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9

The name for the amount $56.25 is certainty equivalent. The expected utility for the individual from taking the bet is calculated as follows: $$E[U]=\frac12U(100+125)+\frac12U(100-100)=75$$ Suppose the individual can pay an amount of money$x$so that she can avoid taking the bet (which leads to expected utility$75$). What's the maximum amount of money$x$... 7 The first order stochastic dominance relation is convex. An easy way to prove this is to use the property that a cdf$F$FOSD another cdf$G$if and only if$F(x)\le G(x)$for all$x$. That is,$F$FOSD$G$if and only if the graph of$F$is never above the graph of$G$. It is then easy to show that$F$is never above any convex combination$H(x)=\alpha F(...

5

Let $u(x) = x$ which is increasing and concave. Then the defining condition of SOSD reads $$\int x\mathrm dF(x)\ge \int x\mathrm dG(x) \implies E_F(X) \geq E_G(X) \tag{1}$$ ..which would contradict the case $E_F(X) < E_G(X)$ that would be permissible under a general "different means" postulate. On the other hand, we see that the "same mean" condition ...

5

To show that transformation is measure preserving you need to show that full preimage $S^{-1}(A)$ of any set A in the Borel $\sigma$-field on [0,1) is again in the same $\sigma$-field, i.e. it is measurable, and that $Pr\{S^{-1}(A)\} = Pr\{A\}.$ As you have correctly pointed out, it will suffice to show it for any base of usual topology, namely, all open ...

5

That's just a question of notation. Note that $$\mathcal{d}F(t) = f(t)\mathcal{d}t$$ Then the proof should be easy to understand.

4

This construction you describe is not fully general. In fact it characterizes strictly stationary time series. You see that it's shift-invariant. This operator $S$ is essentially a shift operator. For comparison, here's the usual definition of, let's say discrete-time, processes: Definition A stochastic process is a sequence $\{ X_t \}$ of Borel measurable ...

4

GDP is a flow (of goods and services) while market capitalization is a stock measure. So I agree, this is not a great comparison. If we want to compare stocks, we'd compare total enterprise value (sum of equity and debt) to national domestic assets. But that's imperfect because much of what makes a country rich is labor income and that's not going to be in ...

4

The first / second seller terminology can be confusing here (does it relate to time or to conditionality?). It's safer to focus on a particular seller. Let the probability that a buyer finds a particular seller, $s$, via single search be $\nu_s$. The probability that a buyer is matched with $s$ and (either before or after) another seller is then (assuming ...

4

The simplest thing is to calculate the expected cost in the two scenarios: Scenario 1: Develop both at the same time $$E[cost]= X + Y$$ Scenario 2: Wait and see if task two is needed $$E[cost]= X + 0.7Z$$ However I would not say expected value calculation represents the "economics view". The general view rather depends on your preferences, often expressed ...

3

I believe there is not a unique price. Say, instead of buying the option you spent 0.5 on a half a unit of the asset $S^2_1$ This asset pays out $[0.4, 0.6, 0.8]$ which first order stochastically dominates the option. So, no matter your probability beliefs about the states, in that setting you'd never pay $0.5$ for the option which pays less in every state....

3

There is a typo in the figure that introduces some confusion in the previous answer, which is basically wrong. Based on the numbers and the figure, the utility is such that $$u=\sqrt{x},$$ so $$E[u]=\frac{1}{2} u(100+125) + \frac{1}{2} u(100−100)= \frac{1}{2} u(225) =\frac{1}{2} \sqrt{225} = 7.5$$. By definition, the risk premium (R) must satisfy the ...

3

I assume that $r(t)$ is continuous. The idea is that, this Poisson process with time-varying parameter $r(t)$, as the limit of Bernoulli trial with time-varying probability of success, is memoryless: For a generic finite partition $\mathscr{P}$ of $[0, T]$ as $\{[0,t_1),[t_1,t_2),[t_2,t_3) \ldots,[t_{n-1},T]\}$, let the $n$th cell following a Poisson ...

3

Well, you cannot take the derivative of $E(n)$ with respect to $n$, because $n$ is an integer variable. More generally, you want to prove a property with respect to $n$. The problem you have is that the corresponding domain is not a convex set: say, for $n$ and $n+1$, $0<\lambda <1$, the value $\lambda n + (1-\lambda) (n+1) = n+1-\lambda$ does not ...

3

After the exchange with the OP in my other answer, let's work a bit with his approach. We have a discrete random variable $X$ with finite support, $X = \{x_1,...,x_k\}$, and probability mass function (PMF), $\Pr(X=x_i)=p_i, i=1,...,k$ The values in the support of $X$ are also inputs in a real-valued cardinal utility function, $u(x_i) > 0\; \forall i$....

3

Before discussion Shannon's entropy, there is another point that should be discussed: it appears that you have in mind cardinal utility rather than ordinal. "Normalized" utility functions can be derived of course in both cases. But the concept of "relative preference" can be defined and measured only in the context of cardinal utility. And the issue ...

