# Tag Info

10

Let me get you started: if $(X,Y)$ is i.i.d., $(Y,X)$ is distributed like $(X,Y)$ hence $Y-X$ is distributed like $X-Y$ hence $E(u(Y-X))=E(u(X-Y))$ for every $u$ such that the expectations exist. For $u(t)=t^3$ (or any odd function), this yields $E((X-Y)^3)=E((Y-X)^3)=-E((X-Y)^3)$ hence... Thus, the answer is "Yes, provided $E(|X|^3)$ is finite" (otherwise ...

5

Since events $\{X \geqslant Y\}$ and $\{X < Y\}$ are disjoint: $$\begin{eqnarray} \Pr(\max(X,Y) > a \min(X,Y)) &=& \Pr(\max(X,Y) > a \min(X,Y), X \geqslant Y) \\ && + \Pr(\max(X,Y) > a \min(X,Y), X < Y) \\ &=& \Pr(X > a Y, X \geqslant Y) + \Pr(Y > a X, X < Y) \end{eqnarray}$$ Now proceed graphically. Since ...

5

The maximum $M_n$ of interest is such that $$(1-2/\sqrt{n})K_n\leqslant M_n\leqslant1,\quad\text{where}\quad K_n=\max\{X_i\mid 1\leqslant i\leqslant\sqrt{n}\}.$$ By independence, $P(K_n\leqslant x)=P(X_1\leqslant x)^{\sqrt{n}}=x^{\sqrt{n}}\to0$ when $n\to\infty$, for every $x$ in $(0,1)$. Hence $K_n\to1$ in probability, $(1-2/\sqrt{n})K_n\to1$ in ...

4

The probability model of quantum mechanics is different from the Kolmogorov model, is due to von Neumann, and actually predates the Kolmogorov model. The basic model goes something like this: events are represented by the lattice of projectors on a Hilbert space. The elementary outcomes are the one-dimenstional projectors. Probabilities are assigned to ...

4

No it doesn't. In fact, if $f$ is strictly convex (e.g. $f(x)=x^2$), and $X$ is not a constant with probability 1, then Jensen's inequality tells us that $$E[f(X)] > f(E(X))$$ One important class of functions where equality does hold is linear functions (e.g. $f(x)=ax+b$)

4

Let the parameters of the two geometrics be $\alpha$ and $\beta$. So these are the probabilities of "success," and the geometrics give the number of trials until the first success. Let $Z=\min(X,Y)$.We have $Z\ge z$ if and only if $X\ge z$ and $Y\ge z$. The probability that $X$ is $\ge z$ is the probability of $z-1$ "failures" in a row. This probability is ...

4

Let $W$ be the maximum of the $X_i$, where $n$ is fixed. Then $W\le t$ if and only if all the $X_i$ are $\le t$. This has probability $\sim(1-t^{-\alpha})^n$. Thus $$\Pr(W\gt t)\sim 1-(1-t^{-\alpha})^n.$$ But for fixed $n$ and large $t$ we have, taking the first two terms of the binomial expansion, $1-(1-t^{-\alpha})^n\sim nt^{-\alpha}$.

