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4

Let the random variables be $X$ and $Y$. Note that $\mathbb{E}(\min \{X, Y \}) = \frac{5}{3}$, and $\mathbb{E}(\max \{X, Y \}) = \frac{10}{3}$. So the average distance is $\frac{5}{3}$. Another approach: I will replace $5$ with $1$; we will just multiply $5$ at the end. Suppose person 1 chooses $x$. Then there is a probability $x$ of person 2 choosing ...


3

Your attempt is not correct, and I think it is better here to switch from PDF to CDF. The CDF $F$ that corresponds with PDF $f$ satisfies: $$F\left(x\right)=\int_{-\infty}^{x}f\left(u\right)du$$ leading here to: $$F\left(x\right)=1-x^{-3}\text{ if }x\geq1\text{ and }0\text{ otherwise.}$$ Then for $u\in\left(0,1\right)$ we find: $$h\left(u\right)=\inf\left\{...


2

If $(X,Y,Z)$ is uniformly distributed in $T=\{(x,y,z)\in[0,1]^3 : x+y+z=1\}$, then $(X,Y)$ is uniformly distributed in $T'=\{(x,y)\in[0,1]^2 : x+y\leqslant 1\}$, and $Z=1-X-Y$ a.s., thus $$\mathbb{E}f(X,Y,Z)=2\iint_{T'}f(x,y,1-x-y)\,dx\,dy.$$ A simplification by symmetry, $\mathbb{E}\max\{X,Y,Z\}=6\,\mathbb{E} Z[X\leqslant Y\leqslant Z]$, gives $$\mathbb{E}\...


1

If we weren't told $\sum_{i=1}^nX_i=t$, the joint PDF of the $X_i$ would be $\exp\left(-\sum_i x_i\right)$ on $[0,\,\infty)^n$. Over the $X_i\ge0$ region of the hyperplane plane $\sum_iX_i=t$ for some $t>0$, this PDF is $\exp(-t)$; and this region is a simplex of measure $\frac{t^n}{n!}$, so the probability density of $t$ is $\frac{t^n}{n!}\exp(-t)$ on $[...


1

Or alternatively : \begin{align*} \mathbb E[\max(X,Y)] &= \mathbb E\left[\frac{X+Y+|X-Y|}{2}\right]\\ &= \frac{1}{2} (\mathbb E[X]+\mathbb E[Y]+\mathbb E[|X-Y|])\\ &= \frac{2}{3}\\ \mathbb E[\min(X,Y)] &= \mathbb E[X+Y-\max(X,Y)]\\ &= \frac{1}{2}+\frac{1}{2}-\frac{2}{3}\\ &= \frac{1}{3} \end{align*}


1

Mean square error (MSE) of an estimator $\hat\theta$ for estimating $\theta$ is defined as $$\operatorname{MSE}_{\theta}(\hat\theta)=\operatorname E_{\theta}(\hat\theta-\theta)^2=\operatorname{Var}_{\theta}(\hat\theta)+(\text{bias}(\hat\theta))^2$$ Both your estimators are unbiased for $\theta$, so MSE here is just variance. Now, $$\operatorname{Var}_{\...


1

Suppose the biased samples are given by $(G_i)_{i\in\mathbb N}$. Then recursively define a new sequence $(H_i)_{i\in\mathbb N}$ by setting $H_1=G_1$ and $$ H_n=G_n-\frac{H_1+\cdots+H_{n-1}}{n-1}+\frac{1}{2}. $$ (Note that for small $n$, the $H_n$ can fall outside the interval $[0,1]$ with low probability. You may want to clamp it to $[0,1]$, which will have ...


1

As correctly pointed out in the comments by David Peterson, you're looking for $E(|X-Y|)$ where $X, Y \sim U(0,5)$. Now since the two random variables are independent, their PDF is given by the product of their marginal distributions, i.e., $$f_{X,Y}(x,y) = f_{X}(x) f_{Y}(y)$$ Now in order to calculate, $E(|X-Y|$, $$E(|X-Y|) = \int_{-\infty}^{\infty} \...


1

For real numbers $X$ and $Y$: There are two regions in the $(x,y)$ plane and the formula for distance is different in the two regions (unless you use an absolute value, which is tricky to cope with in an integral). The total region in $(x,y)$ space is $5 \times 5 = 25$ and the answer becomes $$\frac1{25} \left( \int_{x=0}^5 \int _{y=0}^x (x-y) dy\,dx+ \...


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