In single dimension, the probability that a random variable $X$ is greater than some value $x$ is easily related to the cumulative distribution(c.d.f.) as $Pr(X &gt; x) = 1 - F(x)$ if only $Pr[X \leq x] = F(x)$ is known.  If instead we have a random vector $[X_1, X_2, \cdots, X_n]$, could you please let me know if there is a direct relation between the c.d.f and the probability $Pr[X_1 &gt; x_1, X_2 &gt; x_2, \cdots, X_n &gt; x_n]$ if I only know $F(x_1, x_2, \cdots, x_n) = Pr[X_1 \leq x_1, X_2 \leq x_2, \cdots, X_n \leq x_n]$? I know it is not $1 - F[x_1, x_2, \cdots, x_n]$ in this case.
25755
Suppose I have two discrete-support random variables, $X$ and $Y$.   They have joint CDF $F(X,Y)$.  If I want to find  $\Pr(a \leq X \leq b , c \leq Y \leq d)$.  It is obviously not:  $F(b ,d)-F(a-1 ,c-1)$.  What is it then, in simplest terms with respect to the CDF?  I believe, for two dimensions, it is:  $F(b ,d)-F(a-1 ,d)-F(b,c-1)+F(a-1,c-1)$.  What is it for an arbitrary $d$ dimensions?
22199
Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then $$ E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x). $$  of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.  Why do the medians solve $$ \min_{c \in \mathbb{R}} E(|X-c|) \, ? $$
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I'm learning probability, specifically transformations of random variables, and need help to understand the solution to the following exercise:      Consider the continuous random variable $X$ with probability density function $$f(x) = \begin{cases} \frac{1}{3}x^2 \quad  -1 \leq x \leq 2, \\ 0 \quad \quad \text{elsewhere}. \end{cases}$$ Find the cumulative distribution function of the random variable $Y = X^2$.   The author gives the following solution:   For $0 \leq y \leq 1: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-\sqrt y \leq X \leq \sqrt y) = \int_{-\sqrt y}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{2}{9}y\sqrt y.$  For $1 \leq y \leq 4: F_Y(y) = P(Y \leq y) = P(X^2 \leq y) \stackrel{?}{=} P(-1 \leq X \leq \sqrt y) = \int_{-1}^{\sqrt y}\frac{1}{3}x^2\, dx = \frac{1}{9} + \frac{1}{9}y\sqrt y.$  For $y &gt; 4: F_{Y}(y) = 1.$    Previous to this exercise, I've managed to follow the solutions of two similar (obviously simpler) problems for a strictly increasing and strictly decreasing function of $X$, respectively. However in this problem, I don't understand the computations being done, specifically:    How does the three intervals $0 \leq y \leq 1$, $1 \leq y \leq 4$ and $y &gt; 4$ are determined? In the two previous problems I've encountered, we only considered one interval which was identical to the interval where $f(x)$ was non-zero. In the case where $0 \leq y \leq 1$, why does $P(X^2 \leq y) = P(-\sqrt y \leq X \leq \sqrt y)$ and not $P(X \leq \sqrt y)$? I have put question marks above the equalities that I don't understand.   I think I have not understand the theory well enough. I'm looking for an answer that will make me understand the solution to this problem and possibly make the theory clearer.
31422
Suppose that $X\sim N(0,1)$, $Y\sim N(0,1)$ with correlation $(X, Y) =ρ$ where $ρ ∈ (−1, 1)$.  Show the following,   Correlation $(Φ(X),Φ(Y))=\dfrac6π \arcsin\dfrac ρ2 $.  Here $Φ(X)$, and $Φ(Y)$ denote the CDF of Random variables $X$ and $Y$ respectively.  What I know so far:  $$\text{Cov}( Φ(X),Φ(Y)) =E(Φ(X)Φ(Y)) - E(Φ(X)) ×E(Φ(Y)) $$  We also know since  $X \sim N(0,1)$ and $Y\sim N(0,1)$;  $Φ(X)\sim\text{unif}(0,1)$, $Φ(Y)\sim\text{unif}(0,1)$. Hence $$E(Φ(X)) =E(Φ(Y)) =\frac12.$$  I am stuck in finding $E(Φ(X)Φ(Y))$ [I tried to find it using double expectation].  What I have written may be the wrong way of attempting the question. So I am thankful for any help.
