Hot answers tagged

47

We see that in probability, we represent the event $A$ as a set of elements in our sample space, and $\neg A$ as the complement of $A$ in our sample space. Thus, in probabilistic terms, $$P(A \wedge \neg A) = P\left(A \cap \overline{A}\right) = P(\emptyset)$$ by the definition of the complement. And by the Kolmogorov Axioms, we see that $$P(\emptyset) = ...


12

We have this sum: $$ \sum_{k=0}^n k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} \tag 1 $$ First notice that when $k=0$, the term $k\dfrac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k}$ is $0$, and next notice that when $k\ne0$ then $$ \frac k {k!} = \frac 1 {(k-1)!} $$ so that $$ k\frac{n!}{k!\,(n-k)!}p^k(1-p)^{n-k} = \frac{n!}{(k-1)!(n-k)!} p^k (1-p)^{n-k}. $$ The two expressions ...


11

He probably meant that either: 1. The space of continuous functions which are nowhere monotonic has probability one using Wiener measure. I.e. a continuous function is "almost surely" nowhere monotonic using the standard probability measure for that space. 2. That set of nowhere monotonic continuous functions is of the first category (Baire category ...


8

I think that's a misunderstanding. The passage you quote is a bit informally phrased. By "the mean time to failure of some disk", they don't mean the mean time to failure of each individual disk but the mean time it takes until any one of the $100$ disks fails. If the failures form a Poisson process, then having $100$ disks instead of $1$ disk will increase ...


7

The trick is the using the identity $k { n \choose k} = n {n-1 \choose k-1}$. $$\begin{align*} &\sum_{k=1}^n k { n \choose k } p^k (1-p)^{n-k}\\ &= \sum_{k=1}^n n { n-1 \choose k-1} p^k (1-p)^{n-k}\\ &=np \sum_{k=1}^n {n-1 \choose k-1} p^{k-1} (1-p)^{(n-1)-(k-1)}\\ &=np (p+(1-p))^n\\ &=np \end{align*}$$


6

The answer in the link above using the concept of sigma-algebra is the most general approach and is the best. If you have not studied that yet, here I provide another approach involving the CDF and law of total probability. Note that $X, Y, Z$ are mutually independent if and only if $$ \Pr\{X \leq x, Y \leq y, Z \leq z\} = \Pr\{X \leq x\}\Pr\{Y \leq ...


6

$$\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}\varphi_X(t_k-t_{\ell})=E\left(\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}e^{i(t_k-t_{\ell})X}\right)=E\left(\left|\sum_k\xi_ke^{it_kX}\right|^2\right)\geqslant0$$


5

$S_{n}=n\bar{X}_{n}$ where $\bar{X}_{n}:=\frac{1}{n}\left(X_{1}+\cdots+X_{n}\right)$ and $P\left(\bar{X}_{n}\to\mu\right)=1$ according to the strong law of large numbers. If $\bar{X}_{n}\left(\omega\right)\to\mu>0$ then $S_n(\omega)=n\bar{X}_{n}\left(\omega\right)\rightarrow+\infty$. So: $$\left\{ \bar{X}_{n}\to\mu\right\} \subseteq\left\{ ...


5

You should stay with your initial choice until $N-2$ doors have been opened. Then switch to the single door you can switch to. With this strategy you are sure to win except when your initial choice happened to be the prize door. In other words, your chance of winning is $\frac{N-1}{N} = 1-\frac1N$. If you switch any earlier, your chance of having the ...


5

One way to understand this is that nowhere monotonic functions form a "prevalent" set in the space of all continuous functions on, say, an interval, in the sense of Hunt, Sauer and Yorke.


5

Here is one possible interpretation of what he meant: every continuous, real-valued function on an interval can be approximated as well as you like by continuous, nowhere monotonic functions. In other words, these functions are dense. In more detail, consider the space $C([0,1])$ (for example) of continuous functions $[0,1]\to\mathbb R$. To talk about the ...