3

The way hierarchies of beliefs are specified, beliefs on the same events are encoded in different places. The basic idea is actually quite simple. You have two players, Ann and Bob, say. Ann's first order beliefs specify how likely she thinks each strategy choice of Bob. Ann's second order beliefs specify how likely she thinks each combination of a ...

3

Presumably here the null hypothesis is $H_0:$ You are not pregnant the alternative hypothesis is $H_1:$ You are pregnant so being pregnant would be the positive result. You take a pregnancy test if the pregnancy test gives a positive result when you are not pregnant then this is a false positive, a Type I error when the null hypothesis $H_0$ is in fact ...

2

Here's another way, using the same notation as Adam, to get to the same result: $$P(\text{another match | being matched}) = P(\text{searching twice | being matched})\cdot P(\text{another match | searching twice})$$ Now, $$P(\text{searching twice | being matched}) = \frac{P(\text{searching twice and being matched})}{P(\text{being matched})} \\ = \frac{\... 2 The augmentation of the filtration generated by B is typically just called the augmented filtration and is defined using the below construction. It is more commonly referred to simply as the standard Brownian filtration. The collection \mathcal{C} of all sets of probability 0 in the sigma-field \sigma\{B_{s} \: : \: s \leq t\} The collection of ... 2 (If urns are vacancies and balls are unemployed, what distinction between unemployed workers does the Red/Green dichotomy reflects?) Each ball has in front of it an identical box, each with the exact same lottery tickets, its ticket has a number on it, and each number corresponds to an urn. We say "Go!" and each ball draws "randomly" (i.e. with equal ... 2 Regardless of period if we know the cost 100%, it is simply cost. If there is a degree of uncertainty on what the cost will be (such as in the future) we regard it as expected cost. Regarding your question specifically about how the cost of a project can change in a future period, this falls under an area of study known as cost-benefit analysis. Your ... 2 (1) What value an item has to you depends on your preferences, not scarcity. (2) The market clearing price of an item will depend on the preferences of market participants and the supply schedule ("scarcity") of the item. The way you seem to be thinking about value is very "non-economic", if that's a word. 2 The second question of the OP is an important one, because it asks for clarification on a matter that I have never seen explicitly clarified. The Independence Axiom is defined over simple lotteries. A simple lottery is a set of probabilities that add up to one, and the fixed values ("outcomes") associated with each probability. But these together form a (... 2 First Question: Yes, you can obtain P \succsim P' \Rightarrow P \succsim a P' + (1-a)P \forall a\in (0,1) from setting P''=P in the axiom. Thus, your new condition is a special case of the independence axiom as stated. Second Question: Statistical independence of P,P',P'' is not assumed in the independence axiom. The notion of statistical ... 2 I heard in a lecture at my university, that measure theory is applied in finance. This field also operates with negative probabilities a lot. Anyways, I don't see the practical intuition behind using negative probabilities in a bond-pricing model. You can check this article by Burgin and Meissner: Negative Probabilities in Financial Modeling (Wilmott ... 2 As you say (how did \sin t in the initial problem become \cos t?), \mathbb{Q} is a measure under which W_t becomes \tilde{W}_t - \sin t, where \tilde{W}_t is a \mathbb{Q}-Brownian motion. So Girsanov's theorem would say$$ L_t = E[\frac{ d \mathbb{Q} }{ d \mathbb{P} }|\mathcal{F}_t] = e^{ - \int_0^t \sin s dW_s - \frac{1}{2} \int_0^t \sin^2 s ...

2

This problem is similar to first-price (sealed-bid) auctions with independent private values for the buyers. You can read about those for exaxmple here. Your first question was already ansewered in the comments: given others' efforts the probability of getting the reward is or 1. Let us assume that $e_1 \geq e_2 \geq \dots$ are the other players' efforts ...

2

Your first question was addressed in another answer. For the second question (why $\Phi(b) \ne 1$), you are confusing the cdf and the density. It is true that $\int_{a}^{b}{f(t)dt}=1$ but not that $\int_{a}^{b}{F(t)dt}=1$. Think for instance of $[a,b]=[0,1]$ and the uniform distribution ($f(t)=1,F(t)=t$ for all $t \in [0,1]$).

2

The set $Q^*$ is the set of probability distributions over the maximizers of $u$ for each value of $v$. So for a fixed $v$ all the values of $Y$ that have positive probability must give the same utility. That is, if $P_Y(y|v)>0$ and $P_Y(y'|v)>0$ then $u(y,v)=u(y',v)\geq u(\tilde y,v)$ for all $\tilde y\neq y, y'$. Therefore, convex combinations of ...

1

Strictly speaking, infinitely different utility functions can give the same preferences over lotteries. If person A has utility function $f(x)$ then person B with utility function $g(x)=A+B\cdot f(x)$, $B>0$ will make identical decisions over all lotteries. There would be no way to know which individual was actually using utility function $f(x)$ and which ...

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