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2

Consider $X_k^+ := \max(X_k,0)$. Then, \begin{align*} P\left(\limsup \frac{X_n}{n} < \infty\right)=1 &\Rightarrow P\left(\limsup \frac{X_n^+}{n} < \infty\right)=1 \\ &\Rightarrow \exists A: P\left(\frac{X_n^+}{n} > A \text{ i.o.}\right)=0 \text{ a.s.}\\ &\Rightarrow \sum_{i=1}^n P\left(\frac{X_i^+}{i} > A\right) < \infty \text{ ...


3

Starting where you stopped, assume that $E(X^+)$ is infinite, then $\sum\limits_{n=1}^\infty P(X_n\geqslant xn)=\sum\limits_{n=1}^\infty P(X\geqslant xn)$ diverges for every $x\gt0$ hence Borel-Canteli lemma implies that $X_n\geqslant xn$ for infinitely many $n$, almost surely, thus $\limsup\limits_{n\to\infty}\frac{X_n}n\geqslant x$, almost surely. This ...


1

If the random variables are sampled from a density $f$ that is bounded and strictly positive in a neighborhood of zero, show that $\text{E}|\bar{X}_n^{-1}|=\infty$ for every $n$. This is wrong: try $f$ the standard exponential density, then the density of $S_n=X_1+\cdots+X_n$ is proportional to $x^{n-1}\mathrm e^{-x}\mathbf 1_{x\gt0}$ hence $E(\bar ...


0

Note that $$(t,\omega) \mapsto \int_0^t B_s \, dW_s(\omega) =: M(t,\omega)$$ is a stochastic integral, hence a martingale. Therefore, it follows from the optional stopping theorem and the dominated convergence theorem that $$\mathbb{E} M(\tau,\cdot) = \lim_{t \to \infty} \mathbb{E}M(\tau \wedge t,\cdot) = \mathbb{E}M(0,\cdot)=0.$$


0

$\left\{ Z>0\right\} =\cup_{n=1}^{\infty}\left\{ Z>n^{-1}\right\} $ and consequently $\sum_{n=1}^{\infty}P\left\{ Z>n^{-1}\right\} \geq P\left\{ Z>0\right\} >0$. Then $P\left\{ Z>n^{-1}\right\} >0$ must be true for some $n$. Taking $b>0$ small enough (e.g. $b=\frac{1}{2}P\left\{ Z>n^{-1}\right\} $) we also have $P\left\{ ...


1

Suppose otherwise, that $P(Z>a)=0$ for all $a>0$. If your cdf is continuous, then $$0=\lim_{a \to 0}P(Z>a)=P(Z>0) \ne 0,$$ a contradiction.


0

Sometimes one uses the notation $\operatorname{supp}(X)$, the support of $X$, or the support of the probability distribution of $X$, to mean the closure of the set of all points in $\mathbb R$ for whose every open neighborhood the probability that $X$ is in that neighborhood is positive. For a discrete distribution, that is simply the set of all values that ...


1

If I understand your question, then I believe the answer you're looking for is YES! Proposition: Let $(\Omega, \mathcal{F},P)$ be a probability space and suppose $X: \Omega \to \Bbb{R}$ is $\mathcal{F}/\mathcal{B}$-measurable. Suppose further $Y : \Omega \to \Bbb{R}$ is $\sigma(X)/\mathcal{B}_{\Bbb{R}}$-measurable. Then there is some measurable function $f ...


0

You can calculate the marginal probability of $ Y=1$ as following: $ P (Y=1) = \int_0^1 P (Y=1|X=x) f_x (x) dx = \int_0^1 x dx = 0.5$. We used the fact that $ x $ comes from uniform distribution. Then it is $ P (Y=0) = 1-P (Y=1) = 0.5$.


0

The proper definition of "sequence $B_i$ summed from 1 to $n$ converges almost surely to $\mu$ as $n \rightarrow \infty$" is: For any specified value $\epsilon > 0$ and probability $\alpha > 0$, there exists an $n_0$ such that for all $n > n_0$ $$ \mbox{Prob } \left( \left| \frac{1}{n}\sum_1^n B_i - \mu \right| > \epsilon \right) < \alpha $$ ...


4

If in a box there are $n$ marbles, $k$ of them being red and $n-k$ being black, with $$\sum_{i}\binom{n-k}{i}\binom{k}{k-i}$$ you are counting the ways to select $i$ black marbles and $k-i$ red marbles, for any possible $i$. This is just the number of ways to select $k$ marbles out of $n$, i.e. $\binom{n}{k}$.


