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 Mar30 comment If $X|Y$ and $Y$ are both normal, is $X|Y>y$ normal as well? Ah, yes of course. Thank you @Did! Mar24 asked If $X|Y$ and $Y$ are both normal, is $X|Y>y$ normal as well? Mar24 accepted How do I show that this game played on a Markov chain has a unique Nash equilibrium? Mar24 answered How do I show that this game played on a Markov chain has a unique Nash equilibrium? Feb14 comment How do I show that this game played on a Markov chain has a unique Nash equilibrium? @Xoff The second part is not necessarily true though. It depends on the value of the $\lambda$'s as I tried to illustrate in my above comments. If you look at the table in the original question and fix player 2's strategy as $H$ you'll see that $E$ is better for player 1 iff $\lambda_1 > 2\lambda_2$, i.e., not always. Feb14 comment How do I show that this game played on a Markov chain has a unique Nash equilibrium? @Xoff Yes, I know that. But I want to show that the Nash equilibrium is unique, so I also need to consider cases where other players are playing $H$ for some stages. Feb14 comment How do I show that this game played on a Markov chain has a unique Nash equilibrium? @Xoff And if $\lambda_1 > \lambda_2$, the latter is bigger. Feb14 comment How do I show that this game played on a Markov chain has a unique Nash equilibrium? That depends on the strategy doesn't it? Suppose player 1 has just finished the first stage. Then her expected utility if she is playing $E$ is $1 + \lambda_1/\lambda$, and her expected utility if she is playing $H$ is $2\cdot(1+\lambda_2/\lambda)\lambda_1/\lambda$. Feb14 comment How do I show that this game played on a Markov chain has a unique Nash equilibrium? @Xoff It is a benefit because the probability that another player will do it first is lower. Consider the case $n = 2$, $k = 2$. If player 1 finishes the first stage first and claims utility, her probability of finishing the second stage first is $\lambda_1/\lambda$. But if she doesn't claim, the other player needs to finish stage 1 first, so player 1's probability of finishing the second stage first is then $(1 + \lambda_2/\lambda)\lambda_1/\lambda$. [The gender of the players is of course irrelevant. I usually use female pronouns.] Feb11 asked How do I show that this game played on a Markov chain has a unique Nash equilibrium? Apr11 comment Conditions on r.v. $X$ s.t. $\Pr(X\ge n\mid X\ge n/2)$ gets small? Thanks @KevinCostello that is very helpful. I see that all of the moments of $X$ are bounded in your example but can you say a little bit more about why the desired conclusion fails? Apr10 asked Conditions on r.v. $X$ s.t. $\Pr(X\ge n\mid X\ge n/2)$ gets small? Apr8 awarded Scholar Apr8 accepted Proof that $E(X)<\infty$ entails $\lim_{n\to\infty}n\Pr(X\ge n) = 0$? Apr8 comment Proof that $E(X)<\infty$ entails $\lim_{n\to\infty}n\Pr(X\ge n) = 0$? Thanks so much @Alex. Just to be clear: you meant to type $2^{k(n)}a_{2^{k(n)}}$ rather than $2^{k(n)}a_{k(n)}$ right? Apr8 awarded Supporter Apr8 awarded Student Apr8 asked Proof that $E(X)<\infty$ entails $\lim_{n\to\infty}n\Pr(X\ge n) = 0$?