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1

Many links at https://marginalrevolution.com/?s=pandemic+model (MR is Tyler Cohen's blog. There isn't much else to say, but my reply needs to be at least 30 characters long.)


0

Each of the aspects mentioned by Kyle points to a useful definition of "liquidity." Tightness of the market refers to the bid-ask spread, the price difference between the highest bid price across all venues and the lowest offer price across all venues. Theory by Black (1971) suggests that the size at the inside bid and offer will, however, be small....


2

There is no apriori reason for the Best Response map to be a contraction in general. Here's a simple example (since Battle of Sexes has been my go-to for the past few days): $$ \begin{array}{c|lcr} \text{Player1/Player 2$\rightarrow$} & \text{F} & \text{T} \\ \hline \text{F} & 3,1 & 0,0 \\ \text{T} & 0,0 & 1,3 \\ \end{array} $$ Denote ...


1

To prove that that some given combination of strategies is a Nash equilibrium you don't need to use a fixed-point theorem (such as Brouwer's or the fixed-point theorem for contractions on Banach spaces). What you do have to do, is check that they are best responses to one another. This true for mixed and pure strategies. You also seem to be asking how you ...


3

In the second period, the buyer accepts any offer $s_{2}$ $\leq$ $0.7$ if he has rejected the first offer and any offer $s_{2}$ $\leq$ $0.3$ if he has accepted the first offer. Given this, there are only two offers that may be optimal for the second seller: either $s_{2}$ $=$ $0.3$ or $s_{2}$ $=$ $0.7$. Let $\mu$ denote the probability that the second seller ...


2

Unless I misunderstand the question these seem to be complementing events with probabilities $p$ and $1-p$.


0

Following the answer of user28714, I tried the following. First, substituting for the FOC, I rewrite $V_2$ as \begin{align*} rV_2 &= f(k_t)-i_t - \tau k_t + i_t-\delta k_t \\ &= f(k_t - \tau k_t - \delta k_t \end{align*} Thus, we get $$ V_2 = \frac{1}{r}\left(f(k_t) - k_t(\tau + \delta) \right)$$ Substituting in $V_1$, we get $$ rV_1 = \max_{i} \left\...


2

I would leave this as a comment but I cant. You are on the right track. Once you know $V_2(k)$ then you can plug that into to the first hjb and solve. To solve for $V_2$ you need to find the optimal $i$ as a function of $k$. Then plug $i(k)$ into the 2nd HJB. That will give you a second order ode. Solving that will give you $V_2(k)$ and you go to 1.


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