4

By the 1st Borel Cantelli lemma, $P(X_i \neq Y_i \,\,i.o.)=0$. So for almost every $\omega$, $\exists \,N(\omega)$ such that for $n \geq N(\omega)$, $X_n(\omega)=Y_n(\omega)$. Since the convergence and divergence of any series of real numbers depends only on the tail, for almost every $\omega$, $\sum_{i=1}^\infty X_i(\omega)$ and $\sum_{i=1}^\infty ... 4 I'll get you started: $$E(X-Y)^3 = E(X^3)-E(3X^2Y)+E(3XY^2)-E(Y^3) = E(X^3)-3E(X^2)E(Y)+3E(X)E(Y^2)-E(Y^3)$$ The last step followed since$X^2$is independent of$Y$, and the same for$X$and$Y^2$. Now, how can you simplify this further? 4 Let us write$S_{k}=\xi_{1}+\dots \xi_{k}$an$S=\max_{i}(|S_{i}|)$. We fix$t$for the the rest of the proof. A way to prove Kolmogorov's inequality is to consider a stopping time variable, here the random variable$\tau:\Omega \mapsto \{0,1,2,\dots,n\}$defined by$\tau(\omega)=k$where$k$is the smallest integer such as$|S_{k}|\geqslant t$, or ... 4 By symmetry,$P(P_1 \gt P_2) = P(P_2 \gt P_1)$. It's easy to see that$P(P_1 = P_2) = \frac{1}{k+1}$. So, \begin{eqnarray*} 1 &=& 2P(P_1 \gt P_2) + P(P_1 = P_2). \\ && \\ \therefore P(P_1 \gt P_2) &=& \dfrac{1-P(P_1 = P_2)}{2} \\ &=&\dfrac{k}{2(k+1)}. \end{eqnarray*} 4 Hint: Consider for each$n \in \Bbb{N}$the set $$M_n := \{\beta \in B \mid P(A_\beta) \geq 1/n\}.$$ Show that each of these sets is in fact finite. Why does this help you? 3 Let$\left(\Omega,\mathcal{A},P\right)$be a probability space. Then a random variable on it is a function$X:\Omega\rightarrow\mathbb{R}$such that$X^{-1}\left(B\right)=\left\{ \omega\in\Omega\mid X\left(\omega\right)\in B\right\} \in\mathcal{A}$for each Borelset$B$. Denoting the collection of Borelsets on$\mathbb{R}$by$\mathcal{B}$we state that ... 3 One of the sets$\{\beta | P(A_\beta)>\frac{1}{n}\}$would have to be uncountable if$B$is uncountable and the assumption holds, by Pidgeonhole. 3 The game where one looses \$1 when the toss is heads and one wins \$2 when the toss is tails corresponds to a random walk with positive bias. These have positive probability to never hit \$0. The game where one looses \$1 when the toss is heads and one doubles one's fortune when the toss is tails, is at least as favorable as the first one at every step such ... 3 Let Y has cdf$G(y)$and also assume$X$is a continuous random variable. Then$G(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq \sqrt{y})-P(X<-\sqrt{y})=F(\sqrt{y})-F(-\sqrt{y})$3 A little correction to the previous answer by janak: Let Y have cdf$G(y)then: \begin{align}G(y)&=P(Y\leq y)\\ &=P(X^2\leq y)\\ &=P(-\sqrt{y} \leq X\leq \sqrt{y})\\ &=P(X\leq \sqrt{y})-P(X< -\sqrt{y})\\ &=F(\sqrt{y})-F_{-}(-\sqrt{y}),\end{align} whereF_-(a)=\lim_{x \rightarrow a^-} F(x)$. 3 Hint: the assumptions imply that$X$is square integrable. Compute$\mathbb E[(X-Y)^2\mid\ X]$. 3 Actually you can assume that you have a very large number of apples and therefore sampling from them does not affect the percentage of blemished/not-blemished apples. This gives you a constant percentage of picking a blemished apple in each of the independent samplings and therefore you can use the Binomial Distribution. Indeed, the number$X$of blemished ... 3 Toss two coins, set$\xi = 1$if the first coin is a head, 0 otherwise,$\eta = 1$if the second coin is a head, 0 otherwise,$\zeta = 1$if the coins show different symbols, 0 if they are the same. Then$\xi$,$\eta$and$\zeta$are pairwise independent (hence pairwise uncorrelated) random variables with$E(\xi) = E(\eta) = E(\zeta) = \frac 12$but ... 3 Here is basic lemma from analysis: Let$(x_n)$and$(y_n)$denote some sequences such that there exists some finite$N$such that$y_n=x_n$for every$n\geqslant N$. Then,$\frac1n\sum\limits_{k=1}^nx_k\to\mu$if and only if$\frac1n\sum\limits_{k=1}^ny_k\to\mu$. Can you see why this answers your question? To prove the lemma, use triangular inequality ... 3 The essential result here is$x^r = r\int_0^x t^{r-1}dr = r\int_0^\infty t^{r-1} 1_{\{t | x>t\}}(t)dt$. Then we have$Y^r(\omega) = r\int_0^\infty t^{r-1} 1_{\{t | Y(\omega)>t\}}(t)dt$. Then \begin{eqnarray} E[X Y^r] &=& \int X(\omega) Y^r(\omega) \mu(d\omega) \\ &=& \int X(\omega)r\int_0^\infty t^{r-1} 1_{\{t | ... 3 If$Q$is always non-zero then$P = 0$iff$P/Q = 0$, so there is no need to phrase a new lemma. If$Q$can be zero then you need to explain how you interpret$P/Q$at values at which$Q = 0$. We can say that the probability the fraction of zeroes and poles of$P/Q$is at most$(\deg P + \deg Q)/|S|$. 3 Fact 1: If$X\sim N(0,a)$then$\mathbb{P}(|X|>\epsilon)\leq \frac{1}{\epsilon}\sqrt{\frac{2a}{\pi}}\exp[-\epsilon/2a]\leq \exp[-\epsilon/2a]$, whenever$a\leq \epsilon^{2}\pi/2$. Fact 2: If$t\to 0$then$t^{-b}e^{-c/t}\to 0$for any fixed$b,c>0$. In particular$e^{-c/t}<t^{b}$for sufficiently small$t$. We know that$a_{n}\to 0$. So there ... 3 Disclaim : I'm not a specialist of probability; thus, this is only an HINT. See John Walsh, Notes on Elementary Martingale Theory : Conditional probability of$B$given$A$:$P(B | A) = P(A \cap B) /P(A)$Conditional expectation of$X$given$A$: If$B$is an event then,$I_B$is a random variable with$P(I_B = 1 | A) = P(B | A)$, and$P(I_B = 0 ...

