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The proof to follow is taken from a research paper in risk analysis which is in reviewing. I have been tasked with repeating the results in the paper, under a different risk measure, but I am not sure if the main argument in the paper is correct as it is presented, although I believe the main result should hold.
The context is the following. We are optimizing a risk function over a set of parameters $(a_i{)}_{i=1}^m,(b_i{)}_{i=1}^m$, with particular choices of $(a_i{)}_{i=1}^m,(b_i{)}_{i=1}^m$ denoted as $a^{\ast <!--\; \ast \; -->},b^{\ast <!--\; \ast \; -->}$ respectively. The aim is to establish a criterion for optimality of the parameters.
The argument in the paper is as follows:
Begin by assuming that $a^{\ast <!--\; \ast \; -->}$ is chosen such that $(1)$ holds, where $(1)$ is a condition on $a^{\ast <!--\; \ast \; -->}$. It is then possible to show that there exists some $b^{\ast <!--\; \ast \; -->}$ such that a condition $(2)$ on $b^{\ast <!--\; \ast \; -->}$ holds. We then show that any $b^{\ast <!--\; \ast \; -->}$ satisfying $(2)$ is optimal given that $a^{\ast <!--\; \ast \; -->}$ is chosen according to $(1)$. Assume next that $b^{\ast <!--\; \ast \; -->}$ is chosen such that $(2)$ holds. Then, one shows that $a^{\ast <!--\; \ast \; -->}$ can only be optimal parameters if $(1)$ holds. The conclusion is that choosing $b^{\ast <!--\; \ast \; -->}$ such that $(2)$ holds is sufficient for obtaining optimal $(a^{\ast <!--\; \ast \; -->},b^{\ast <!--\; \ast \; -->})$.
There are two things that make me uneasy about this argument. Firstly, as we begin by assuming a condition on $a^{\ast <!--\; \ast \; -->}$, my instinct is to look next at what happens if $a^{\ast <!--\; \ast \; -->}$ does not satisfy $(1)$.
Secondly, it is easy to find $b^{\ast <!--\; \ast \; -->}$ such that $(2)$ does not hold. While existence of $b^{\ast <!--\; \ast \; -->}$ satisfying $(2)$ given $a^{\ast <!--\; \ast \; -->}$ such that $(1)$ holds, it seems to me like we assume existance in the second part of the proof. Then, we seemingly use the assumed existence of $b^{\ast <!--\; \ast \; -->}$ such that $(2)$ to show that we must choose $a^{\ast <!--\; \ast \; -->}$ such that $(1)$, and hence $b^{\ast <!--\; \ast \; -->}$ will exists such that $(2)$. Is this not a circular argument?

Hence, $v(6,000)<v(4,000)+v(2,000)$ and $v(-<!--\; -\; -->6,000)>v(-<!--\; -\; -->4,000)+v(-<!--\; -\; -->2,000)$. These preferences are in accord with the hypothesis that the value function is concave for gains and convex for losses.
What this means and by how we can know if these align with convex and concave functions?

What are the outputs for a business risk analysis?

In risk analysis, several types of uncertainties are generally analyzed. However, what is the fundamental difference between risk analysis and sensitivity analysis to address uncertainties?
a) Risk analysis shows correlation of an input's value of a project with the value of its output while sensitivity analysis provides a functional form of that relationship
b) The use of probabilities to quantify uncertainties in risk analysis and absence of it in sensitivity analysis
c) The use of probabilities to quantify uncertainties in sensitivity analysis and absence of it in risk analysis
d) While sensitivity analysis is associated with one-by-one change in inputs' values, risk analysis is associated with the change of all inputs as a set
e) Sensitivity analysis is based on the concept of expected value while risk analysis is associated with the concept of variance

