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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 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} ...


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 ...


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 ...


2

So we have $S_n \thicksim^{iid} \ ?$ $\bar{s} = \frac{1}{n}\sum{s_n}$ Exponential Suppose $S_n$ follows the exponential distribution. $$f(s|\beta) = \frac{1}{\beta} e^{-\frac{s}{\beta}} \quad , \quad 0 \leq s < \infty \quad , \quad \beta > 0$$ Take the simple bivariate case. Say $Z = \frac{S_1 + S_2}{2}$ and $W = S_1$. So $S_2 = 2Z - W$ and $S_1 ...


2

The best introduction to measure-theoretic probability for economics is probably: Chapter 7 Measure Theory and Integration, Recursive Methods in Economic Dynamics by Stokey and Lucas. Its presentation is along mathematical lines but with judicious (and numerous) omissions for economics audience. For example, a measure space is defined but no real concrete ...


2

If you want a good book with emphasis on rigorous then An Introduction to Probability: Theory and Application, by William Feller is good source. The book starts completely from the first principles and covers also a lot of applications in statistics. Arguably the book is more suited to graduate as it has a steep curve - the content difficulty increases ...


1

The game has positive expected value, so you should play (assuming constant utility of each dollar, which tends to break down somewhat for very large sums of money). If you start with \$X, after 1 round you have a 50% chance of having \$1.5X, and a 50% chance of having \$0.6X, for a total expected value of \$1.05X. That is, for every dollar you bet, on ...


1

In relation to the problem of heteroskedasticity in linear probability models, the following Ben Lambert video is a useful link: youtube.com/watch?v=pgPhbVEbYqw


1

I cannot really follow your formulas, what is the logic behind them? Seems to me there is no way to divine three state risk-neutral probabilities from 1 financial instrument's prices. The equation $$ p_1 288 + p_2 180 + (1 - p_1 - p_2) 120 = 180 $$ is underdetermined, there are infinitely many solutions to it. Now if you solved the equation system $$ \...


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