2

It is simply a matter of considering the coefficient of $x^{k}$ on both sides of $$ (1 + x)^{k} (1 + x)^{n-k} = (1 + x)^{n}. $$


2

No, it is not a misprint. By definition, $$\mu(E) := p_t(x,E)$$ defines a measure for each fixed $x$ and $t \geq 0$. The integral $$\int_K p_t(x,dy) p_s(y,E)$$ is the integral of the mapping $y \mapsto p_s(y,E)$ with respect to the measure $\mu$, i.e. $$\int_K p_t(x,dy) p_s(y,E) = \int_K p_s(y,E) \mu(dy).$$


0

When it comes to convergence in probability it is to be shown that $P\left\{ \left|\xi_{n}V_{n}\right|\geq\epsilon\right\} $ converges to $0$ for each $\epsilon>0$. Suppose not. Without essential loss of generality we could say that $P\left(\left|\xi_{n}V_{n}\right|\geq1\right)\geq\frac{1}{2}$ for each $n$. For a fixed positive integer $k$ we have ...


1

Convergence in probability: Fix $K>0$. It is not difficult to see that $$\begin{align*} \mathbb{P}(|\xi_n V_n| \geq \varepsilon) &\leq \mathbb{P}\left( |V_n| \geq \frac{\varepsilon}{K} \right)+ \mathbb{P}(|\xi_n| \geq K). \end{align*}$$ Since $V_n \to 0$ almost surely, we get $$\limsup_{\varepsilon \to 0} \mathbb{P}(|\xi_n V_n| \geq \varepsilon) ...


0

Yes, the result is known as Lévy's modulus of continuity: Let $(B_t)_{t \geq 0}$ a (one-dimensional) Brownian motion. Then $$\mathbb{P} \left( \limsup_{h \to 0} \frac{\sup_{0 \leq t \leq 1-h} |B(t+h)-B(t)|}{\sqrt{2h |\log h|}} = 1 \right)=1.$$ See e.g. René Schilling/Lothar Partzsch: Brownian motion - An introduction to stochastic processes for a proof ...


2

Hint: $$(X^{-1}(M) \setminus Y^{-1}(M))\cup (Y^{-1}(M) \setminus X^{-1}(M))$$ $$= \{\omega \mid X(\omega)\in M, Y(\omega)\notin M\}\cup \{\omega \mid X(\omega)\notin M, Y(\omega)\in M\} $$ $$\subset \{\omega \mid X(\omega)\not= Y(\omega)\} $$


0

you will use the given numbers as probabilities and the result for $a)$ is obvious since $P(A)=1/2$ and it is repeated independently for $3$ times and you get $(1/2)^3=1/8$ Part $b)$ is also simple you have binomial distribution with success probability of $1/6$ then you simply have $\binom{3}{2}(1/6)^2(5/6)=5/72$.


1

I think you're supposed to assume (perhaps rather strange assumption) that the results of the previous matches can be used to state the probability of outcomes of future matches. So in any given match, $A$ wins with probability $\frac{6}{12}=\frac12$; $B$ wins with probability $\frac13$; and the match ends in a tie with probability $\frac16$. I believe ...


0

Let $f$ be the p.d.f. of $X$ (exponential distribution) and let $g$ be the p.d.f. of $Y$ (beta distribution). The c.d.f. of $W=XY$ is then: $$\int_0^1 \int_0^{w/y} f(x)g(y) \, dx \, dy$$ The bounds of integration come from the support of $X$ and $Y$ (i.e. X has a exponential distribution so $x \ge 0$, and $Y$ has a beta distribution so $0 \le y \le 1$).


0

First you should have, given the measure $\mu$, a sequence to be your candidate. Then you prove that this candidate is unique. Every set $E \in \mathcal{P}(\mathbb{N})$ can be written as the union of his elements (note that this union is enumerable and disjoint). So, write $E = \{e_1,..,e_n,... \}$, then, using that $E = \cup_n \{e_n\} $ and the properties ...


1

The indicator function $\mathbf 1_A$ is a random variable that can take on values $0$ and $1$ only. As the name implies, its value is $1$ exactly when the event $A$ occurs. Since $\mathbf 1_A$ takes on value $1$ with probability $P(A)$, its _expected value is$$E[\mathbf 1_A] = 0\times (1-P(A)) + 1\times P(A) = P(A).$$ So, begin by showing that $1 - \mathbf ...


