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I'm trying to determine which strategy is stochastically stable.

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In my lecture notes, we assign probabilities $a$, $b$ $1-a-b$ to player 2 strategies $T,M,B$ resp.

Hence we have the payoffs

$$u(t)=2a+1-a-b$$ $$u(m)=b+2-2a$$ $$u(b)=4-2b-3a$$

And it is inferred from this that $b,b$ is stochastically stable. I tried using the utilities to derive a system of three equations, and showing that $B$ yields the highest payoff for the largest range of $a$ and $b$, but this did not seem to succeed.

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As a caveat, disclaimer, I am not too familiar with Evolutionary Game Theory and it is quite outside my research area/main areas of expertise. However, I can show here that range of $(a,b)$ for which $B$ is a best response is strictly greater than for either of the other two strategies. To that end,

You can characterize the range of $(a,b)$ for which $B$ is a (strict) best response by solving the system

$$\begin{split} 4−2b−3a &\> b+2-2a\\ 4−2b−3a &> 1+a-b\\ \end{split}$$

Or,

$$\begin{split} 2 &> 3b + a\\ 3 &> 4a + b\\ \end{split}$$

Or,

$$\begin{split} a &> 7/11, b<3-4a\\ a &\leq 7/11, b<\frac{2-a}{3}\\ \end{split}$$

Range for Strategy B

We can solve for the area of this quadrilateral:

$$\begin{split} B &= \int_{0}^{7/11}\frac{2-a}{3}da + \int_{7/11}^{3/4}(3-4a)da\\ &= \frac{259}{726} + \frac{25}{968} = \frac{101}{264} \approx .382\bar{57} \end{split}$$

Through similar analysis for $M$, we obtain

$$\begin{split} a &> 7/11, b>\frac{3a-1}{2}\\ a &\leq 7/11, b>\frac{2-a}{3}\\ \end{split}$$

Range for Strategy M

Hence,

$$\begin{split} M &= 1 - \int_{0}^{7/11}\frac{2-a}{3}da - \int_{7/11}^{1}\frac{3a-1}{2}da\\ &= 1 - \frac{259}{726} - \frac{32}{121} = \frac{25}{66} \approx .3\bar{78} \end{split}$$

Finally,

$$T=1-\frac{25}{66}-\frac{101}{264} = \frac{21}{88} \approx .238\bar{63}$$

Since $$\begin{split} \frac{101}{264} &> \frac{25}{66}\\ \frac{101}{264} &> \frac{21}{88} \end{split}$$ the result is shown.

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