75088
Say I have two independent random variables $X \sim N(u_1, \sigma_1)$ and $Y \sim N(u_2, \sigma_2)$. I want to get the conditional distribution of X given whether X is bigger than Y or not.  $P(X|X&lt;Y)$ = ... and  $P(X|X&gt;Y)$ = ...  I am thinking solving this in this way:  \begin{align} P(X|X&lt;Y) &amp;= \frac{P(Y&gt;X|X)P(X)}{P(Y&gt;X)} \\ &amp;= \frac{(1-\Phi(\frac{x-\mu_2}{\sigma_2}))N(\mu_1,\sigma_1)}{\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})}\\ P(X|X&gt;Y) &amp;= \frac{P(Y&lt;X|X)P(X)}{P(Y&lt;X)} \\ &amp;= \frac{\Phi(\frac{x-\mu_2}{\sigma_2})N(\mu_1,\sigma_1)}{1-\Phi(\frac{\mu_2-\mu_1}{\sqrt{\sigma_2^2+\sigma_1^2}})} \end{align}  My questions are:   (1) Whether above solution is correct  (2) How to get the mean and sd for $P(X|X&lt;Y)$ and $P(X|X&gt;Y)$ if they are still normal?
22029
$$P(X_1 &lt; X_2 &lt; X_3)= \frac{(1-p_1)(1-p_2)p_2p_3^2}{(1-p_2p_3)(1-p_1p_2p_3)}$$   To be frank I do not know where to start with this question, I would like an idea to get me going, or better yet an answer. This was on an earlier exam, I would like to be ready if something like this were to come up again tomorrow. What makes this a little bit more difficult than usual is the fact that geometric distribution is  a discrete variable..
3975
Suppose $X$ is a random variable with pdf $f_X(x)$. Then the random variable $Y=X^2$ has the pdf   $$f_Y(y)=\begin{cases}\frac{1}{2\sqrt{y}}\left(f_X(\sqrt{y})+f_X(-\sqrt{y})\right) &amp; y \ge 0 \\ 0 &amp; y \lt 0\end{cases}$$   I understand the calculus behind this. But I'm trying to think of a way to explain it to someone who doesn't know calculus. In particular, I'm trying to explain why the factor $\frac{1}{\sqrt{y}}$ appears out front. I'll take a stab at it:  Suppose $X$ has a Gaussian distribution. Almost all the weight of its pdf is between the values, say, $-3$ and $3.$ But that maps to 0 to 9 for $Y$. So, the heavy weight in the pdf for $X$ has been extended across a wider range of values in the transformation to $Y$. Thus, for $f_Y(y)$ to be a true pdf the extra heavy weight must be downweighted by the multiplicative factor $\frac{1}{\sqrt{y}}$  How does that sound?  If anyone can provide a better explanation of their own or link to one in a document or textbook I'd greatly appreciate it. I find this variable transformation example in several intro mathematical probability/stats books. But I never find an intuitive explanation with it :(
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This theorem seems to hold great potential for an intuitive explanation, yet I'm still struggling for it.  Basically, it is saying that if $y = F(x)$, then y would have distribution $Unif[0,1]$. Intuitively, why?  Many thanks!
51403
Let $X\sim N(0,1)$ and let $Z$ be a random variable independent of $X$ such that:  \begin{equation} \Pr(Z=z) = \begin{cases} \frac{1}{2} &amp; \mbox{if $z = -1$ or $z=1$}, \\ \\ 0 &amp; \mbox{otherwise}. \end{cases} \end{equation}   Define $Y = ZX$. Show that $Y \sim N(0,1)$. Determine $\operatorname{Cor}(X,Y)$. Are $X$ and $Y$ independent?   Progress:  I'm basically stuck at the beginning. I got as far as $P(ZX &lt; y) = P(Z &lt; y/X)$, which I'm not even sure is correct. I'm also not clear on what it means by defining $Y = ZX$ - does it mean I need to state the probability density function or something? And how do I show that $Y \sim N(0,1)$?