2

If I understand correctly, your D is an expectation here. Here are the key steps: 1) $\max(a,b) = \frac{a+b+|a-b|}{2}$ 2) $|c^2-d^2| = |c+d||c-d|$ 3) Cauchy Schwarz gives $E|UV| \leq \sqrt{E[U^2]}\sqrt{E[V^2]}$ This is a sketch. Just substitute appropriately to get your answer.


1

First assume that $X$ is nonnegative. Let $I_n$ be the indicator function for $\bigcup_{k=0}^n A_k$. Then $$ \sum_{k=0}^n E(X;A_k)=\int_{\cup_{k=0}^n A_k} X(\omega)dP(\omega)=\int_\Omega I_n(\omega)X(\omega)dP(\omega). $$ The first equality above uses the disjointness of the $A_k$'s. The sequence $\{I_nX\}_n$ is increasing and converges to $I_AX$ where $I_A$ ...


0

Let us consider angles $$ \varphi_{ij}=\arccos r_{ij}. $$ It is the great-circle distance between points $i$ and $j$ on the unit $N$-dimensional sphere. Then for any $k$ points $i_1,i_2,\ldots,i_k$ you have to satisfy inequality $$ \varphi_{i_1i_k} \le \sum_{l=1}^{k-1} \varphi_{i_li_{l+1}}. $$


0

Define $S_{n}=\sum_{k=1}^{n}X_{k}$ and $S=\sum_{k=1}^{\infty}X_{k}$. Then $S_{n}\uparrow S$ monotonically and Lebesgues monotone convergence theorem tells us that $\mathbb{E}S_{n}\uparrow\mathbb{E}S$. Here $\mathbb{E}S_{n}=\sum_{k=1}^{n}\mathbb{E}X_{k}$ so that $\mathbb{E}S_{n}\uparrow\sum_{k=1}^{\infty}\mathbb{E}X_{k}$ Combination of these facts give: ...


5

All you need is the monotone convergence theorem. Note that if you write $\mathbb E(\cdot)$ as $\int (\cdot) d\mathbb P $ and also write $\sum_{i=1}^\infty(\cdot)$ as $\lim_{n\to\infty}\sum_{i=1}^n(\cdot)$, you can rewrite your desired statement as $$\int \lim_{n\to\infty} \sum_{i=1}^n X_n \ d\mathbb P = \lim_{n\to\infty} \int \sum_{i=1}^n X_n \ d\mathbb ...


1

Actually the Markov property for Itô diffusions rather reads $E[f(X_{t+h})\mid F_t] = E[f(X_{t+h})\mid X_t]$. (In a second phase, the stationarity of the transitions of these processes then yields $E[f(X_{t+h})\mid X_t] = E^{X_t}[f(X_h)]$, that is, more explicitely, $E[f(X_{t+h})\mid X_t] = g(X_t)$ where, for every $x$, $g(x)=E^x[f(X_h)]$.) This holds for ...


1

It looks like you are looking for Tonelli's theorem.


0

Hints: $X_n^-\leq X_1^-$, $X_n^+\leq X_{n+1}^+$ and $X_n^\pm\rightarrow X^\pm$. Also, the result does not hold true in the generality that you state it. You need an assumption such as integrability of $X_n^-$ for some $n$ for instance.


1

What is used in this solution is called pull out and works for any integrable random variable (it does not need to be bounded): Let $X \in L^1$ and $Y \in L^1(\mathcal{G})$ (i.e. integrable and $\mathcal{G}$-measurable) such that $X \cdot Y \in L^1$. Then $$\mathbb{E}(X Y \mid \mathcal{G}) = Y \cdot \mathbb{E}(X \mid \mathcal{G}).$$ Proof: Let us ...


0

Take a discrete random variable $X$ with the distribution $P(X=\log y_m)=p(m)$. Since $\sum_{m=1}^n p(m)=1$ and each $p(m)\geq 0$, this is a probability distribution. Jensen's inequality states that $$\phi(\mathbb{E}[X])\leq \mathbb{E}[\phi(X)]$$ whenever $\phi$ is convex. Choose $\phi=e^x$. Plugging into the equation yields $$ \begin{align*} ...


