Okay so this might be a basic question/stupid question, but I cannot find an answer for it.   So let's say we have a continuous random variable that returns a value in 0, 1. I know that the probability to observe any single given point in this interval is zero (one point over infinite). How come I can observe this point that has no probability to be observed ?
123763
Can an observed event in fact be of zero probability?  Of course, I know that there exist non-empty events of zero probability. Mu question is the reverse: given that we have observed an event (and we have no other information about it, just the fact that it has been observed), is it possible that the event has in fact zero probability? Or does an observation necessarily mean that the probability is strictly positive?    Example context for the question:  Suppose $x_i$ (countably many) are i.i.d on $[0,1]$, but we do not know the distribution they come from. It may be uniform on $[0,1]$, may be discrete, may be any legitimate distribution (discrete or continuous). Just imagine we have some sort of a machine that shows us one by one a list of randomly drawn numbers between $0$ and $1$. We are comparing the observed numbers one by one to some special number chosen beforehand, for example $\frac12$. Now, given that at some iteration we have observed that special number at least once, does that mean that $\frac12$ has some positive probability under that (unknown) distribution? And if the probability can be zero, can we nevertheless say that we will necessarily observe $\frac12$ again later, if we continue the experiment ad infimum?  Also, disregard the "real world limitations" such as an inability to produce truly uniformly distibuted numbers, or rounding errors or any such thing.
62327
On googling I got     Random numbers are numbers that occur in a sequence such that two conditions are met: (1) the values are uniformly distributed over a defined interval or set, and (2) it is impossible to predict future values based on past or present ones. Random numbers are important in statistical analysis and probability theory.   What if I want to pick a random integer from the number line ? The number line consists of infinitely many integers. My favorable event is to get a number (random) and the number of events possible is the complete number line i.e. infinity. I won't be considering any limits here as our number line doesn't tend to infinity but does contain infinitely many elements.  Now the probability of getting a random number  P(E) = 1 / ∞ = 0   So I won't be able to get a random number, ever. This implies, never would I ever be able to get a random number.  That's absurd as many of our science topics do depends on things being random.
32582
What is the intuitive difference between a random variable converging in probability versus a random variable converging in distribution?  I've read numerous definitions and mathematical equations, but that does not really help. (Please keep in mind, I am undergraduate student studying econometrics.)  How can a random variable converge to a single number, but also converge to a distribution?
76579
Is there a way to resolve probability of an event, given another event that never happens? Mathematically speaking the problem is:  Given that $P(B) = 0$,  $$P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{0}{0}$$  Is this probability vacuously $0$ of $1$? Can we show that it's one or the other?
6196
I would like to learn how to calculate the expected value of a continuous random variable. It appears that the expected value is $$E[X] = \int_{-\infty}^{\infty} xf(x)\mathrm{d}x$$ where $f(x)$ is the probability density function of $X$.  Suppose the probability density function of $X$ is $$f(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}$$ which is the density of the standard normal distribution.  So, I would first plug in the PDF and get  $$E[X] = \int_{-\infty}^{\infty} x\frac{1}{\sqrt{2\pi}}e^{\frac{-x^{2}}{2}}\mathrm{d}x$$ which is a rather messy looking equation. The constant $\displaystyle\frac{1}{\sqrt{2\pi}}$ can be moved outside the integral, giving  $$E[X] = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} xe^{\frac{-x^{2}}{2}}\mathrm{d}x.$$  I get stuck here. How do I calculate integral? Am I doing this correctly this far? Is the simplest way to get the expected value?
56775
In my probability class the terms "sums of random variables" is constantly used. However, I'm stuck on what exactly that means?  Are we talking about the sum of a bunch of realizations from a random variable? If so, doesn't that add up to a single number? How does a sum of random variable realizations lead us to a distribution, or a cdf / pdf / function of any kind? And if it isn't random variable realizations, then what exactly is being added?
76998
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.
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?
29888
Wikipedia defines convergence in probability as  A sequence ${X_n}$ of random variables converges in probability towards the random variable $X$ if for all ε > 0 $$\lim_{n\rightarrow\infty} P(|X_n-X|&gt;\varepsilon) = 0$$  I wonder why the limit $X$ is a random variable. I think $X$ as a r.v. can only be a constant, and defining convergence to a number $a$ instead of a r.v. $X$  $$\lim_{n\rightarrow\infty} P(|X_n-a|&gt;\varepsilon) = 0$$  makes the definition more clear.
46087
Given that $X_1, X_2...$ are mutually independent random variables. For each $i$ with $1\le i \le n$   the variable $X_i$ is equal to either $0$ or $n+1$ $E(X_i)$ = $1$   also.. if $X_i$ is equal to either $0$ or $n+1$, doesn't that mean that all $X_i$ need to be $0$?  Thanks
11813
Let $X$ and $Y$ be independent and identically distributed.    Show that if $X$ and $Y$ are continous, then $P(X&lt;Y) = 1/2$ Give an example of two IID RVs $X$ and $Y$ such that $P(X&lt;Y)\neq 1/2$   I can't make sense of ncmathsadist's answer. Can somebody clarify?
42263
Let $X$ be a real valued random variable.  Let $\operatorname{sgn}(x)$ be $1$ when $x&gt;0$, $-1$ when $x&lt;0$ and $0$ when $x=0$.  Why  are $|X|$ and $\operatorname{sgn}(X)$ independent, when the density function of $X$ is symmetric with respect to $0$?  Are $|X|$ and $\operatorname{sgn}(X)$ independent, when the density function of $X$ is not necessarily symmetric with respect to $0$?  Thanks!
34282
I have random variables $X_{1}, X_{2}, \dots X_{n}$ that are each Unif$\{1, \dots, n\}$, and I want to calculate $$P(\{X_{1}, X_{2}, \dots X_{n}\}=\mathbb{N})$$  My intuition is that this means each $X_{n} = n$, which occurs with probability $\frac{1}{n!}$.  Does this seem correct?
70398
Let $X$ be a continuous random variable defined on the interval $[0,1]$ with density function $f_X(u)=c(1−u^2)$ for some suitable constant $c$.   Find: (a) the value of $c$, (b) (i) the expectation and (ii) the variance of $X$. Also find (c) the density function of the random variable $Y = X^{2}$.   Now I've already attempted (a) by integrating $(1-u^{2})$ between $0$ and $1$ (like $\pi$ in the Cauchy density function) and got $c=\frac{2}{3}$ and also for (b) (i) I integrated $uc(1-u^{2})$ between $0$ and $1$ which came out to be $c/4 = 1/6$.  Have I done these first $2$ parts correctly? If so how do I go about doing the rest and if not same question but also where did I go wrong?
68381
Question: If $n^r(X_n-X)$ converges in distribution for some $r&gt;0$, do we have that $X_n$ converges to $X$ in probability?    As $P(n^r|X_n-X|&gt;\varepsilon)\to P(|Z|&gt;\varepsilon)$ (where $Z$ is the random variable to which $n^r(X_n-X)$ converges), can we say something about $P(|X_n-X|&gt;\varepsilon)$?
56137
There is a probabilistic method to solve it. But I am not familiar with probability. I am trying to compute it by analytic method, such as using L Hospital's rule or Stolz formula, but they are not working.
7579
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!