5

This would be true if $g$ were continuous with no other assumptions. If you make $g$ discontinuous then "generically" this convergence fails. To be more specific, here is a simple approach: assume $X_n$ converges a.s. to some fixed constant $c$, but that the $X_n$ are never equal to $c$. Then define $g(c)$ to have one value and $g(x)$ to have some other ...


5

In discrete time at least, the definitions I'm familiar with are fairly straightforward. Given an increasing filtration $\{\mathcal{F}_n\}_{n=0}^{\infty}$, a process $\{X_n\}_{n=0}^{\infty}$ is adapted if each $X_n$ is $\mathcal{F}_n$-measurable. For predictable processes, the random variables are measurable with respect to slightly smaller ...


4

Write $\cos(a_j t) = (\exp(i a_j t) + \exp(-i a_j t))/2$, and expand the product. We get $$ 2^{-n} \sum_{x \in \{-1,1\}^n} \exp \left( i \sum_{j=1}^n x_j a_j t\right) $$ Now note that $$\dfrac{1}{2\pi} \int_0^{2\pi} \exp(ikt)\; dt = \cases{0 & if $k$ is a nonzero integer\cr 1 & if $k = 0$\cr}$$ Thus, assuming the $a_j$ are all integers, your right ...


4

Compare it to lottery. If you own one ticket, the probability to win is very low. If you own 100 tickets it's much more likely that you win (something). Likewise, the probability that one disk fails is quite low but the probability that any of the 100 disks fails is higher, and therefore the mean time to failure is lower.


4

Note that $Y=|X|$ is a random variable that only takes on non-negative values. So the lemma holds for the random variable $Y$. Because $\lim_{y\to\infty}y\Pr(Y\gt y)=0$, there is a $B\ge 1$ such that if $y\ge B$ then $\Pr(Y\gt y)\le \frac{1}{y}$. Let $p=1-\epsilon$. By the lemma, $$E(Y^{1-\epsilon})=\int_0^B (1-\epsilon)y^{-\epsilon}\Pr(Y\gt y)\,dy+ ...


4

This is a very valid question, and one that I didn't really touch until my second time learning undergraduate probability. I am by no means an expert; this is merely the intuition I've developed. $X$ is a function. What they don't tell you in undergraduate-level probability is that $$X: \Omega \to \mathbb{R}$$ is a "random variable," where $\Omega$ is the ...


4

Change variables $s=tu$: $$\int_0^t e^{B_s} \, ds = \int_0^1 e^{B_{t u} }t \, du=(*)$$ Use scaling: $(B_{tu}:u \ge 0) \overset{\mbox{dist}}{=}\sqrt{t}( B_{u}:i \ge 0)$. Therefore $$ (*) \overset{\text{dist}}{=} \sqrt{t} \int_0^t e^{\sqrt{t} B_u} \, du $$ Take $\sqrt{t}$-th root: $$t^{1/\sqrt{t}} \left(\int_0^1 e^{\sqrt{t} B_u} \, ...


4

NOTE: Referring to the note at the end of the General Proof Section of THIS ARTICLE, L'Hospital's Rule states that given two functions $f$ and $g$ that are differentiable in an open neighborhood of $\xi$ with $g'(x)\ne 0$ in that neighborhood, such that $\lim_{x\to \xi}|g(x)|=\infty$, then if the limit $\lim_{x\to \xi}\frac{f'(x)}{g'(x)}$ exists, ...


4

The second player has 8 choices for his first move, 6, for the second, etc., so in any game he has a total of $8 \times 6 \times 4 \times 2 = 384$ possible sequences of moves. If only one choice is correct at each point, which is an underestimation, then in 5 million games he would be expected to draw at least $\lambda = 5 \times 10^6 / 384 \approx 1.3 ...


