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$Y|X=x$ has the same distribution of $f(x,U)$ for some measurable $f$ (take $f(x,.) $ the inverse of the CDF of $Y|X=x$ for an 'explicit' choice). Then $(X,Y), (X,f(X,Y))$ have the same distributions.


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Let $X \sim Uniform(0,1)$ with pdf $f(x)$: Let $X_n$ and $X_{n-1}$ denote the largest and second largest order statistics in a sample of size $n$. Then, the joint pdf of $(X_{n-1},X_n)$, say $g(x_{n-1},x_n)$, is: where I am using the OrderStat function from the mathStatica package for Mathematica to automate, or do it manually following: Wiki -- ...


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The probability that $n-1$ of them are less than $x$ is $nx^{n-1}(1-x)+x^n$, so the PDF of the second-highest is the derivative of that. If the second-highest is $x$, then the expected value of the sum of the top one is $(1+x)/2$, and the sum of the top two is $(1+3x)/2$. Integrate the PDF of the second-highest, multiplied by $(1+3x)/2$.


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Note that \begin{align} Y:\Omega&\longrightarrow \mathbb R\\ \omega&\longmapsto \max\{X_1(\omega),X_2(\omega)\} \end{align} Then $F_Y(y):=\mathbb P(Y\leq y) =\mathbb P(\max\{X_1,X_2\}\leq y) =\mathbb P(X_1\leq y, X_2 \leq y)$. If they are independent then $$F_Y(y)=\mathbb P(X_1\leq y)·\mathbb P(X_2\leq y) = F_{X_1}(y)·F_{X_2}(y).$$


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$P(Y < t) = P (\max(X_1, X_2) < t) = P (X_1 < t ~\wedge X_2 < t) = P(X_1<t)P(X_2<t) $ by independence. Calculate this cumulative distribution function and it should be easy to find the density


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If $Y \in (0, b)$ (for $b \in [0, 1]$), then we know that both $X_1 \in (0, b)$ and $X_2 \in (0, b)$. Then, $$\begin{align*} \mathrm{Pr}(Y \leq b) &= \mathrm{Pr}(X_1 \leq b \cap X_2 \leq b) \\ &= \mathrm{Pr}(X_1 \leq b) \cdot \mathrm{Pr}(X_2 \leq b) \tag{by independence} \\ &= (b)(b) \\ \mathrm{Pr}(Y \leq b) &= b^2. \end{align*}$$ Then we ...


2

By the Jacobian formula, $$f_Y(y)=\frac{\mathbf 1_{|y|\lt1}}{\pi\sqrt{1-y^2}}.$$ This is the so-called Arcsine distribution, which famously appears in probability theory and in number theory, as shown by two mathematical giants from the 20th century: $\qquad\qquad\qquad\qquad$


2

The support of the distribution is of course $[-1,1]$. For $y\in[0,1]$, we have \begin{align} \Pr(Y\le y) & = \Pr(0\le X\le\arcsin y\text{ or }2\pi\ge X\ge \pi-\arcsin y) \\[10pt] & = \frac{\arcsin y + (2\pi-(\pi-\arcsin y))}{2\pi}. \end{align} The density is the derivative of that, and that is not a constant function, so it's not uniformly ...


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This paper has an explanation about the uniformity of infinite coin tosses. Also, there is a detailed explanation given in Santosh Venkatesh's Probability Theory: Explorations and Applications. (starting at p. 144...I linked to the Google preview). I will not reproduce them here, as these links are excellent.


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X is uniformly distributed, $X\in \langle 0,10\rangle \Rightarrow E(x) = \frac{0+10}{2}=5,\,D(x)=\frac{(10-0)^2}{12}=\frac{25}{3}$ Average values ​​at 100 experiments: $\mu_{100}=E\left(\frac{x_1+x_2+\cdots+x_{100}}{100}\right)=E(x)=5$ $D_{100}=D\left(\frac{x_1+x_2+\cdots+x_{100}}{100}\right)=\frac{D(x)}{100}\doteq 0.08333\Rightarrow \sigma \doteq 0.2887$ ...


1

$\forall x_i\in [-1,1]$   When given $X=x_i$ then $Y$ is certainly equal to $x_i^2$, so we have a probability mass with a support of one point: $$\begin{align} \mathsf P[Y=y\mid X=x_i] & = \begin{cases} 1 & : y=x_i^2 \\ 0 & : y\neq x_i^2\end{cases} \\[4ex] \mathsf E_{Y\mid X}(Y\mid x_i) & = \sum_{y=x_i^2} y\, \mathsf P(Y=y\mid X=x_i) ...


