The random variables $X_1, X_2, . . . , X_n, Y_1, Y_2, . . . , Y_n$ are iid $\mathcal{U}(0, a)$. Determine the distribution of $$Z_n = n \log\bigg(\frac{\max\{X(n), Y(n)\}}{\min\{X(n), Y(n)\}}\bigg)$$ Should I find the joint distribution of $\max$ and $\min$ and then find distribuation of $Z_n$, since we have a two different random variable I do not know how to do that!
122555
The random variables $X_1,X_2,...X_n,Y_1,Y_2,...Y_n $ are independent and $U(0,a)$-distributed. Determine the distribution of $$Z_n=n\cdot \log\left(\frac{\max\left(X_{(n)},Y_{(n)}\right)}{\min\left(X_{(n)},Y_{(n)}\right)}\right)$$ where $X_{(n)},Y_{(n)}$ denotes $n$th order variable.  This problem is giving me a headache. The answer is that $Z_n$ is $\text{Exp}(1)$-distributed. But i can't derive that conclusion. I'd like to show you my derivation, but its so long and contains so many integrals and fractions that it would take a while to write it here. But basically my approach was to derive the distribution of $M=\max(X_{(n)},Y_{(n)})$ and $L=\min(X_{(n)},Y_{(n)})$. I get the following cdf's: $F_M=(F_X\cdot F_Y)^n=(F_X)^{2n}$ and $F_L=1-(1-(F_X)^n)\cdot (1-(F_Y)^n)=1-(1-(F_X)^n)^2$ ,  correct? $F_X $ stands for the cdf of $X_i$  Then i define the random variable $T=\frac{M}{L}$ and compute its cdf to finally be able to derive the distribution of $Z_n=n\cdot \log(T)$. But thats where i think everything goes wrong. Any help appreciated, thanks.
65360
Assume that $X$ and $Y$ are independent random variables with $X \sim \exp(\lambda)$ and $Y \sim \exp(\mu)$. It is impossible to obtain direct observations of $X$ and $Y$. Instead, we observe the random variables $Z$ and $W$, where $Z = \min\{X,Y\}$ and $W =1$ if $Z=X$ and $W=0$ if $Z=Y$.  How to find the joint distribution of $Z$ and $W$?
14828
Question Suppose that a random variable $X$ has the property that, for any $a &gt; 1$, the conditional distribution of $\frac 1 a X$ given that $X &gt; a$ is the same as the distribution of $X$. Specify all possible distributions of $X$. My thoughts I do know that $f_{X\mid Y}(x\mid y) = \frac {f_{X, Y}(x,\ y)} {f_Y(y)}$ and I suppose I have to make use of this relationship, but I have no idea how to even begin - When I find conditional distributions, I am usually given a specific value, not something like $X &gt; a$. Any proposed solutions with intuitive explanations will be greatly appreciated :)
21152
I am trying to find distribution of random variable:  $$\frac{X_1+X_2 X_3}{\sqrt{1+X_3^2}}$$  where $X_i \sim \mathcal{N}(0,1)$ and independent.  My thinking so far is to use random variable algebra and to find distributions step by step. However, I don't know what to do with the variable: $\sqrt{1+X_3^2}$ (it seems like some variant of $\chi$-distribution).   Moreover, I'm worried that my thinking is to complicated and that the solution should be more elegant.
84761
Assume ${{X}_{1}}$, ${{X}_{2}}$, ${{X}_{3}}$,...,${{X}_{n}}$  are i.i.d. samples from distribution with density f, and cdf F.  Let V=min( ${{X}_{1}}$, ${{X}_{2}}$, ${{X}_{3}}$,...,${{X}_{n}}$)  To find the minimum $${B}_{v}={\{V\ge{v}\}}=\{{X}_{1}\ge{v}\}\cap\{{X}_{2}\ge{v}\}\cap...\cap\{{X}_{n}\ge{v}\}$$  So, $$P(V\ge{v})=[1-F(v)]^{n}$$  The cdf is  $$F_{v}(V)=1-[1-F(v)]^{n}$$   The density is $${f}_{v}(V)=\frac{d}{dv}[1-F(v)]^{n}=nf(v)[1-F(v)]^{n-1}$$   To find the density, why did we use $[1-F(v)]^{n}$ instead of $1-[1-F(v)]^{n}$, shouldn't we used the cdf to derive the pdf?
