</span> (or <span class="math-container">$\underline{p}$</span>) on the right hand side of our indifference equation and then solve for <span class="math-container">$F$</span>.</p>

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2019-04-14T09:21:23.547Z
2019-04-14T09:21:23.547Z
2020-06-17T09:40:54.643Z
Ubiquitous
<p>I am not sure if you are looking for a very abstract, intuitive trick for finding mixed Nash equilibrium, or just an example of how to possibly find one. </p>

<p>I will assume the latter.</p>

<p>Suppose we have the following game:</p>

<p><span class="math-container">$$
L \ \ \ \     \ \ |\ \ R 
\\
U (1,1) | (0,0)
\\
D (0,0) | (1,1)
$$</span></p>

<p>Clearly <span class="math-container">$(U,L)$</span> and <span class="math-container">$(D,R)$</span> are NE (they are pure NE). However, there is a mixed strategy NE too. How can we find this?</p>

<p>We do the following:</p>

<p>Suppose Player 1 mixes, going <span class="math-container">$U$</span> with probability <span class="math-container">$\alpha$</span>, and <span class="math-container">$D$</span> with probability <span class="math-container">$1-\alpha$</span>. (So <span class="math-container">$\alpha \in [0,1]$</span>) That means Player 2 must have the <em>same expected outcome</em> whether they go <span class="math-container">$L$</span> or <span class="math-container">$R$</span>. </p>

<p>So, we can express Player 2's strategy like this:</p>

<p>If they went <span class="math-container">$L$</span>, their expected pay-off is:</p>

<p><span class="math-container">$$1\times \alpha + 0 \times (1-\alpha) = \alpha$$</span></p>

<p>If they went <span class="math-container">$R$</span>, their expected pay-off is:</p>

<p><span class="math-container">$$0 \times \alpha + 1 \times (1-\alpha) = 1 -\alpha$$</span></p>

<p>Remember, their expected outcome from <span class="math-container">$L$</span> and <span class="math-container">$R$</span> must be equal, thus:</p>

<p><span class="math-container">$$\alpha = 1-\alpha \Rightarrow 2\alpha = 1 \Rightarrow \alpha = \frac{1}{2}$$</span></p>

<p>What we just did is solve for Player 1's mixed strategy- going <span class="math-container">$U$</span> with probability <span class="math-container">$\frac{1}{2}$</span>, and <span class="math-container">$D$</span> with probability <span class="math-container">$1-\frac{1}{2} = \frac{1}{2}$</span>. </p>

<p>We can find Player 2's probabilities in a similar way. Note that the process to do this is different depending on the type of game- but their main "trick" here is noting in mixed strategy equilibrium, expected outcomes over the strategies must be equal.</p>

CC BY-SA 4.0
2019-03-27T17:53:08.043Z
The OP asks specifically about *continuous* games, but your example features a game with *finite* strategy spaces.
Herr K.
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2019-03-27T18:06:53.150Z
Sorry I misread- I thought they meant a continuous probability space, my mistake.
user43395
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2019-03-27T03:12:13.767Z
2019-03-27T03:12:13.767Z
user43395
<p>In continuous games, the probability distributions over the players' strategy spaces are infinite. How then is it even possible to then derive a mixed-strategy nash equilibrium?</p>

<p>One would have to show that the mixed-strategy of one player is the best possible mixed-strategy response to the strategy of another player. In the continuous case, how is this even possible? How do you maximize over "all possible probability distributions".</p>

<p>Can anybody provide a clue?</p>

CC BY-SA 4.0
2019-03-27T17:50:45.980Z
1. If a mixed strategy $\sigma_i$ is a best response to other players' strategies $\sigma_{-i}$, then all pure strategies that are played with positive probability in $\sigma_i$ must lead to the same payoff when played against $\sigma_{-i}$. 2. Even in games with finitely many pure strategies, the probability distributions over them are also infinite.
Herr K.
CC BY-SA 4.0
2019-03-27T17:56:18.430Z
See [this post](https://economics.stackexchange.com/q/8935/42) for an example of a mixed strategy NE in a game with continuous strategy spaces.
Herr K.
<game-theory><nash-equilibrium>
What trick can be used to calculate mixed-equilibria?
PaliPali
# What trick can be used to calculate mixed-equilibria?

(Asked by: username_0 on 2019-03-26)

In continuous games, the probability distributions over the players' strategy spaces are infinite. How then is it even possible to then derive a mixed-strategy nash equilibrium?

One would have to show that the mixed-strategy of one player is the best possible mixed-strategy response to the strategy of another player. In the continuous case, how is this even possible? How do you maximize over "all possible probability distributions".

Can anybody provide a clue?

**Tags:** game-theory, nash-equilibrium
> **Comments:**
> **username_1**: 1. If a mixed strategy $\sigma_i$ is a best response to other players' strategies $\sigma_{-i}$, then all pure strategies that are played with positive probability in $\sigma_i$ must lead to the same payoff when played against $\sigma_{-i}$. 2. Even in games with finitely many pure strategies, the probability distributions over them are also infinite.  
> **username_1**: See [this post](https://economics.stackexchange.com/q/8935/42) for an example of a mixed strategy NE in a game with continuous strategy spaces.  

---

## 2 answers

### Answer by username_2 on 2019-04-14. Score: 2

The "trick" is to identify an equilibrium action where you know (or can pin-down) the payoff. You can then impose indifference between this action's known payoff and the payoffs from other actions (which will depend on the mixing probabilities) to determine those equilibrium mixing probabilities.