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.
73237
This is what i know about symmetric distributions: The distribution of rv (random variable) $X$ is symmetric about $a$ iff $$  P ( X \le a - x ) = P ( X \ge a + x )  \qquad \forall x  \in \mathbb{R} $$ I just want to confirm ,.,is this " iff " correct ? i mean it is clear to me that if $x$ is symmetric about $a$ it will satisfy above equation, is the converse also true ?
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If $x_1,x_2,x_3$ and $x_4$ are independent identically distributed  random variable with continuous distribution .what is the probability of $$P=p(x_1 &lt; x_2&lt; \max(x_3,x_4))$$ I know the exhaustive number of cases are $p(x_1&lt;x_2&lt;x_3)+p(x_1&lt;x_2&lt;x_4)$ and  every $x$ are iid so both have same probabiltiy so is $2p(x_1&lt;x_2&lt;x_3)$ is right? But  i get the $p(x_1&lt;x_2&lt;x_3) = 1/6$ and then my answer is $1/3$ but the answer is $1/6.$ Can we write $P$ as $p(x_1&lt;x_2&lt;x_3&lt;x_4)+p(x_1&lt;x_2&lt;x_4&lt;x_3)$?
62709
Let $Z$ be a random uniformly distributed variable on $[0,1]$. Show that the random variables $X = \sin 2\pi Z$ and $Y = \cos 2\pi Z$ are not independent, but nevertheless Cov$[X,Y]=0$.  This is a homework assignment, but I'm a bit stuck.  My thoughts  We can see that $X$ and $Y$ are not independent, since both depend on $Z$. If we want to show this explicitly, then we need to show that $$f_{X,Y}(a,b) \neq f_X(a)\;f_Y(b),$$ where $f_{X,Y}(a,b)$ is the joint probability distribution function. But how can I find the (joint) probability distribution function(s) $f_X, f_Y$ and $f_{X,Y}$?   If I can find these functions, I can also solve the covariance problem. Is this the right way? Or is there a 'better' way to solve this problem?
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... and why ?  Assuming $X_1$,$X_2$ are independent random-variables with mean $\mu_1,\mu_2$ and variance $\sigma^2_1,\sigma^2_2$ respectively. My basic statistics book tells me that the distribution of the $X_1-X_2$ has the following properties:    $E(X_1-X_2)=\mu_1-\mu_2$ $Var(X_1-X_2)=\sigma^2_1 +\sigma^2_2$   Now let's say $X_1$, $X_2$  are t-distributions with $n_1-1$, $n_2-2$ degrees of freedom. What is the distribution of $X_1-X_2$ ?   This question has been edited: The original question was "What are the degrees of freedom of the difference of two t-distributions ?". mpiktas has already pointed out that this makes no sense since $X_1-X_2$ is not t-distributed, no matter how approximately normal $X_1,X_2$ (i.e. high df) may be.
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I have two independent variables: $X$ follows from standard Gaussian distribution $N(0,\sigma^2)$; $Y$ follows from Rademacher distribution, i.e., $Y$ can be either $-1$ or $1$ with the same probability.  My question is:  For a new random variable of $Z=XY$, What is the probability density function?   Or can we have some concentration inequalities for $Z$ (e.g., show how $Z$ approaches its expectation, etc.)?
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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.)
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Let $X$ be a random variable that follows $Uniform(-1,2)$ distribution.   Then what will be the cdf of $X^2$?   What i did was $$P(X^2\le y)=P(X\le y^{0.5})= \int_{-1}^{y^{0.5}}(1/3)dx=(1/3)(y^{0.5}+1) $$ $0\lt y \lt 4$ But this cdf doesnt give the probability $0$ at $y=0$ also the pdf which i get doesn't integrate to 1 What is wrong can somebody explain?? Am i doing something wrong??