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The $1-\alpha$ confidence interval for $\hat p$ is $$\large{\left[ \hat p-z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} , \hat p+z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} \right]}$$ $z_{(1-\frac{\alpha}{2})}$ is the z-value of the standard normal distribution. $1-\alpha$ is the confidence level. In your case ...


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Hint:$$\Phi^{-1}(U)\leq x\iff U\leq \Phi(x)$$ Here $\Phi$ denotes the CDF corresponding with standard normal distribution.


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Use the inverse transform method: With $X \sim N(0,1)$ and CDF $F$. $1.$ Generate a random number $u$ from $U$ in the interval $[0,1]$ $2.$ Compute the value $x$ such that $F(x) = u$ $3.$ Take $x$ to be the random number drawn from the distribution described by $F$


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If $$Y = X_{(n)} = \max_i X_i,$$ then $$\Pr[Y \le y] = \Pr\left[\max_i X_i \le y\right] = \Pr\left[\bigcap_{i=1}^n (X_i \le y) \right] \overset{\text{ind}}{=} \prod_{i=1}^n \Pr[X_i \le y].$$ This is because the largest observation is at most $y$ if and only if each observation is at most $y$. Since $$\Pr[X_i \le y] = F_X(y) = \frac{y}{\theta}$$ for $0 \le ...


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The only (somehow) meaningful origin that I can think of, is an estimator of $EX = \frac{a+b}{2} = \frac{\theta}{2}$, that was modified appropriately. I.e., clearly $$ \hat{\theta}_n = \frac{1}{2}(X_{(1)} + X_{(n)}) $$ is consistent estimator of $EX$ (which for some cases can be unbiased as well). I don't know if it has any special name, but I would call it ...


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Let's start from the beginning. The joint probability of the sample $\boldsymbol x = (x_1, \ldots, x_N)$ where each $$x_i \sim \operatorname{DiscreteUniform}(1,n)$$ is IID, is given by $$f(\boldsymbol x \mid n) = \prod_{i=1}^N f(x_i \mid n) = n^{-N} \mathbb 1(1 \le x_{(1)} \le x_{(N)} \le n).$$ Consequently, the joint likelihood is proportional to ...


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Guessing that $G^*$ denotes the set of non-zero elements of $G$, and hoping that the question has finally stabilized: If $X$ and $Y$ are independent then $X+Y$ is uniform on $G$. Without assuming independence I don't see how you can expect to say anything about $X+Y$.


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Let $U$ be uniform continuous on $[0,1]$ and $V$ be uniform discrete on $\{0,1,\cdots,m\}$. Then $U+V$ is uniform continuous on $[0,m+1]$. More generally, assume $U$ to be uniform continuous on $[0,1]$ and $V$ to be another random variable such that $U+V$ is uniform continuous on some other interval. By translation, we can assume that this interval is ...


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Hint. The probability density function of $X$ is $$ f(x)=\begin{cases} \frac{1}{2\pi} & \mathrm{for}\ -\pi \le x \le \pi, \\[8pt] 0 & \mathrm{for}\ x<-\pi\ \mathrm{or}\ x>\pi \end{cases} $$ then you have $$ P(X\leq0)=\int_{-\infty}^0f(x)\:dx $$ $$ P\left(X\leq \frac{\pi}{2}\right)=\int_{-\infty}^{\pi/2}f(x)\:dx. $$ Can you take it from ...


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$W$, the conditional expectation of $Y$ given $X$ will be a piecewise function partitioned on $X$'s enumeration. $$\begin{align}W~=~& \mathsf E(Y\mid X) \\[2ex]~=~& \sum\limits_{\omega\in X^{-1}(X)} Y(\omega)\cdot\mathsf P^{Y\mid X}(\omega) & :~ \mathsf P^{Y\mid X}(\omega) \mathop{:=} \Pr(Y{=}Y(\omega)\mid X{=}X(\omega)) \\[2ex] =~& \tfrac ...


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This can be determined analytically. P(Bomb hits the target)=P(correct X) P(correct Y) P(correct Y)=30/55 easily, since Y follows a uniform distribution P(correct X) is a difference of the cumulative distribution, given by $$P_x=1-e^{150/75}=1-e^2$$ So the expected number of bombs hitting the target is given by $$E=20 \cdot 30/55 \cdot (1-e^2)$$


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I am trying to use simulation methods to estimate the expected number of bombs hit by the twenty aircraft, given 20 pairs of random numbers from [0,1]. Details will depend on the programming language, but the idea is to have a procedure that simulates twenty random points $(x,y)$ and determines how many of those twenty points, say $N_{20}$, fall in ...


