# Tag Info

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

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 $\sum_{i=1}^\infty ... 4 Your$Z=X-Y$will not be a "shifted binomial" unless$p=\frac12$, or the trivial cases where at least one of$n$and$m$is zero. For the case$p=\frac12$,$m-Y$has the same distribution as$Y$so$X+Y$and$X-Y+m$have the same distribution, which is indeed binomial. In general consider the means and variances of the distributions:$X$has mean$np$... 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 You get that by summing a geometric series. 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$Y=X. How you planned to apply the lemma is a mystery. 3 I have some partial results; perhaps someone can continue/correct my work. \begin{align*} &\phantom{{}={}}P\left(1-\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}>\epsilon\right)\\ &=P\left(1-\epsilon>\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}\right)\\ &=\prod_{i=1}^{n/2} P\left(1-\epsilon ... 2 Hint:X_1 + \ldots + X_n < t$iff there exist rational numbers$r_1, \ldots, r_n$such that$r_1 + \ldots+ r_n < t$and all$X_i \le r_i$. 2 \sum _{k=m}^n {p_1}^k {p_2}^m \binom{k}{m} \binom{n}{k} (1-{p_2})^{k-m} (1-{p_1})^{n-k} =\\ p_1^mp_2^m\sum _{k=m}^n {p_1}^{k-m} \frac{k!}{m!(k-m)!}\frac{n!}{k!(n-k)!} (1-{p_2})^{k-m} (1-{p_1})^{n-k} =\\ p_1^mp_2^m\sum _{k=m}^n \frac{(n-m)!}{(n-k)!(k-m)!}\frac{n!}{m!(n-m)!} (p_1(1-{p_2}))^{k-m} (1-{p_1})^{n-k} =\\ p_1^mp_2^m\sum _{k=m}^n ... 2 Put$p=p_1, q=p_2$to avoid messy subscripts. Midway, put$r=k-m$and$N=n-mto simplify notation. \begin{align} &\sum _{k=m}^n \color{green}{{p_1}^k {p_2}^m} \color{orange}{\binom{k}{m} \binom{n}{k}} (1-{p_2})^{k-m} (1-{p_1})^{n-k}\\ &=\sum_{k=m}^{n}\color{green}{p^kq^m}\color{orange}{\binom nk \binom km}(1-q)^{k-m}(1-p)^{n-k}\\ ... 2 Already posted n times. The desired formula for \mathbb E(X) is the integration of the pointwise identityX=\sum_{i=1}^\infty\mathbf 1_{X\geqslant i},$$valid for every nonnegative integer valued random variable X. 2 Assuming P is the probability density associated with X, note that it is identical to the density of a normal random variable,$$ P\left(x;\mu,\sigma\right)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{\left(x-\mu\right)^{2}}{2\sigma^{2}}} $$with \mu=2 and \sigma=6 (verify this as a simple exercise). This should answer your first question. As for the ... 1 Only if the travel times of AB and BC are uncorrelated. For general random variables: Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y). 1 I understand that \frac{N−1}N is the probability of picking any type except 1 particular type i. And hence (\frac{N−1}N)^n is picking types distinct from type i for each of n picks, In other words, the probability of picking zero amount of type i. \mathsf P(X_i=0)=(\frac{N-1}N)^n ... but I don't really understand what the complement is. Is ... 1 First of all, your second and third paragraphs. Recall that if \xi is F-measurable and F\subseteq G then \xi is G-measurable: that follows from the definition of measurability. Now, if \Bbb F = (F_n)_{n\geq 0} is a filtration and \xi = (\xi_n)_{n\geq 0} is \Bbb F-adapted, then for each n we have \xi_n is F_n-measurable. Since ... 1 Hint: for a fixed a, the events (U_{a,n})_{n\geqslant 1} are independent. The equivalent given in the opening post allows us to determine the convergence of the series \sum\limits_{n=1}^\infty\mathbb P(U_{a,n}). Then we use the \fbox{B___-__} lemma. For part b), see here. 1 As yourself this: If I define f_X (1) = 1, will it be true that$$ \int_{-\infty}^x f_X(t) ~ dt = F_X(x) ?$If the answer's "yes", then you've got a PDF for$X$. Corollary to the result you'll get: the value of the PDF at any particular point doesn't matter. Why? 1 Fix$\omega \in (0,1]$, then for$n > 1/\omega$you have$X_n(\omega)=0$. So we have pointwise convergence on$(0,1]$. Since$\{ 0 \}$has probability zero, we have almost sure convergence. 1 The measure of the set where$X_n$converges to$0$can be calculated as follows: Let$x \in (0,1]$. Then choose$N$such that$\dfrac{1}{N} < x$. Then,$X_n(x)=0$for all$n>N$. So, the set of points where$X_n$converges to$0$has measure$1$, and hence you get almost sure convergence. 1 Does$S_n/n$converge almost surely? Not necessarily, consider deterministic random variables such that$X_i=1$if$4^k\leqslant i\lt2\cdot4^k$and$X_i=-1$if$2\cdot4^k\leqslant i\lt4^{k+1}$, for every$k\geqslant0$. The first values of$(X_i)_{i\geqslant1}$are$+1$once then$-1$twice then$+1$four times then$-1\$ eight times, and so on, hence ...

Only top voted, non community-wiki answers of a minimum length are eligible