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Mar
30
comment If $X|Y$ and $Y$ are both normal, is $X|Y>y$ normal as well?
Ah, yes of course. Thank you @Did!
Mar
24
asked If $X|Y$ and $Y$ are both normal, is $X|Y>y$ normal as well?
Mar
24
accepted How do I show that this game played on a Markov chain has a unique Nash equilibrium?
Mar
24
answered How do I show that this game played on a Markov chain has a unique Nash equilibrium?
Feb
14
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.
Feb
14
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.
Feb
14
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.
Feb
14
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 $2\cdot(1+\lambda_2/\lambda)\lambda_1/\lambda$.
Feb
14
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.]
Feb
11
asked How do I show that this game played on a Markov chain has a unique Nash equilibrium?
Apr
11
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?
Apr
10
asked Conditions on r.v. $X$ s.t. $\Pr(X\ge n\mid X\ge n/2)$ gets small?
Apr
8
awarded  Scholar
Apr
8
accepted Proof that $E(X)<\infty$ entails $\lim_{n\to\infty}n\Pr(X\ge n) = 0$?
Apr
8
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?
Apr
8
awarded  Supporter
Apr
8
awarded  Student
Apr
8
asked Proof that $E(X)<\infty$ entails $\lim_{n\to\infty}n\Pr(X\ge n) = 0$?