I am having a small doubt regarding maximum of random variables. I have $$Z= \max\{ X_1, X_2,\dots X_p, \dots X_N\}$$ where all $X_i$ are independent, identically distributed. Now, If for sure, I know that $$X_p&gt;X_{p+1}&gt;\dots&gt;X_N$$ the can i say that the CDF of $Z$,  $$\begin{align*}P(Z&lt;z)&amp;= P\left(\max\{ X_1, X_2,\dots X_p, \dots X_N\}&lt;z\right)\\&amp;=P\left(\max\{ X_1, X_2,\dots X_p\}&lt;z\right)\\&amp;=\left[P(X_1&lt;z)\right]^{N-p-1}\end{align*}$$  How to find CDF of $Z$ in this case ? Is there any theorem that reduce the size of total variables for certain condition  ?
121789
I am having a small doubt regarding maximum of random variables. I have $$Z= \max\{ X_1, X_2,\dots X_p, \dots X_N\}$$ where all $X_i$ are independent, identically distributed. Now, If for sure, I know that $$\{X_1,\dots,X_p\}&gt;X_{p+1}&gt;\dots&gt;X_N$$ the can i say that the CDF of $Z$,  $$\begin{align*}P(Z&lt;z)&amp;= P\left(\max\{ X_1, X_2,\dots X_p, \dots X_N\}&lt;z\right)\\&amp;=P\left(\max\{ X_1, X_2,\dots X_p\}&lt;z\right)\\&amp;=\left[P(X_1&lt;z)\right]^{N-p-1}\end{align*}$$  How to find CDF of $Z$ in this case ? Is there any theorem that reduce the size of total variables for certain condition  ?
83684
I've seen the statement:     It's possible that random variables $X_i, X_j$ are independent for $i≠j$, but $X_1, X_2, X_3$ are dependent.   I haven't been able to find examples of this though.  Any examples?
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I'm trying to answer the following question from the book :  Let $X_1,X_2,\dots$ be a sequence of sub-gaussian random variables, which are not necessarily independent. Show that   $E\bigg[ \max_i \frac{|X_i|}{\sqrt{1 + \log i}} \bigg] \le CK$,  where $K = \max_i \|X_i\|_{\psi_2}$. Deduce that for ever $N \ge 2$ we have   $E\bigg[ \max_{i \le N} |X_i| \bigg] \le CK \sqrt{\log N}$.  I've tried to figure out what is the distribution of the maximum of Gaussians, but I'm reaching only inequalities that that don't help me answer the question.   I've also seen a similar question .  Does anyone have a clue or something to start with in order to answer this question?  Thanks!
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This question is related to  , with the difference being that $F_X$ (the probability distribution function of random variable $X$) is an arbitrary continuous distribution function, not necessarily strictly increasing.  I think the proof is similar, but we have to take care of the possibility that $F_X$ may not be $1$-to-$1$.  I list my attempt below, and would appreciate it if someone can confirm if it's correct, and, in particular, if it can be improved.  Thanks a lot!  The goal is to show $F_Y(y)=y$ for any $y \in [0,1]$.  To do so, note that $F_Y(y)\triangleq\mathbb P(\{Y\le y\})$, and  $\{Y\le y\}=\{F_X(X)\le y\}=\{X\in F_X^{-1}([0, y])\}.$  Since $F_X$ is continuous, $F_X^{-1}([0,y])$ must be closed.  So it follows that   $\sup F_X^{-1}([0, y])=\max F_X^{-1}([0, y])=\max F_X^{-1}(\{y\}),$ which let's denote by $a$.    Therefore, $F_Y(y)=\mathbb P(\{X\le a\})=F_X(a)=y.\quad$  (Q.E.D.)
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?
