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If $X\sim\mathcal{N}(6,4)$, then using the properties of the normal distribution, we have $\frac{X-6}{\sqrt{4}}\sim\mathcal{N}(0,1)$. Thus $P(X>a)=0.4\iff P\left(\frac{X-6}{2}>\frac{a-6}{2}\right)=0.4\iff P\left(\frac{X-6}{2}\leq \frac{a-6}{2}\right)=0.6$ As $\frac{X-6}{\sqrt{4}}\sim\mathcal{N}(0,1)$, this is equivalent to ...

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There are two ways of looking into this problem. For this we need to define what uniform distribution means. I will alter and observe the interval $[0,1]$ and adopt $x = [x]+\{x\}$ for clarity. If we take that uniform distribution is corresponding to uniform distribution of random numbers then $[\pi^k]$ does not correspond to uniform random distribution. ...

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Maybe this would get better answers at stats.stackexchange.com than here? Maybe this isn't extremely "real-world", but it's a place where the Cauchy distribution appears where people were not thinking about Cauchy distributions: A Brownian motion in the $(x,y)$-plane starts at $(0,1)$. The $x$-coordinate of the point at which it first reaches the $x$-axis ...

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We have that \begin{align} \operatorname E\mathrm{sgn}(X-Y) & =-1\cdot\Pr(X<Y)+0\cdot\Pr(X<Y)+1\cdot\Pr(X>Y) \\ & =\Pr(X>Y)-\Pr(X<Y) \end{align} and \begin{align} \operatorname E\mathrm{sgn}^2(X-Y) & =(-1)^2\cdot\Pr(X<Y)+0^2\cdot\Pr(X<Y)+1^2\cdot\Pr(X>Y) \\ & =\Pr(X>Y)+\Pr(X<Y) \end{align} using the law ...

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The answer above is called the probability integral transform. Aside from using that, we can directly compute the expectation with integration by parts. First, note that the pdf is the derivative of the cdf. \begin{align*} \mathbb{E}\left[\Phi\left(X\right)\right] &= \int_{-\infty}^{\infty} \Phi\left(x\right)\phi\left(x\right)\, dx\\ &= ... 3 Observe that an element k of a permutation of S_n can participate in zero, one, two etc. up to k-1 inversions. Hence we obtain the following generating function of permutations of S_n classified according to inversions:G(z) = 1\times (1+z)\times (1+z+z^2)\times\cdots\times (1+z+z^2+\cdots+z^{n-1}).$$This is$$G(z) = ...

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Assuming you want to estimate $\sigma$, you need $c=c_n$ to satisfy $\sigma=E(T)$ in order for $T$ to be unbiased. Using rules of expectation and variance, you should get expectation $$E(T)=E(c\sum|X_i|) = cnE|X_1|,$$ since the $X$'s are identically distributed, and variance $$V(T)=V(c\sum|X_i|)=c^2V(\sum|X_i|) \stackrel{(1)}= c^2\sum V(|X_i|) ... 0$$F_X(n)=P_x(X\le n)=\sum_{n_i\le n} 2^{-n_i}= 1-2^{-n}1-2^{-n} \geq t\\ 1-t \geq2^{-n} \\ \log_2(1-t) \geq -n \\ n \geq - \log_2(1-t) = \log_2\left( \frac{1}{1-t} \right)  F_X^{-1}(t) = \left\lceil \log_2\left( \frac{1}{1-t} \right) \right\rceil $$where \lceil x \rceil is the least integer greater than or equal to x function. 2 I will illustrate the process of determining what you call the "luck percentile" by a specific case: Sergeant Joe fires ten shots, with success probability of 3/4 on each shot. He hits nine of the ten targets. Well, if you fire ten shots with probability 3/4 of success on each, the number of hits will be distributed according to the binomial ... 0 You measure the ratio r$$ r = \frac{\text{hits}}{\text{shots}} \to p = \frac{P}{100} $$which approaches the probability p, expressed as percentage P. 3 Hint: Show that$$\mathcal{D} := \{B \in \mathcal{B}(\mathbb{R}); X^{-1}(B) \in \mathcal{F}\}$$is a Dynkin system. Conclude from the fact that$$\mathcal{G} := \{(a,b]; a<b\}$$is contained in \mathcal{D} and that \mathcal{G} is a \cap-stable generator of \mathcal{B}(\mathbb{R}) that$$\mathcal{D} = \sigma(\mathcal{G}) = ...

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This problem relates to empirical distribution theory. Here, a complete treatment of the subject: EMPIRICAL PROCESSES: Theory and Applications https://www.stat.washington.edu/jaw/RESEARCH/TALKS/Delft/emp-proc-delft-big.pdf

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This is just the definitions. Distribution $\mu$ is absolute continuous wrt distribution $\nu$ means for any (measurable) set $A$, $\nu(A)=0$ implies $\mu(A)=0$. Does that hold for your examples? No, $A=[3,5]$ has measure zero under $Unif[0,3]$ but 1/2 under $Unif[1,5]$. $\mu$ is singular wrt $\nu$ means for any set $A$, $\mu(A)>0$ implies $\nu(A)=0$. ...