A nice example can be found in a Varian's "[A Model of Sales](https://www.jstor.org/stable/1803562?seq=1#page_scan_tab_contents)". Let me describe a simple version of that model before discussing what it teaches us about finding continuous mixed equilibria.

---

**Set up**

Two firms each set a price, $p\_i$, for a homogeneous product they product at zero cost. There is a continuum of consumers, each willing to pay up to $v$ for the product.

A fraction $a$ of the consumers (call them "shoppers") will buy from whichever firm sets the lowest price (breaking ties randomly). A fraction $\frac{1-a}{2}$ of the consumers (call them "1's captives") always buy from firm $1$ (as long as $p\_1). The final $\frac{1-a}{2}$ of the consumers (call them "2's captives") always buy from firm $2$ (as long as $p\_2).

You should be able to quickly check that this game has no pure strategy equilibrium (think about why we can't have an equilibrium with $0, $0, or $0=p\_1=p\_2$).

Thus, we look for a mixed strategy equilibrium where prices are drawn from the distribution $F$. Let the support of $F$ (i.e., the range of prices over which firms mix) be $[\underline{p},\overline{p}]$.

No firm will ever choose $p\_i>v$ because it then serves no consumers, but could earn a profit of $v(1-a)/2$ by serving its captives at a price of $v$. Thus, $\overline{p}\leq v$.

A firm's profit if it sets price $p$ is
$$\frac{1-a}{2}p+ap[1-F(p)].$$
The first term is the profit from captives (who the firm serves for sure). The second term is the profit from shoppers, who are served only if the rival firm sets a higher prices (which happens with probability $1-F(p)$).

**Finding the mixed-strategy equilibrium**

We know that if the firm is to mix over prices $p$ and $\overline{p}$ then it must be indifferent:
$$\frac{1-a}{2}p+ap[1-F(p)]=\frac{1-a}{2}\overline{p}+a\overline{p}[1-F(\overline{p})].$$
Notice that if we could eliminate the $\overline{p}$ terms from the right-hand side then we would have an equation that could be solved for the equilibrium strategy, $F$.

> The next part is the key to being able to find the equilibrium, so I'll put it in a box:
>
> Let's think about what happens when a firm sets $p\_i=\overline{p}$. Because $\overline{p}$ is the highest price ever charged in equilibrium, we know that the firm never serves the shoppers at that price (i.e., $F(\overline{p})=1$). Moreover, since the firm serves only its captives who will buy at any price below $v$, it must be the case that $\overline{p}=v$.

We therefore know that:

* $\overline{p}$ is in the support of $F$, meaning firms must be indifferent between $\overline{p}$ and any other price
* $\overline{p}$ results in profit $\frac{1-a}{2}\overline{p}$
* $\overline{p}=v.$

Putting these facts together with the indifference condition, we have
$$\frac{1-a}{2}p+ap[1-F(p)]=\frac{1-a}{2}v,$$
which implies
$$F(p)=1-\frac{\frac{1-a}{2}(v-p)}{ap},$$
which is our mixed strategy equilibrium. The only part of the equilibrium we haven't solved is the value of $\underline{p}$. But this is easily found by solving $F(\underline{p})=0$ for $\underline{p}$ now that we know $F(\cdot)$.

---

**Final remarks**

The trick used in the above example is pretty useful more generally.

Unless the equilibrium strategy, $F$, has an atom, we know by definition that $F(\underline{p})=0$ and $F(\overline{p})=1$. Thus, if we can use *economic reasoning* to determine the value of either $\overline{p}$ or $\underline{p}$ (as in the box above) then we can eliminate all of the terms that depend on $\overline{p}$ (or $\underline{p}$) on the right hand side of our indifference equation and then solve for $F$.


---

### Answer by username_3 on 2019-03-27. Score: 0

I am not sure if you are looking for a very abstract, intuitive trick for finding mixed Nash equilibrium, or just an example of how to possibly find one.

I will assume the latter.

Suppose we have the following game:

$$
L \ \ \ \ \ \ |\ \ R
\\
U (1,1) | (0,0)
\\
D (0,0) | (1,1)
$$

Clearly $(U,L)$ and $(D,R)$ are NE (they are pure NE). However, there is a mixed strategy NE too. How can we find this?

We do the following:

Suppose Player 1 mixes, going $U$ with probability $\alpha$, and $D$ with probability $1-\alpha$. (So $\alpha \in [0,1]$) That means Player 2 must have the *same expected outcome* whether they go $L$ or $R$.

So, we can express Player 2's strategy like this:

If they went $L$, their expected pay-off is:

$$1\times \alpha + 0 \times (1-\alpha) = \alpha$$

If they went $R$, their expected pay-off is:

$$0 \times \alpha + 1 \times (1-\alpha) = 1 -\alpha$$

Remember, their expected outcome from $L$ and $R$ must be equal, thus:

$$\alpha = 1-\alpha \Rightarrow 2\alpha = 1 \Rightarrow \alpha = \frac{1}{2}$$

What we just did is solve for Player 1's mixed strategy- going $U$ with probability $\frac{1}{2}$, and $D$ with probability $1-\frac{1}{2} = \frac{1}{2}$.

We can find Player 2's probabilities in a similar way. Note that the process to do this is different depending on the type of game- but their main "trick" here is noting in mixed strategy equilibrium, expected outcomes over the strategies must be equal.

> **Comments:**
> **username_1**: The OP asks specifically about *continuous* games, but your example features a game with *finite* strategy spaces.  
> **username_3**: Sorry I misread- I thought they meant a continuous probability space, my mistake.  

---