4

This is easiest if we treat the people as being distinguishable. There are $7$ floors each person can go to, and since the floors are chosen independently, there are $7^5$ total ways for the people to get off of the elevator. There are $\binom{7}{5} = 21$ ways to choose different floors for all $5$ people, and having chosen these floors, there are $5!$ ...


4

Calling the persons $P_i$ there is a probability of $7^{-5}$ that person $P_i$ steps out on floor $i$ for every $i\in\{1,2,3,4,5\}$. There are $5!$ rearrangements for the persons that result in a step out of exactly one person at the floors $1,2,3,4,5$. Moreover there are $\binom75$ ways to choose $5$ floors. So we end up with probability ...


4

The answer is YES. Let me introduce some notations first. Suppose $X=(X_1,\cdots,X_n)$ is an $\mathbb{R}^n$-valued random variable (i.e., a random vector). The characteristic function for $X$, denotes as $\varphi_X(u)$, is a function from $\mathbb{R}^n$ to $\mathbb{R}$: $$\varphi_X(u):=E(e^{iu\cdot X}),\quad u\in\mathbb{R}^n$$ where $u\cdot X$ is the ...


3

Note that $X$ is a finite set. So given $x\in X$, the set $B_x= \{ S \in \mathcal{F} : x\in S\}$ is finite. Since $\mathcal{F}$ is a $\sigma$-algebra, we have $X\in B_x$ ($B_x$ is not empty) and $$A_x=\bigcap_{S\in B_x} S \in \mathcal{F}$$ Clearly $x\in A_x$, and $A_x$ is the smallest element in $\mathcal{F}$ containing $\{x\}$. Note that if $y\in X$ and ...


3

"Theory of Stochastic Processes", Gusak et al., Springer, 2010: Problem 1.10: Prove that it is impossible to construct on the probability space $\Omega = [0, 1]$, $\mathcal{F}=\mathcal{B}([0,1])$, $\mathsf{P}=\lambda$ a family of independent identically distributed random variables $\{\xi_t, t\in[0,1]\}$ with a nondegenerate distribution. ($\lambda$ ...


3

This follows from a generalized dominated convergence theorem. Let $\zeta_n:=X_n^k$ and $\zeta:=X^k$. Since $\mathbb{E}|\zeta_n|\le c_k[\mathbb{E}|X_n-X|^k+\mathbb{E}|\zeta|]$ and $k\le p$, $\zeta_n\in L^1$ ($\zeta\in L^1$ by assumption). In addition, you've already shown that $\mathbb{E}|\zeta_n|\to\mathbb{E}|\zeta|$. Thus, for any subsequence $n_m$, there ...


3

The mean time to failure of some disk is not the mean time to failure of those disks: it is the mean time before at least one of those 100 disks fails. In other words, the time to failure of some disk $T$ is the minimum of the time to failure of all disks, i.e. $$T = \min_{i \leq i \leq N} \{X_i\}.$$ A standard of model is that aging has no memory ...


3

$$ P\{|X_n-Y_n|>\epsilon\}\le P\{|X_n-X|+|Y_n-Y|+|X-Y|>\epsilon\} \\ \le P\{|X_n-X|>\epsilon/3\}+P\{|Y_n-Y|>\epsilon/3\}+P\{|X-Y|>\epsilon/3\} $$


3

You meant $\{X_n\}$ are uniformly integrable if $$ \lim_{M\to\infty} \sup_n E[ |X_n|\chi_{\{|X_n|>M\}}]=0$$ (e.g. https://en.wikipedia.org/wiki/Uniform_integrability) Now for your question. Answer is NO. Why ? In general the random variables are not in $L^p$. Easy example to remember $X_n = X$, where $X$ is a random variable in $L^1$ but not in ...


3

Partial answer: Simple cases: If we generate only two points, this is like fixing one as north pole and generating the other randomly. Here, the expected angular distance is "clearly" $\frac\pi2$ because points at angular distances $\theta$ are just as likely as points at distance $\pi-\theta$. The same argument holds for the $m$th closest neighbour when ...



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