# Tag Info

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(...

7

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 ...

6

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

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, $u(x_i) > 0\; \forall i$....

5

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 ...

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.

5

Why don't you just take a weighted average? Suppose you have ten years $t \in \{1,...,10\}$ and year $t$ has $N_t$ observations such that in total you have $\sum_t N_t=N$ observations. Let the year-$t$ CDF be $F_t$ with support $[\underline w_t,\overline w_t]$. You can then define a weighted average CDF as $$\overline F (w) = \sum_t \frac{N_t}{N} F_t(w).$$ ...

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

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 ...

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

(Looking at the question and notation used more closely, the formulation seems to be problematic in couple places.) General Fact Let $W$ be standard Brownian motion with respect to filtration $( \mathscr F_t )_{t \in [0,T]}$. Consider $(L_t)_{t \in [0,T]}$ defined by $$\frac{dL_t}{L_t} = \psi_t dL_t, \; L_0 = 1.$$ In general, $L_t = e^{\int_0^t \psi_s ... 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 ... 4 The answer by @Baysiean proposed to compute a weighted average of the per-period empirical distribution functions$EDF_t(w)$(where$w$is the value in the support of a random variable$W$), a value at which we evaluate the$EDF_t$of$W$. Let's see what that may mean. The$EDF_t(w)$expression is, for each value$w$in the support, $$EDF_t(w) = \frac 1{N_t} ... 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 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 nth 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 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 I'm sorry, this is probably better a comment than an answer, but I don't have sufficient points: In the diagram you've included, Type I and Type II errors are more properly conditional probabilities. \alpha = Prob( Reject H_0 | H_0 ) (= probability of saying not not-pregnant, conditional on actually not-pregnant)) \beta = Prob( Fail to reject H_0... 3 If \alpha \sim U, then how come there is no expectation in your profit function? The \alpha is unknown and which \alpha-types the firm gets depends on salary v. This should be reflected in the profit function. Next, your n(v) seems to assume that \alpha \sim U[0,1], but you set \alpha \sim U[0,2]. I assume this is a typo and I edited your ... 3 Hint: Simply apply integration by parts to the integral on the LHS. Simplify and you should arrive at the following expression: $$(1-R)-\int_{R-k}^1G(\theta)\mathrm d\theta.$$ Add and subtract k to obtain: (1-R+k-k)-\int_{R-k}^1G(\theta)\mathrm d\theta = (1-(R-k))-k-\int_{R-k}^1G(\theta)\mathrm d\theta. \end{... 2 Here's another way, using the same notation as Adam, to get to the same result:$$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 ...

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