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For every positive $y\lt\theta$, the event $[Y\lt y]$ is the intersection of $n$ independent events $[X_k\lt y]$, each with probability $y/\theta$, hence $P_\theta(Y\lt y)=(y/\theta)^n$ and $Y$ has density $$f_Y(y;\theta)=ny^{n-1}\theta^{-n}\mathbf 1_{0\lt y\lt\theta}.$$ Now, assume that $g$ is a measurable function such that $E_\theta(g(Y))=0$ for every ...

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You have $P(Y>x) = (P(X>x))^n$. Now differentiate with respect to $x$ gives: $$f_Y(x,\theta) = -\frac d{dx} P(Y>x) = -\frac d{dx}(P(X>x))^n = n(P(X>x))^{n-1} f(x;\theta)$$

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We are counting the number of pairs $(A, t)$ where $A = (A_1, \dots, A_N)$ is a sequence of $N$ numbers in the range $[-M, M]$ and $t$ is a subset of $\{1, \dots, N\}$ of size $3$ such that $\sum_{i\in t} A_i = 0$. We first pick a triple (this can be done in ${N \choose 3}$ ways), then pick three values from $[-M, M]$ that sum to $0$ (this can be done in ...

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As in your calculation, let $X_i=1$ if Player $i$ is a recordist, and $0$ otherwise. Since we are dealing with a continuous distribution, the probability of a tie is $0$. Since all permutations are equally likely, the probability $i$ is a recordist is $\frac{1}{i}$. Thus the expected number is ...

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If I understand your question correctly then the answer is $\frac {1764} {720} = \frac {49} {20}$. To each person assign a ranking. The person who draws lowest has ranking 1, the person who draws the largest number has ranking 6. List the rankings of a drawing in order as an array, $S$, by the order of draw; for example, one $S$ could be $\{3, 4, 1, 6, 5, ... 0 Just to verify your answer (not to disprove it) using your definitions and notations we find that: $$P(X_1=1)=1$$ that is player nr.1 is for sure a recordist. Then for player nr2. we have that $$P(X_2=1)=P(N_2>N_1)$$ where with$N_i$I denote the number drawn by the player$i$. For all$i$we have according to the definition that$N_i \sim U[0,1]$. So, by ... 3 $$E(U^\alpha)=\int_a^bu^\alpha\,\frac{\mathrm du}{b-a}=\frac1{\alpha+1}\,\frac{b^{\alpha+1}-a^{\alpha+1}}{b-a}$$ 1 We are given that$X_1, X_2, X_3 \sim U[0,1]$Hint: Show$X_1 + X_2 \sim G$, where the probability distribute function is$g(x) = \begin{cases} x & 0\leq x\leq 1 \\ 2-x & 1 < x \leq 2 \\ 0 & \text{otherwise} \end{cases}$Hint: Evaluate the cumulative distribution function$G(x) = \int^x g(y) \, dy$. Hint: Hence,$P(X_1 + X_2 \leq X_3) = ...

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Re "concepts": every $X_i$ has density $f=\mathbf 1_{(0,1)}$ hence, by independence, $$P(X_1+X_2\leqslant X_3)=\iint\!\!\!\int f(x_1)f(x_2)f(x_3)\mathbf 1_{x_1+x_2\leqslant x_3}\mathrm dx_1\mathrm dx_2\mathrm dx_3.$$ Re computations: a shortcut is to note that, for every $x$ in $(0,1)$, $$P(X_1+X_2\leqslant x)=\iint f(x_1)f(x_2)\mathbf 1_{x_1+x_2\leqslant ... 1 I'd like to add some details to Did's answer, for my own future reference. PART I: SOME RELEVANT DEFINITIONS AND RESULTS FROM THE LITERATURE Definition 1: A separable metric space ([R], Definition 1.41, p. 19) A metric space \left(M,d\right) is separable iff it contains a countable, dense subset. (The empty set is regarded as separable.) Definition 2: ... 2 The stationary distribution of the Markov chain on [0,1] with transition x→x+z (mod 1) where z is standard normal, is uniform hence the empirical frequency of every digit at every place converges to 10%. Likewise the empirical frequency of the digits at places from 123,456 to 123,459 being 2014 converges to 0.01%. 4$$P(B(U)=0)=\int_0^1P(B(u)=0)\,\mathrm du=0$$0 We need to assume independence. Draw the rectangle with corners (0,0), (4,0), (4,7), and (0,7). Draw the line y=x. We want to find the probability that the pair (X,Y) lands below the line y=x. So we want to find the probability (X,Y) lands in a certain triangle. That probability is the area of the triangle, divided by the area of the whole ... 2 We have$$F_Y(y)=\Pr(Y\le y)=\Pr(F_X^{-1}(U)\le y).$$Because F_X is strictly increasing,$$\Pr(F_X^{-1}(U)\le y)=\Pr(U \le F_X(y))=F_X(y).

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