statement of the proof was that $S$ was closed from below, and bounded from below. It is fine when they say that $B$ is bounded below which is also obvious from the definition. So, there should be a g.l.b. for the set $B$, which is denoted by $b_0$.
Now why there should exist a sequence of points $\left({\mathbf{y}}^{\mathbf{(}\mathbf{n}\mathbf{)}}\right)$ in the risk set $S$ such that $\sum <!--\; \sum \; -->p_j{y}_j\to <!--\; \to \; -->b_0$? Is $b_0$ a limit point of $B$? Is $B$ closed ? Even if this happen why $b_0$ which is greatest lower bound of $B$ should belong to $B$?
Next, I guess they apply Bolzano Weierstrass theorem to say that ${\mathbf{y}}^{\mathbf{0}}$ is a limit point of the sequence $\left({\mathbf{y}}^{\mathbf{(}\mathbf{n}\mathbf{)}}\right)$. But why the last step $\sum <!--\; \sum \; -->p_j{y}_j^0=b_0$?

Organizations typically use both quantitative and qualitative risk analysis techniques when analyzing the risk to the organization and determining the appropriate counter-measures. Compare and contrast quantitative and qualitative risk analysis. Describe a situation when a qualitative risk analysis method is most appropriate, and describe a situation when a quantitative risk analysis method is most appropriate.

The following risks are given:
1. $X_3$
2. $X_2\sim <!--\; \sim \; -->Bin(15,\frac{1}{3})$
3. $X_3\sim <!--\; \sim \; -->15I,\text{where}I\sim <!--\; \sim \; -->Bernoulli(\frac{1}{3}$
Do any docaiou makers with (increasing) ulility function agree aboul prceferring risk$X_1$ to $X_2$? given an exponential utility function which risk does a decision makerpreder: $X_1$ or $X_3$?
The last task is related to eponential ordering of risk denoted by $\le <!--\; \le \; -->e$.

What is a 'critical value' in statistics?
The raw material needed for the manufacture of medicine has to be at least $97\mathrm{\%<!--\; \%\; -->}$ pure. A buyer analyzes the nullhypothesis, that the proportion is ${\mathrm{\mu <!--\; \mu \; -->}}_0=97\mathrm{\%<!--\; \%\; -->}$, with the alternative hypothesis that the proportion is higher than $97\mathrm{\%<!--\; \%\; -->}$. He decides to buy the raw material if the nulhypothesis gets rejected with $\mathrm{\alpha <!--\; \alpha \; -->}=0.05$. So if the calculated critical value is equal to $t_{\mathrm{\alpha <!--\; \alpha \; -->}}=98\mathrm{\%<!--\; \%\; -->}$, he'll only buy if he finds a proportion of $98\mathrm{\%<!--\; \%\; -->}$ or higher with his analysis. The risk that he buys a raw material with a proportion of $97\mathrm{\%<!--\; \%\; -->}$ (nullhypothesis is true) is $100\times <!--\; \times \; -->\mathrm{\alpha <!--\; \alpha \; -->}=5\mathrm{\%<!--\; \%\; -->}$

Which of the following is / are TRUE about risk analysis? Choose all that apply.
$\mathrm{◻<!--\; ◻\; -->}$ Risk analysis involves a detailed consideration of the uncertainties
$\mathrm{◻<!--\; ◻\; -->}$ Purpose of analysis may influence the detail and complexity of risk analysis
$\mathrm{◻<!--\; ◻\; -->}$ Risk analysis involves a detailed consideration of controls
$\mathrm{◻<!--\; ◻\; -->}$ Availability of information may influence the detail and complexity of risk analysis

If you were given a choice to conduct a quantitative risk analysis or a qualitative risk analysis, which option would you choose and why? How would you go about completing the risk analysis?

