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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

\begin{equation} \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 $N=n-m$ to 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 identity $$X=\sum_{i=1}^\infty\mathbf 1_{X\geqslant i},$$ 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 ...



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