64127
We have $X_1,...$ indepdent random variables with common distribution $F$ and $N$ is a geometric random variable independent of $X_i$'s. Let $M = \max(X_1,...X_N)$. Im trying to convince myself that   $$ P( M \leq x \mid N &gt; 1 ) = F(x) P(M \leq x ) $$  I understand that since $X_1$ is independent of all $X_i's$ and from $N$, when we can pull it out of the probability and we have   $$ P(X_1 \leq x) P( \max(X_2,....,X_N ) \leq x \mid N &gt; 1 ) $$  And then we have  $$ F(x)  P( \max( X_1,...,X_{N-1} \leq x \mid N &gt; 1 ) $$  How does    $$ P( \max( X_1,...,X_{N-1} \leq x \mid N &gt; 1 )  = P( M \leq x ) $$   ??
32235
Please. I am trying to understand the proof, that cdf of minimum of $n$ random variables is $1-[1-F(x)]^n$  If I have $n$ independent random variables $X_1, \dots, X_n$, all of them have the same CDF $F(x)$, what is the point of searching for the $\min=\{X_1, \dots, X_n\}$? Shouldn't $X_1, \dots, X_n$ have the same value for every x? (In the meaning: $P (\min \{X_1, \dots, X_n\}≤x) = P (\bigcup_{i=1}^n X_i\}≤x) \cdots$)  Thanks.
53110
Could anybody help me with solving such problem:  You are given two independent random variables $X$ and $Y$ with continuous uniform distribution. You are to find expected value $E$ of random variable $Z = \max(X,Y)$.
61924
Let $X_{1},X_{2},....,X_{n}$ be independent $N(0,1)$ random variables and set $\bar{X}=\frac{1}{n}\sum_{i=1}^{n}X_{i}$ show that $\bigg(\bar{X},\frac{1}{n}\sum_{i=1}^{n}(X_{i}-\bar{X})^{2}\bigg)$ and $\bigg({X}/\sqrt{n},\frac{1}{n}\sum_{i=1}^{n-1}(X_{i})^{2}\bigg)$ have the same distribution. This shows that the sample mean and sample variance are independent in this model.  Thoughts I tried showing that they have the same characteristic function but couldn't get it to work that way. I think maybe I could do something involving the CLT but I'm not sure how.
97066
Suppose $X$ and $Y$ are random variables that are independent also $X$ and $Y$ are uniformly distributed on the interval $(0, 1)$. If $Z=\max \{ X, Y \}$. Then find the probability that $Z \leq z$ and then determine the probability density function of $Z$.  Okay so I must admit I an trumped by this question but I think it's mainly to do with not actually understanding what I'm meant to do. Here are my issues:  1) Okay so $X$ and $Y$ are uniformly distributed on $(0,1)$ does that mean that it is $0$ elsewhere?  2) What exactly is $Z$ is it just the maximum value that either $X$ or $Y$ can take and how exactly do I interpret this? At the moment I'm just thinking of two graphs and $Z$ being the highest point that is taken on both graphs in $X$ or $Y$ since this is a uniform distribution then it will be the same highest point for all values in the interval on either $X$ or $Y$ whatever is highest. This is just intuition again I'm not sure.  3) What is  $P(Z \leq z)$ actually meaning? I'm guessing this is because I'm confused on what $Z$ is and so I can't get this either.  If anyone could help me on this it would be greatly appreciated thanks!
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I have an answer but I'm not sure if I'm doing it right: $f(x)$ is the pdf $f(x)=2e^{-2x}$, then, if the random variable $Y=\ln(X)$, then we simply substitute:  $P(Y \leq y)$=  $P(2e^{-2X} \leq y)$  $P(-4X \leq y)$ since $y=ln(x)$  $P(X \leq - \dfrac {1} {4} y)$  This is the distribution function, right?  Thanks!
27672
Let $X_1,\ldots,X_n$ be independent random variables with common cumulative distribution function $F$. I am trying to find the joint cumulative distribution function of $$U=\min(X_1,\ldots,X_n)\quad\text{and}\quad V=\max(X_1,\ldots,X_n).$$ The result that I get is $$F_{U,V}(u,v)=F(v)^n-(F(v)-F(u))^n,$$ but this seems wrong because letting $u\to\infty$ gives the marginal CDF $$F_V(v)=F(v)^n-(F(v)-1)^n,$$ while it should be $F_V(v)=F(v)^n$. I obtained the result as follows $$\begin{align} F_{U,V}(u,v) &amp;= P(U\leq u,V\leq v) \\ &amp;= P(V\leq v)-P(U&gt;u,V\leq v) \\ &amp;= P(X_1\leq v,\ldots,X_n\leq v)-P(u&lt;X_1\leq v,\ldots,u&lt;X_n\leq v) \\ &amp;= F(v)^n-(F(v)-F(u))^n. \end{align} $$ But $$F_V(v)=P(V\leq v)=P(X_1\leq v,\ldots,X_n\leq v)=P(X_1\leq v)\cdots P(X_n\leq v)=F(v)^n.$$ What is wrong with these computations?