Which of the following will assess key political refulations susceptible to affect business condcted in a specific country?
$\circ <!--\; \circ \; -->$ Multidimensional political risk analysis
$\circ <!--\; \circ \; -->$ Micro political analysis
$\circ <!--\; \circ \; -->$ Macro political analysis
$\circ <!--\; \circ \; -->$ Standard political analysis

Suppose, for example, that a decision maker can choose any probabilities $p_0$, $p_1$, $p_2$ that he or she wants for specified dollar outcomes
$D_0$
and that they have a given expected value
$p_0{D}_0+p_1{D}_1+p_2{D}_2=k$
For example, if $D_0<0$ were the price of a lottery ticket with possible prizes $D_1$ and $D_2$, then $k=0$ would define a “fair” lottery, while $k<0$ would afford the lottery organizer a profit. We may arbitrarily let the utilities of $D_0$ and $D_2$ be $u_0=0$ and $u_2=1$; then the utility of $D_1$ is $u_1\in (0,1)$. For a typical lottery, |$D_0$| is quite small as compared to $D_1$ and $D_2$. With $k\le 0$, this implies that feasible $p_1$ and $p_2$ are small, with $p_1+p_2$ well under $0.5$, and therefore with $p_0$ well over $0.5$.
Questions:
1. If $D_0$ is the price of a lottery ticket, how could it possibly be less than zero?
2. Why include the price of a lottery ticket in an EV calculation? The prizes $D_1$ and $D_2$ have a probability associated with them, that makes sense when calculating expected value. But the price of a lottery ticket? What does it mean for a ticket price to have a probability "well over $0.5$"
3. For $k\le 0$, it only makes sense that $D_0$ must be negative, but again, how could the price of a lottery ticket be negative?

I passed the real analysis course at my college and now, after 3 years, im not that good anymore at integrating, solving big complicated limits etc. But I still remember the concepts behind them. Now, since I'm trying to self-study some topics that I'm not familiar with (topology, galois theory etc), as an aspiring engineer, I was wondering if I shouldn't move to the next topic until I'm very good at doing every exercise (“calculate the integral of” “calculate the limit of” etc) with the risk of one day forgetting almost everything or if i should focus more on understanding the concepts and use a computer for the calculations.

Example of identify risks, perform qualitative risk analysis and perform quantitative risk analysis

Risk analysis is typically a two-step process: qualitative risk analysis and quantitative risk analysis.
As a systems analyst, for which sorts of project management decisions would you use the results from qualitative risk analysis? From the quantitative risk analysis?

What sort of qualitative behaviour does a stock following a process of the form
$d{S}_t=\alpha (\mu -<!--\; -\; -->S_t)dt+S_t\mathrm{\sigma <!--\; \sigma \; -->}d{W}_t$
exhibit? What qualitative effects do altering $\mu$ and $\alpha$ have? What effects do they have on the price of a call option?
My try:
Assume that $\mathrm{\alpha <!--\; \alpha \; -->}<0,$. If $S_t<\mathrm{\mu <!--\; \mu \; -->},$, then $\mathrm{\alpha <!--\; \alpha \; -->}(\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->S_t)>0.$. Similarly, if $S_t>\mathrm{\mu <!--\; \mu \; -->},$, then $\mathrm{\alpha <!--\; \alpha \; -->}(\mathrm{\mu <!--\; \mu \; -->}-<!--\; -\; -->S_t)<0.$. This means that if $\mathrm{\alpha <!--\; \alpha \; -->}<0,$, then $S_t$ is mean-reverting.
Conversely, if $\mathrm{\alpha <!--\; \alpha \; -->}<0,$, similar analysis above implies that $S_t$ is trend following.
In the derivation of Black-Scholes call option price, mean rate of a geometric Brownian motion does not affect the price as it will be changed into risk-free rate.
As the volatility is the same, $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\mu <!--\; \mu \; -->}$ do not have any effect on the price of a call option.
Am I correct?

Which of the following is not a risk management process?
A. Perform Actuarial Risk Analysis.
B. Perform Qualitative Risk Analysis.
C. Perform Quantitative Risk Analysis.
D. Implement Risk Responses.