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Question  Suppose we have a multinomial distribution with $N$ possible outcomes, with probabilities $p_1,\ldots,p_N$. We sample this $n$ times, and denote the observed frequency of the $i$th outcome as $\xi_i$. In [1] the author claims that the distribution of the $\xi_i$ in the limit of large $n$ is:  $$p(\xi_1,\ldots,\xi_N)\propto\exp\left(-\frac{n}{2}\sum_{i=1}^N\frac{(\xi_i-p_i)^2}{p_i}\right).\;\;\;\;\;(1)$$  We can see immediately that this must be an approximation, as it assigns nonzero probabilities for $\xi_1+\cdots+\xi_N&gt;1$. However we can see that these have vanishing probability in the limit $n\rightarrow\infty$. My question is how do we derive (1) from the multinomial distribution, and show that they match in the $n\rightarrow\infty$ limit?  My thoughts  My first thought would be to appeal to the central limit theorem. The multinomial distribution has mean $\mu_i=p_i$ and covariance matrix $\Sigma_{ij}=\delta_{ij}p_i-p_ip_j$, so we would expect this in the large $n$ limit to be described by a multivariate Gaussian with mean $\mu$ and covariance $\frac{1}{n}\Sigma$. However, things are complicated by the fact that the multinomial covariance is singular (since $\xi_N$ is determined by the other $\xi_i$s), and so the multivariate Gaussian is not defined.  To address this, we may try and consider only the first $\xi_1,\ldots,\xi_{N-1}$, which have a non-singular covariance matrix and hence well-defined multivariate Gaussian distribution. Let's take the Binomial distribution $N=2$. The frequency $\xi_1$, this has mean $p_1$ and variance $p_1(1-p_1)$, so this would be described the the Gaussian: $$\propto\exp\left(-\frac{n}{2}\frac{(\xi_1-p_1)^2}{p_1(1-p_1)}\right).\;\;\;\;\;(2)$$ The expression (1) gives: $$\propto\exp\left(-\frac{n}{2}\left(\frac{(\xi_1-p_1)^2}{p_1}+\frac{(\xi_2-p_2)^2}{p_2}\right)\right).\;\;\;\;\;(3)$$ If we substitute $\xi_2\rightarrow 1-\xi_1$, $p_2\rightarrow 1-p_1$ into (3), we can verify that this gives the same answer as (2). I have verified that this also works for $N=4$.   I'm sure that if I just bashed out the algebra for general $N$ we would get agreement between the central limit theorem and (1) when we restrict the latter to $\xi_1+\cdots+\xi_N=1,p_1+\cdots+p_N=1$. However, how can we start with the multinomial distribution and derive (1) as a limit which is valid everywhere? One idea would be to say that (1) goes to zero as $n\rightarrow\infty$ when you are not on that plane, however I am a bit uncomfortable with this as it goes to zero everywhere except the mean as $n\rightarrow\infty$, so I don't know if that argument is good enough.  [1] Wootters, William K. "Statistical distance and Hilbert space." Physical Review D 23.2 (1981): 357.
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I came across this question as an exercise, had a brief idea, but didn't know how to proceed.     Let $X \sim N(0, 1)$. What is the p.d.f of $|X|$ ?   I know the final p.d.f looks just like the right half of the original pdf, but extended vertically for a factor of 2.  Could someone please show mathematically what the p.d.f of variable |X| is going to be, and explain briefly? Thank you.
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Let $T_n$ be a random variable with $T_n=X_1+...+X_n$ where the $X_i$'s are iid. Further we set $N(t)=max\{ n: T_n\leq n\}$ with the property $\Pr(N(t)&lt;\infty)=1$.  I want to prove that $\mathbb{E}[N(t)]&lt;\infty$.   I have done this so far. Let $F_n$ be the distribution function of $T_n$. $$\mathbb{E}[N(t)]=\sum_{n=1}^{\infty} n P(N(t)=n)=\sum_{n=1}^{\infty} n (F_n(t)-F_{n+1}(t))=\sum_{n=1}^{\infty} F_n(t)$$  Further I know that $0 \leq F_n \leq 1$. But does this implies $\mathbb{E}[N(t)]&lt;\infty$ ? I don't think so... I don't know what I can do further...   Can anybody help me how I can proof it?   Thank you for any help!!!
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Suppose $X,Y$ are random variables. I'm trying to understand why $\max\{X,Y\}$ and $\min\{X,Y\}$ are also random variables. The proof in the book that I'm using states that for each $t$,      $\{ \max\{X,Y\} \leq t \} = \{X \leq t \} \cap \{Y \leq t \}$,   and     $\{ \min\{X,Y\} \leq t \} = \{X \leq t \} \cup \{Y \leq t \}$.   I can't make the mental leap. Could someone explain this more explicitly?  Edit:   We use the following definition for a random variable: A random variable is a function on  $X: \Omega \rightarrow \mathbb{R}$ such that      $X^{-1}((-\infty,t]) := \{\omega \in \Omega : X(\omega) \leq t\} \in \mathcal{F}$   for all $t\in\mathbb{R}$, where $\mathcal{F}$ is the set of all events.  The obvious first step that I made was      $\{ \max\{X,Y\} \leq t \} = \{ \omega : \max\{X(\omega),Y(\omega)\} \leq t \}$.   I just don't see why the next equality follows.  Edit2: Fixed typo.
