5

As pointed in the comments this was done by Ragnar Frisch. At least Barten and Böhm. (1982) as well as Johansen (1969) attribute these axioms to one of these two publications: Frisch, Ragnar (1926). "Sur un problème d'économie pure [On a problem in pure economics]". Norsk Matematisk Forenings Skrifter, Oslo. 1 (16): 1–40 Frisch,(1926). "...


4

Decision under uncertainty is sometimes called a "game against chance", and can thus be modeled as a two-player normal form game: the decision-maker vs Nature/Chance. The possible states would form a set of pure strategies for Nature, and Nature commits to a publicly known mixed strategy that randomizes over those pure strategies (assume Nature is ...


3

As mentioned in the comments this comes down to stylistic choices, since as you correctly pointed out: $$u(c_t)+\beta E_t[u(c_{t+1})]=E_t[u(c_t)+\beta u(c_{t+1})]$$ However, in principle both expressions are correct. The first expression states that the $U_t(c_t, c_{t+1})$ is a composite function of present utility of consumption $u_t(c_t)$ and expected ...


3

The terminology in microeconomics is not completely unified but typically differs slightly from the mathematical one. For a real-valued random (outcome) variable $X$, the mathematical expected value $\mathbb{E}(X)$ would rather be called the expectation of $X$. The utility $u(x)$ of an outcome $x$ is understood to be given by a (Bernoulli) utility function ...


2

Yes. In general not. Let's say the individual has initial wealth $W$ and the gamble $g$ has payouts $0$ and $G$, each with probability $1/2$. As you say, the certainty equivalent $C$ of the gamble is the amount $C$ with $$u(W+C)=(u(W)+u(W+G))/2.$$ Now the same individual would be willing to pay at most $P$ to enter the gamble, where $$u(W)=(u(W-P)+u(W+G-P))/...


2

This is trivially not true. Consider simple example of utility: $$u(x) = x^{1/2}$$ Expected utility $E(u(x)) = E[x^{1/2}]$ Inverse utility is $u^{-1} \implies x = u^2 $ clearly generally $E(u) \neq u^{-1} $.


1

You have the right idea, make a 4 by 3 payoff matrix with fruit at the top and at the side weather/water yes/no, giving 12 different cells, then put the respective dollar payoffs. You could also make another table and find the expected payoff by multiplying each payoff by its probability, although since all the probabilities are the same, the table wouldn’t ...


1

Using $\mathbb E$ for the expected value symbol, $E_R$ and $V_R$ for the mean and the avriance of returns $R$, for a utility function of the form $$U(R) = \ln(1+R)$$ the second-order Taylor expansion gives (ignoring the remainder) $$U(R) \approx \ln(1+E_R) + \frac{1}{1+E_R} (R- E_R) -\frac{1}{2(1+E_R)^2}(R-E_R)^2.$$ Then $$\mathbb E[U(R)] \approx E_R + 0 - \...


1

I realized my mistake whilst writing this, so I will just share the answer as well. Essentially my mistake was, that I assumed this aspect of the definition ($E[z|x_g] = 0\ \forall\ x_g$) to mean that the expected value of z has to be 0 given, that we are in the lottery that is given by the random variable $x_g$. This reading was however wrong. It actually ...


1

AA approach appears to follow two stages: first, "nature" provides which event obtains, which results in the given lottery; second, the lottery is resolved, therefore revealing the intrinsic probabilities. AA suggests to resolve by backward induction. Under usual conditions for vNM utility, agent may define a function for her utility. I found a ...


1

"Is this reasoning correct" No. The reason why this reasoning is incorrect is that the expected value here depends on the probabilities that envelope 2 contains $2x$ or $x/2$. You have assumed that this is 50/50, but this cannot possibly be the case for all $x$. To see this, suppose that you open envelope 1 and find \$10. You assume that envelope ...


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