Can you find a real-valued random variable $X$ that is almost surely rational but also such that, for any $a, b ∈ \mathbb{Q} \cap [0, 1]$ with $a &lt; b$, $p(X ∈ [a, b]) = b - a$? If not, can you prove that no such $X$ exists?
121480
I just had an (intellectual) panic attack.     A continuous random variable that follows a uniform in a closed interval $U(a,b)$: a comfortably familiar statistical concept.   A continuous uniform r.v. having support over the extended reals (half or whole): not an r.v. proper, but a basic Bayesian concept for an improper prior, useful and applicable. A discrete uniform taking a finite number of values: let's throw a geodesic dome, no big deal.     But what about a function that has as its domain all the rationals that are included in a closed interval with integer bounds (start with the $[0,1]$ if you wish)? And we want to use it in a probabilistic framework, requiring that each possible value has equal probability with all the others?   The number of possible values are countably infinite (which characterizes a lot of discrete distributions), but how to express the probability of a single value given that we want probabilities equal?  Can we say-show-prove that such an entity is (is not) a random variable?    If not, is this another incarnation (perhaps already well-known) of an "improper prior"?  Is it possible that this entity is in some well-defined sense, however special,  "equivalent" to a continuous uniform r.v.? Or I just committed a cardinal(ity) sin?  It appears that the fact that the domain is a closed interval does not let me let go. Bounded things are usually manageable.  The questions are many in order to be indicative of the internal maelstrom- I am not asking to get answers to each one of them.    At any time that I may come up with any insights, I will update.    UPDATE: the present question just acquired a constructivist sequel
49133
Show that: $$P(N(x)&gt;n)= \frac{x^n}{n!}$$ I was thinking that we can take $X_1+...+X_n$ to be a single random random variable $S_n$ and that the $n$ in the $P(N(x)&gt;n)$ is not the same as the $n$ in the definition of $N(x$). But unsure how to actually start proceeding to prove this. Any help appreciated, thank you.
115817
I am curious about whether $|x|^p$ with $0&lt;p&lt;1$ satisfy $|x+y|^p\leq|x|^p+|y|^p$ for $x,y\in\mathbb{R}$.   So far my trials show that this seems to be right... So can anybody confirm whether this is right or wrong and give a proof or counterexample?
70947
I know that for continuous variable $P[X=x]=0$.   But i can't visualize that if $P[X=x]=0$, there is infinite number of possible $x$'s. And also why do their probabilities get infinitely small ?
53948
Doing a little reading over the break (The Probabilistic Method by Alon and Spencer); can't come up with the solution for this seemingly simple (and perhaps even a little surprising?) result:     (A-S 1.6.3) Prove that for every two independent identically distributed real random variables $X$ and $Y$,   $$Pr[|X-Y| \leq 2] \leq 3 Pr[|X-Y| \leq 1].$$
112356
If $Y$ and $Z$ are two distinct random variables with the same distribution (for example maybe $Y$ is constant equal to $1$ and $Z$ is equal to $1$ almost everywhere), then surely any sequence $X_n$ which converges in distribution to $Y$ also converges to $Z$. This seems to conflict with the fact that convergence in distribution is metrizable, and therefore limits should be unique.  What about convergence in probability, or almost sure convergence?
90558
The full question is,     If X is a Geometric random variable with parameter p, what value of k ∈ {1, 2, . . .}   maximizes P(X = k) = (1 − p)^(k-1) * p? This is referred to as the mode of the distribution.   Through trial and error (plugging in random values), I found that p = 1/2 yields the highest value of P(X=k).  Is there a way to figure this out without trial and error?