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Assume the standard situation, that is, let $X_1, \ldots , X_n$ be independent and identically distributed with $X_k ∼ P_\theta(x; \theta)$ , where $P_\theta(x; \theta) = 2x/\theta^2$  if $0\le x \le\theta$ and $0$ otherwise. It is required to estimate $\theta$. Show that the maximum likelihood estimator for $\theta$ is $\hat{\theta} = \max{[X_1, \ldots , X_n]}$ and then show that the cumulative distribution function of $\hat{\theta}$ is $F_\theta(z) = z^{2n}/\theta^{2n}$. Here's what i did so far: Maximum Likehood estimator: $L_x(θ) = \prod_{i = 1}^{n} P_θ(x_k)$ Here we have $P_θ(x_1,...,x_n;θ) = P_θ(x_1,θ).P_θ(x_2,θ)...P_θ(x_n,θ)$ Likehood = $L_{x,θ}(θ) =P_θ(x_1,θ).P_θ(x_2,θ)...P_θ(x_n,θ)= 2x_1/θ^2. 2x_2/θ^2.. 2x_n/θ^2 = [2^n.\prod_{i = 1}^{n}x_i]/θ^{2n}$ Log-likehood: $\sum_{i = 1}^{n}log(P_θ(x_1,...x_n;θ))= \sum_{i = 1}^{n}log(2x_i/θ^2)$ Is this correct so far? Im still not sure how to get to $\hat{θ} = max{[X_1, . . . , X_n]}$ As for the the cumulative distribution part to show $F_θ(z) = z^{2n}/θ^{2n}$: $F(z) = P(max({x_k})&lt;z) = P(x_1&lt;z).P(x_2&lt;z)..P(x_n&lt;z) = 2x_1/θ^2.2x_2/θ^2...2x_n/θ^n = 2^n.\prod_{i = 1}^{n}x_i/θ^2n$ Not sure if this is correct. Would really appreciate some help. Edit: From the answers below, we can deduce the the estimator is biased. What estimator would be unbiased? How can i find it?
72927
Let $\mathbb P(X=1) = \mathbb P(X=-1) = 1/2$. Define       $$X_n = \begin {cases} X &amp; \text{with probability } 1- \frac{1}{n}\\  e^n &amp; \text{with probability }  \frac{1}{n} \end {cases}$$      Does $X_n \leadsto X$ (i.e., converge in distribution)? Does $X_n \stackrel{P}{\rightarrow} X$? Does $\mathbb{E}[(X-X_n)^2] \to 0$?   I think $X_n \leadsto X$ because just by picturing the CDF  as n goes to infinity, their CDFs converge. Can I get some confirmation on this?  But for convergence in probability, I'm having trouble with the limits.   To do this, what I'm trying to do is find $\mathbb P(|X_n-X| \leq \epsilon)$ and then just take 1 minus that as shown on this site:   If $X_n = X$, then $|X_n - X|$ is just $0$ so for all $n$, $|X_n -X| &lt; \epsilon$ But if $X_n = e^n$ then I have the two cases for $X$. For $X=1$, $|X_n-X| = |e^n-1| \leq \epsilon$. and For $X=-1$, $|X_n-X| = |e^n +1| = e^n + 1 \leq \epsilon$. This is where I am stuck. I'm trying to find conditions relating $n$ and $\epsilon$ similiar to here the example on the site. But I'm not sure how to get that from $|e^n - 1| \leq \epsilon$ and  $e^n + 1 \leq \epsilon$ .  Am I even on the right track?
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Suppose that  $X$ and $Y$ are two i.i.d. uniform random variables on the interval $[0,1]$  Let $Z=X/Y$, I am finding the cdf of $Z$, i.e. $ \Pr(Z\leq z) $.  Now, I came up with two ways of doing this. One produces a correct answer consistent with the pdf here: , the other does not. Why is the second method wrong?  First Method  $\newcommand{\rd}{\mathrm{d}} \Pr(Z\leq z) = \Pr(X/Y\leq z) = \Pr(X\leq zY) = \int^{1}_{0}\int^{\min(1,zy)}_{0} \rd x \rd y = \int^{1}_{0}\min(1,zy)\ \rd y$ $  = \left\{      \begin{array}{lr}        \int^{1/z}_{0}zy\ \rd y + \int^{1}_{1/z} \rd y&amp; : z &gt; 1\\        \int^{1}_{0}zy\ \rd y &amp; : z \leq 1      \end{array}    \right. $ $  = \left\{      \begin{array}{lr}        1 - \frac{1}{2z} &amp; : z &gt; 1\\        \frac{z}{2} &amp; : z \leq 1      \end{array}    \right. $  This appears correct.  Second Method  $\Pr(X/Y\leq z) = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(zY \geq 1) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(zY &lt; 1) $ by total probability  $ = \Pr(X \leq zY\ |\ zY \geq 1)\Pr(Y \geq 1/z) + \Pr(X \leq zY\ |\ zY &lt; 1)\Pr(Y &lt; 1/z)$  Taking $z&gt;1$ yields $ (1)(1-\frac{1}{z}) + (\int^{1/z}_{0}\int^{zy}_{0} \rd x \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + (\int^{1/z}_{0}zy\ \rd y)(\frac{1}{z}) = 1-\frac{1}{z} + \frac{1}{2z^{2}}$  This is already different. Why is this wrong?  Thanks!