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$\operatorname{E}(X\mid Y)=Y$, so by the law of total expectation, $\operatorname{E}(X) = \operatorname{E}(\operatorname{E}(X\mid Y)) = \operatorname{E}(Y) = 3$. For the Poisson distribution, the variance is the same as the expected value, so $\operatorname{var}(X\mid Y) = Y$. Then the law of total variance tells us that $$ \operatorname{var}(X) = ...


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Let's rescale to $[0,1]$ and a sum of $\frac43$ to make it a bit easier. Fleshing out Brian's suggestion: At least one pair sums to at least $\frac43$ if the maximum pair sums to at least $\frac43$. The maximum voltage has cumulative distribution function $x^3$ and thus probability density function $3x^2$. Conditional on $x$, the second-highest voltage ...


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Conditioned on $Y_{i-1}$ and $Y_{i+1}$, $Y_i$ has uniform distribution on $[Y_{i-1}, Y_{i+1}]$. Thus conditioned on $Y_{i-1}$ and $Y_{i+1}$, both $Y_i - Y_{i-1}$ and $Y_{i+1} - Y_i$ are uniform on $[0, Y_{i+1} - Y_{i-1}]$. Since the conditional distributions given $Y_{i-1}$ and $Y_{i+1}$ are the same, the unconditional distributions are the same.


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Because the joint distribution is so simple: $$f_{X,Y}(x,y) = 1, \quad 0 \le x, y \le 1,$$ it is easier to integrate directly: $$\operatorname{E}[|X-Y|] = \int_{x=0}^1 \int_{y=0}^1 |x-y| f_{X,Y}(x,y) \, dy \, dx = \int_{x=0}^1 \int_{y=0}^x x-y \, dy \, dx + \int_{x=0}^1 \int_{y=x}^1 y-x \, dy \, dx.$$


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Let us write the Marsaglia formulas under the form : $$X = \sqrt{-2\log{U}}\frac{V_1}{\sqrt{R}},Y = \sqrt{-2\log{U}}\frac{V_2}{\sqrt{R}}$$ By comparison with Box-Muller formulas, you see that the cosine and the sine are resp. replaced by $$C=\frac{V_1}{\sqrt{R}},S=\frac{V_2}{\sqrt{R}} \ \ \text{with} \ \ C^2+S^2=1$$ which is exactly the same because the ...


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We can derive the pdf by calculating the cdf and taking the derivative. The probability that $X_{max}\le x$ for $x$ between $0$ and $\theta$ is the probability that all $n$ measurements fall between $0$ and $x$. But for a single measurement, this is $\frac{x}{\theta}$. The probability $n$ independent measurements fall between $0$ and $x$ is then ...


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If so, should I add them together or should I separate the situation and say if x<=z, then the answer is xz and if x>z, then the answer is z^2? The latter.   You have a piecewise function.   Indeed there are a few other cases to consider. $$\begin{align}\mathsf P(U\leq x, T\leq z) ~=~& \begin{cases}\mathsf P(U\leq x, V\leq z) & : ...


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Your mistake is using Bayes' Rule.   Don't. $$\begin{align} \Pr(\lvert X-Y\rvert\leq 2) ~=~& \Pr(Y-2\leq X\leq Y+2) \\[1ex] = ~& \dfrac{1}{5} \end{align}$$ Because whatever the value of $Y$ is in $[6;14]$, the probability that $X$ will lie within the four second interval near that value is: $4/20$.


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You should be integrating over the support of the PDF. The support is the set of all numbers such that the PDF is not zero. Therefore, your integral should be $\displaystyle\int_a^x \frac{1}{b-a} \, dx$. This will give CDF$=\dfrac{x}{b-a}-\dfrac{a}{b-a}=\dfrac{x-a}{b-a}$ just like wikipedia has. The reason we integrate over the support is simple. We'd ...


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For company $B$, let $\{X_n\}$ be the interarrival times, supposing that they are independent and identically distributed with mean $\mathbb E[X_1]=30$. Let $S_n=\sum_{k=1}^n X_k$, then $\{S_n\}$ is a renewal process. Let $N(t)=\sum_{n=1}^\infty\mathsf 1_{(0,t]}S_n$ be the number of renewals up to time $t$, then at a given time $t$, the time until the next ...


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The conditional density of $Y$ given $X=x$ is $f_{Y|X}(y|x)=\frac{1}{x+1}1_{0<y<x+1}$, hence the joint density of $X$ and $Y$ is $$ f(x,y)=f_{Y|X}(y|x)f_X(x)=\frac{1}{x+1}1_{0<x<1,0<y<x+1}$$ The marginal density can then be obtained by "integrating out" the $x$-variable: $$ f_Y(y)=\int f(x,y)\;dx$$



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