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I don't see how you get $E[X^2]=1$. In fact, it is 1/3. Other than that, you are correct: since $\text{cov}(X,X^2)=0$, the best linear predictor is a constant.


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Ad a) The random variable is binomial distributed, $X \sim B(400,0.4)$. Ad b) Now you can approximate the binomial distribution by the normal distribution. You take the mean and the standard deviation of the binomial distributed random variable. $\mu=0.4\cdot 170$ and $ \sigma=\sqrt{ \sigma ^2 } =\sqrt{n\cdot p \cdot (1-p)}=\sqrt{400\cdot 0.4\cdot 0.6}$ ...


2

In this problem, $A$ is to be treated as a constant. The mean of $A+U$ is just $A$, since $U$ is uniform on $[-1/2,1/2]$, and therefore has mean $0$. The variance of $A+U$ is just the variance of $U$, which by a standard formula or by a calculation is $\frac{1}{12}$. For an SNR of $1000$, we want $\frac{A^2}{1/12}=1000$. Solve for $A$. Remark: To answer ...


0

For the uniform distribution, the variance is $\frac{1}{12}(b-a)^2$. So $3$ standard deviation units up from the mean is $\frac{a+b}{2}+\frac{\sqrt{3}}{2}(b-a)$. We can get a similar expression for $3$ standard deviation units below the mean. Since $\frac{\sqrt{3}}{2}(b-a)\gt \frac{b-a}{2}$, the interval from $3$ standard deviation units below the mean to ...


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We want the discriminant to be $\ge 0$, so we want $4X^2-4Y\ge 0$, that is, $Y\le X^2$. Draw the parabola $y=x^2$. The joint density function is $\frac{1}{2}$ on the rectangle $-1\le x\le 1$, $0\le y\le 1$. Let $A$ be the area of the part of the rectangle that is below the parabola. Then our required probability is $\frac{A}{2}$. We can save ourselves the ...


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First, you need to find where on the rectangle, $(X,Y)\in[-1,1]\times[0,1]$, the quadratic equation $t^2+2Xt+Y=0$ has real roots, and which values of $X,Y$ make the roots complex. Hint: Examine the square-root term in the quadratic root formula. When is this not imaginary? Next you need to measure the probability of this, either algebraically or ...


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Hint: to arrive in less than $2.75$ minutes to the first floor, he has to wait the elevator at the second floor for a maximum of $2.25$ minutes. Which is the proportion of the area of the probability function that is identified by this value?


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Think about it. It takes the elevator $2$ to $4$ minutes to arrive (uniformly distributed), then another $0.5$ minutes to reach the destination. You want to know the probability of reaching the destination within $2.75$ minutes of pressing the button. So you want to know the probability that the elevator arrives before what time? Find: $\mathsf P(X \leq ...


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$P(|X_n| > \epsilon) = 0$ as soon as $\epsilon > \frac 1n$. Check your computation again.


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It is a right angled triangle. The legs are formed by the two pieces of the stick. The lengths are x and 4-x. The sum of the lengths of the two pieces is $x+(4-x)=4$. The formula for the area of a right angle triangle is $A=\frac{1}{2} \cdot a \cdot b$. a and b are the legs. The expected value of a uniformly distributed ranodm variable is ...


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If $F$ is the CDF of a random variable then $F(x)$ stands for the probability that $X$ will not exceed $x$, i.e. $F(x)=P(X\leq x)$. If $X$ is distributed over interval $[a,b]$ then it will only take values in $[a,b]$ so that for every $x\geq b$ it is true that $X$ will not exceed $x$. Equivalently you can say that for every $x$ with $x\geq b$ the probability ...


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Look at the definition of CDF. By definition $F(x)=P(X\leq x)$. In this case, the support of the random variable is $[0,2\pi]$. So, $F\left(-\frac{\pi}{6}\right)=P\left(X\leq-\frac{\pi}{6}\right)=0$, since $X$ never attains values less than $-\frac{\pi}{6}$. Furthermore, consider a point to the right of the support of $X$, such as $F(3\pi)=P(X\leq 3 \pi)=1$, ...



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