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Let $X_n$, $X$ random variables on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ and $(c_n)_n \subseteq \mathbb{R}$, $c \in \mathbb{R}$ such that $c_n \to c$ and $X_n \stackrel{d}{\to} X$. Then $c_n \cdot X_n \stackrel{d}{\to} c \cdot X$.   Here is my proof: By Lévy's continuity theorem it suffices to show that $$\forall \xi: \quad \hat{\mu}_n(c_n \cdot \xi) \to \hat{\mu}(c \cdot \xi) \qquad (n \to \infty)$$  where $\hat{\mu}_n$ (resp. $\hat{\mu}$) denotes the characteristic function of $X_n$ (resp. $X$). Using the tightness of the distributions one can show that $\{\hat{\mu}_n; n \in \mathbb{N}\}$ is uniformly equicontinuous. We have $$|\hat{\mu}_n(c_n \cdot \xi) - \hat{\mu}(c \cdot \xi)| \leq |\hat{\mu}_n(c_n \cdot \xi)- \hat{\mu}_n(c \cdot \xi)| + |\hat{\mu}_n(c \cdot \xi)- \hat{\mu}(c \cdot \xi)|$$ The first addend converges to zero as $n \to \infty$ by the uniform equicontinuity and the second one converges to zero since $X_n \stackrel{d}{\to} X$.  I was wondering whether there is an easier proof - my own proof seems rather like overkill to me. Thanks!
61164
The p.d.f. of the random variable $X$ is given by $f(x) =  \begin{cases}   e^{x-2} &amp; \mbox{for $0 \leq x \leq 2$}, \\  e^{-x} &amp; \mbox{for $x &gt; 2$}, \\  0 &amp; \mbox{otherwise,}  \end{cases}$ I need to find $F^{-1}(y)$ to construct a generator for $X$. So I start with calculating the cdf by taking the integral of $f(x)$: $$ F(X)= \begin{cases} \frac{e^x-1}{e^2} &amp; 0 \leq x \leq 2 \\ -e^{-x}+e^{-2} &amp; x \geq 2 \\ 0 &amp; o.w \end{cases} $$ Next, I need to find the inverse of this function: $$y = \frac{e^x-1}{e^2} \implies x= \ln(e^2y+1) \text{ for } y \in [0, 1-e^{-2}]$$ $$y=-e^{-x}+e^{-2} \implies x = -\ln(e^{-2}-y) \text{ for }y \in (0, e^{-2})$$ Now at this point I am confused since the intervals for $F^{-1}(y)$ overlap. How do I break this up into cases? Did I screw up a calculation along the way( I retraced my steps and it seems fine...)?
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Say $X_1, X_2, \ldots, X_n$ are independent and identically distributed uniform random variables on the interval $(0,1)$.  What is the product distribution of two of such random variables, e.g., $Z_2 = X_1 \cdot X_2$?  What if there are 3; $Z_3 = X_1 \cdot X_2 \cdot X_3$?  What if there are $n$ of such uniform variables? $Z_n = X_1 \cdot X_2 \cdot \ldots \cdot X_n$?
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Show that. for any discrete random variable X that takes on values in the range [0,1]. Var[X] $\le$  1/4.  I translate it into a inequality like this: $x_1, x_2, x_3 \cdots ,x_n$ where $0 \le x_i \le 1$, and $p_1, p_2, p_3 \cdots ,p_n$ where $p_1+ p_2+ p_3 \cdots +p_n = 1$, prove that $\sum _1^nx_i^2p_i - (\sum _1^n x_ip_i)^2 \le {1\over 4}$ , how to prove it?  At first I tried cauchy inequatlity, but I fail :(
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Is there a nice limiting distribution of $\max( X_1,X_2,...,X_n) $ as $n$ goes to $\infty$, assuming that they are iid normal distributions with variance $\sigma^2$.   This is almost certainly a well known problem with a clever proof and nice solution, but I've been digging around and haven't found anything.