55963
I'm trying to show that for $b&gt;1$, $x&gt;0$ and $x$ irrational, that $$\sup \{b^p:p\in Q,\;0&lt;p&lt;x\} = \inf\{b^q:q\in Q,\;x&lt;q\}$$ I know this follows immediately if we define $b^x=\exp(x\log(b))$ and use the fact that the exponential function is strictly increasing and continuous over $\mathbb R$, but I'm not allowed to do that. In fact, I only have that $b^{m/n}=(b^{1/n})^m$, where $b^{1/n}$ means the positive $n^{th}$ root of $b$; $b^x$ is not yet defined for irrational $x$.  I know that the supremum and infimum both exist because if $b&gt;1$ and $p&lt;q$, then $b^p&lt;b^q$. Also, by assuming that the inf $&lt;$ sup, I got a contradiction. The part I'm stuck on is deriving a contradiction from assuming that sup $&lt;$ inf. Does anybody have any ideas?  Thank you.
4614
I'm facing the following issue. Let $X$ be an integrable random variable on the probability space $(\Omega,\mathcal{F},\mathbb{P})$ and $\mathcal{G},\mathcal{H} \subseteq \mathcal{F}$ be two sigma-algebras. We assume that $X$ is independent of $\mathcal{G}$, i.e. $\sigma(X)$ is independent of $\mathcal{G}$. Can I say (and can I prove) that $$ E(X \mid \sigma(\mathcal{G} \cup \mathcal{H})) = E(X\mid \mathcal{H}) ?$$  Thank you very much for your help!
97524
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!!!
90494
We consider a finite field $\mathbb{F}_q$ where $q=2p+1$ and $q$ and $p$ are prime numbers.  Let $r_i$ be a value picked uniformly at random from the field such that $r_i&gt;\frac{q}{2}$. Let $z$ be a fixed field element.   Question: Is $v=(\ (r_i)^{-1}\cdot z) \bmod q$, a uniformly random value of the field?
27690
Is there a measurable set $A$ such that $m(A \cap B)= \frac12 m(B)$ for every open set $B$?  Edit: (t.b.) See also  for further answers.
12085
The following is a homework question for which I am asking guidance.     Let $A$, $B$, $C$ be independent random variables uniformly distributed between $(0,1)$. What is the probability that the polynomial $Ax^2 + Bx + C$ has real roots?   That means I need $P(B^2 -4AC \geq 0$). I've tried calling $X=B^2 -4AC$ and finding $1-F_X(0)$, where $F$ is the cumulative distribution function.  I have two problems with this approach. First, I'm having trouble determining the product of two uniform random variables. We haven't been taught anything like this in class, and couldn't find anything like it on Sheldon Ross' Introduction to Probability Models.  Second, this strategy just seems wrong, because it involves so many steps and subjects we haven't seen in class. Even if I calculate the product of $A$ and $C$, I'll still have to square $B$, multiply $AC$ by four and then subtract those results. It's too much for a homework question. I'm hoping there might be an easier way.
95318
I'm trying to show that if $\alpha$ is the square root of a non-perfect square $d$, then for any rational $p/q$ there's a positive real $x$ such that $|\alpha - p/q| &gt; x/q^2$. I'm having trouble starting on the problem, could someone give a hint? Thanks
102600
Finding the closure of A=$\mathbb Q \cap [0,1]$.  My attempt: Between every two rational numbers, there is a rational number. That means that for  any arbitrarily small $\epsilon \gt0$, and some $q\in \mathbb Q \cap [0,1]$  there is $x \notin \mathbb Q$, $x\in [0,1]$ so that $x\in (q-\epsilon, q+\epsilon)$. That means - and I'm not sure if that's correct and how to show it - there is a sequence $\{a_n\}_{n=1}^\infty \subseteq \mathbb Q \cap [0,1]$ so that $\lim_{n \rightarrow \infty} a_n=x$.   Therefore, the closure of A is $[0,1]$.   Is this correct? And how do I show my missing step (if it's valid?).  Thanks for any assistance in advance!
90539
Apparently, the solution to the  is that a uniform probability distribution over the positive real numbers doesn't exist. Can anyone explain why this is the case and what probability distributions can exist over the positive real numbers (it seems that this would be quite limited, given that such a simple distribution is impossible)?
94260
Suppose that $X$ is a real-valued random variable, with $EX^4=1$ and $EX\leq0$. Find an explicit constant $c&lt;1$ such that $EX^3&lt;c$.