3

It is just notation. Think of it as the measure of the 'infinitesimal' slice $dx$. I like the notation in that it is suggestive of a Darboux sum, but it is a little cumbersome. Alternative notations are $\int e^{i(u,x)} d \mu(x)$, or $\int f d\mu$ with $f(x) = e^{i(u,x)}$.

3

$$\mathbb E(X)=\sum_{i=1}^{\infty}i\mathbb P(X=i)=\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}\chi_{[1,i]}(j)\mathbb P(X=i)=\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\chi_{[j,\infty)}(i)\mathbb P(X=i)=\sum_{j=1}^{\infty}\mathbb P(X\geq j)$$ where the next-to-last equality follows by fubini's theorem.

3

Then, are $X$ and $Y$ independent as well? Of course not, consider some nondegenerate random variable $X$, independent sequences $(X_n)$ and $(Y_n)$ i.i.d. distributed like $X$, and $Y=X$. How you planned to apply the lemma is a mystery.

3

As @JMoravitz says, this is a Markov chain problem. But a $69\times 69$ matrix is unwieldy. A good approximation is obtained by saying that, with every score $1,2,3,4,6$ equally likely, on average you will get $5$ scores in every total of $16\; (=1+2+3+4+6)$. E.g. if your first five scores were $1,2,3,4,6$ then you have gotten totals of $1,3,6,10,16$. Thus, ...

3

For every $x$, the limiting distribution is continuous at $x$ hence $P(X_n\gt x)\to1-\Phi(x)$. There exists some finite $N$ such that $a_n\geqslant x$ for every $n\geqslant N$ hence $$\limsup_{n\to\infty}P(X_n\gt a_n)\leqslant\lim_{n\to\infty}P(X_n\gt x)=1-\Phi(x).$$ This holds for every $x$ hence \limsup_{n\to\infty}P(X_n\gt ...

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