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\begin{align} \operatorname{var}(Z) & = \operatorname{var}(\operatorname{E}(Z\mid N)) + \operatorname{E}(\operatorname{var}(Z\mid N)) & & (\text{This is the law of total variance.}) \\[10pt] & = \operatorname{var}(N\mu) + \operatorname{E}(N\sigma^2) \\[10pt] & = \mu^2 \operatorname{var}(N) + \sigma^2 \operatorname{E}(N) \\[10pt] & = ...

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If $T = X_1 + X_2 + \cdots +X_N$, where $X_i$ are iid and $N$ is independent of $X_i$, then a standard result, usually derived via a conditioning argument, is that $E(T) = E(N)E(X)$ and $V(T) = E(N)V(X) + V(N)[E(X)]^2.$ (Search the Internet for proofs under 'Random sum of random variables'; the UNL page seems complete.) A small adjustment will deal with ...

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The sequence is almost surely non-convergent. Assume the sequence was convergent. By the Hewitt-Savage zero-one law the limiting random variable has to be constant. If the limit isn't zero, we can apply the continuous mapping theorem and we get that $\log(R_n)$ converges almost surely to a constant. This is a contradiction, since ...

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Hint: rewrite $$Z = \sum_{n=1}^\infty X_n\mathbb{1}_{\{N+1 \geq n\}}$$ and apply your "expectation of the sum is the sum of the expectations" idea. Following the comment below, more detail. Write $Y_n = \mathbb{1}_{\{N+1 \geq n\}}$, which is independent of $X_n$. Then for the expectation, you have $$\mathbb{E}[Z] = \sum_{n=1}^\infty ... -1 In my experience: There seem to be two times here. t=0 and t=T. A portfolio is a collection of financial instruments. For instance, I could have a portfolio consisting of 3 stocks and 1 bond. Its value today is the sum of the individual values of the instruments today. V(0) is the value of the portfolio at time 0 (today?) V(T) is the value of the ... 2 I agree that @BCLC is right on saying that I have used risk neutral information The Edited Answer is  V(0) = 15\times90+ 10\times25 = 1600 Now compute V(T)$$V(T) = 1800, \text{ if stock goes up}1800 = 30\times 10 + 100\times 15V(T) = 1700, \text{ if stock goes down}1700 = 20\times 10 + 100\times 15$$V(T) = 15\times A(T) + ... 3 By the Borel-Cantelli lemma, if the series$$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} $$converges for each \varepsilon>0, Y_n\to0 almost surely as n\to\infty. Using Chebyshev's inequality,$$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\operatorname E|Y_n|^2 $$since \operatorname EY_n=0. By ... 2 Hint: use the Borel—Cantelli lemma. In more detail: fix any \varepsilon > 0, and let A_n=A_n(\varepsilon) be the event \{ \lvert Y_n\rvert > \varepsilon \}. By Borel—Cantelli, to show that Y_n \xrightarrow[n\to\infty]{\rm a.s.} 0, it is sufficient to show$$ \sum_{n=1}^\infty \mathbb{P} A_n < \infty. $$In even more detail: (place your ... 0 If f is convex, f(\alpha a + \beta b) \le \alpha f(a) + \beta f(b) where two nonnegative coefficents \alpha \ge 0, and \beta \ge 0$$\alpha + \beta = 1$$Now let \alpha = 0.5*(1 - Y_i/c_i) and \beta = 0.5*(1 + Y_i/c_i) and  a = -kc_i, and b = kc_i and f = e^x$$ e^{kYi} = e^{(1-Y_i/c_i)/2 (-kc_i) + (1+Y_i/c_i)/2 (kc_i)}\le \alpha ...

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Note that $$P(X^n\in C_n(t))=P(p(X^n)\ge 2^{-nt})=P\left(-\frac{1}{n}\sum_{1}^n \lg p(X_i)\le t\right)$$ Now, it follows from WLLN, $$\lim_{n\to \infty}P\left(-\frac{1}{n}\sum_{1}^n \lg p(X_i)\le H(X)+\epsilon\right)=1\ \forall \epsilon>0$$ Thus, to have $\lim_{n\to \infty}P[X^n\in C_n(t)]\to 1$, we must have $t\ge \epsilon+H(X)\ \forall ... 2 I think the problem is that the number of attempts that can be used in a numerical simulation$n$is finite. Notice this: if$Y_n=\frac{S_n}{\sqrt{2n\log\log n}}$, by properties of random walk we know$\mathbb{E}[Y_n]=\frac{\mathbb{E}[S_n]}{\sqrt{2n\log\log n}}=0$and$$Var[Y_n]=\frac{Var[S_n]}{2n\log\log n}=\frac{n}{2n\log\log n}=\frac{1}{2\log\log n}\to ... 2 Think of 75% as the probability that the stock goes up, i.e.$0.75$. Then, the trader gains 100 with probability$0.75$, loses 200 with probability$1-0.75=0.25$. On expectation, what is his gain? More formally: let$X$be the random variable representing his gain. Then,$\mathbb{P}\{X=100\} = 0.75$, and$\mathbb{P}\{X=-100\} = 0.25\$. You are asked to ...

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