67801
Let $Y_1, Y_2, ..., Y_n$ be iid random variables and $B_1, B_2, ..., B_n$ be Borel sets. It follows that  $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$...I think?  If so, does the converse hold true? My Stochastic Calculus professor says it does (or maybe misinterpreted him somehow?),  I was  the  that independence of the n random variables was equivalent to saying for any indices $i_1, i_2, ..., i_k$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$.  So, if the RVs are independent, then we can choose $i_j=j$ and k=n to get $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, but given $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$, I don't know how to conclude that for any indices $i_1, i_2, ..., i_n$ $P(\bigcap_{j=i_1}^{i_k} (Y_j \in B_j)) = \Pi_{j=i_1}^{i_k} P(Y_j \in B_j)$, if that's even the right definition.  p.17  seems to suggest otherwise. idk  Help please?  Also this:    or    So, this answer is to use the Omega part to establish pairwise independence and ultimately conclude independence. Without that assumption, we cannot conclude independence. Is that right? Why does that not contradict the definition of independence: $P(\bigcap_{i=1}^{n} (Y_i \in B_i)) = \Pi_{i=1}^{n} P(Y_i \in B_i)$ ?
9750
Let be $X_1,X_2,...,X_n$ be uniformly distributed random variables i.i.d.  a)  Find the cumulative function and the density of $m_n \text{ and } M_n$ , where $m_n=min(X_1,X_2,...,X_n)$ and $M_n=max(X_1,X_2,...,X_n)$.  b)  Let $Z_n=n(1-M_n)$. Show that $Z_n \xrightarrow{d} Z $, where $Z$ is a random variable with cumulative function $F_Z(z)=1-e^{-z}$
75632
Let $Y_1$ and $Y_2$ be independent random variables with   $Y_1\sim N(1,3)$ and $Y_2 \sim N(2,5).$      If $W_1=Y_1+2Y_2$ and $W_2=4Y_1-Y_2$   what is the joint distribution of $W_1$ and $W_2$?   Is my procedure correct?  $E(W_1)=E(Y_1+2Y_2)=E(Y_1)+2E(Y_2)=1+2\cdot2=5$  $Var(W_1)=Var(Y_1+2Y_2)=Var(Y_1)+4 Var(Y_2)=3+4\cdot 5=23$  $f_{W_1}(w_1)=\frac{1}{\sqrt{2\pi}23}\epsilon^{\frac{-1}{2}(\frac{x-5}{23})^2}$  I can't conclude.
64072
Let $X_1(x),X_2(x),\ldots ,X_n(x)$ be $n$ iid random variables. I know that  $$\mathbf{P}(\lim_n X_n(x)=X(x))=1$$ for each $x\in \mathbb{R}$. We also know that the limit random variable $X$ is a.s. constant and considering $X\colon \mathbb{R}\to\mathbb{R}$, it is continuous. I want to show that  $$\mathbf{P}(\lim_nX_n(x)=X(x)\ \text{ for all }x\in \mathbb{R})=1.$$ I thought about using countable additivity of probability on $\mathbb{Q}$ and then applying continuouty to extend the result to the whole $\mathbb{R}$. Since $$\mathbf{P}(\lim_nX_n(x)\neq X(x)\ \text{ for all }x\in \mathbb{Q})=\sum_{x\in \mathbb{Q}}\mathbb{P}(X_n(x)\neq X(x))=0+0+\ldots.$$ Is there anything else I should add to this explanation? Ideas?
29766
Help me to solve this problem please..  Let $Y_{(1)}, Y_{(2)}, Y_{(3)}, Y_{(4)}, Y_{(5)}$ denote the order statistics of a random sample of size 5 from a distribution having p.d.f. $f(y) = e^{(-y)}, 0 &lt; y &lt; \infty$, zero elsewhere. Show that $Z_1 = Y_{(2)}$ and $Z_2 = Y_{(4)} − Y_{(2)}$  are independent. Hint: First find the joint p.d.f. of $Y_{(2)}$  and  $Y_{(4)}$.