0

By applying Hölder's inequality $$ \operatorname E|XY| \le \bigl(\operatorname E|X|^p\bigr)^{1/p}\bigl(\operatorname E|Y|^q\bigr)^{1/q} $$ to the random variables $|X|^r$ and $1_\Omega$, we obtain that $$ \operatorname E|X|^r\le\bigl(\operatorname E|X|^s\bigr)^{r/s} $$ for $0<r<s$.


1

Here's a simple proof of the first inequality, assuming we express $\mathbb{E}$ as a Stieltjes integral with respect to the cumulative distribution function $F$. If $|x| \geq 1$ and $0 < j < k$, then $|x|^j \leq |x|^k$. Therefore, $$\begin{align} \mathbb{E}[|X|^j] &= \int_\mathbb{R} |x|^j dF\\ &= \int_{-1}^{1}|x|^j dF + \int_{|x| > 1}|x|^j ...


2

The second inequality, that trivially implies the first one, is a consequence of Jensen's inequality: $$\varphi(\mathbb{E}[Z])\leq \mathbb{E}[\varphi(Z)]$$ by choosing $\varphi(x)=x^{k/j}$ and $Z=|X|^{j}$.


0

Yes, use linearity of expectation after expanding out the complex-valued function into its real and imaginary parts. Let $h_m(x):=f_m(x)+ig_m(x)$ for real-valued $f_m$ and $g_m$. \begin{align*} \mathbb{E}[h_1(X_1) h_2(X_2)] &=\mathbb{E}[((f_1(X_1)+ig_1(X_1))(f_2(X_2)+ig_2(X_2))]\\ ...


0

To visualize the problem draw the quarter xy plane, that is, the positive x and y axises, on a graphing sheet. Now mark all the points with integer coordinates on and inside the square whose vertices are (0,0),(0,5),(5,0), and (5,5). You can draw small dots to make these marks. So, for example, you would have a dot on your graphing paper at the point (1,1). ...


2

As the hint suggested, to reach $(5,5)$ from $(0,0)$, we will take $10$ consecutive "steps," of which $5$ will be up and $5$ to the right. We can choose any $5$ of these $10$ steps to be the "up" steps. So there are $\binom{10}{5}$ possible paths.


1

We have that $X_1$ and $X_2$ are independent $\Gamma(\alpha_i,1)$ random variables. Note that $X_1$ and $X_2$ are non-negative with values in $[0,\infty).$ Given the transformation, the joint density of $Y_1$ and $Y_2$ is: $$ f_{Y_1Y_2}(y_1,y_2)=f_{X_1X_2}[X_1(y_1,y_2),X_2(y_1,y_2)]|J|= f_{X_1}(y_1y_2)f_{X_2}(y_2-y_1y_2)y_2= ...


1

Here How exactly are the beta and gamma distributions related? you can find in one of the solutions posted. X1/X1+X2 has betta distribution


1

For $i=0,1,2,3,\dots$, you have found that $$\Pr(Y=i+1)=e^{-i}(1-e^{-1}).$$ Let $k=i+1$. You have found that for $k=1,2,3,\dots$, we have $$\Pr(Y=k)=(1-e^{-1})(e^{-1})^{k-1}.\tag{1}$$ That's because $e^{-i}=(e^{-1})^i=(e^{-1})^{k-1}$. Formula (1) is exactly the formula for the probability that $Y=k$, where $Y$ has geometric distribution with parameter ...


1

Theorem (2) states that ${\rm E}[Y\mid X]$ is the unique integrable, $\sigma(X)$-measurable random variable satisfying ${\rm E}[Y Z]={\rm E}[{\rm E}[Y\mid X]Z]$ for all bounded, $\sigma(X)$-measurable $Z$. So let us show that ${\rm E}[Y]$ satisfies these requirements. It's obvious that ${\rm E}[Y]$ is integrable and $\sigma(X)$-measurable (why?). So let ...


0

Let $X \sim Exponential(1)$ with pdf $f_x(x)$ and $Y \sim PowerFunction(a,k)$ with pdf $f_y(y)$: Then the pdf of $Z = X Y$ is: where: The TransformProduct function is from the mathStatica package for Mathematica. As disclosure, I should add that I am one of the authors. ExpIntegralE[n,z] denotes the exponential integral function $E_n(z)=\int ...


1

Since $X$ and $Y$ are independent, the joint density of $X$ and $Y$ is the product of the individual densities, so it is $$\frac{1}{2\pi}e^{-(x^2+y^2)/2}$$ on the entire $x$-$y$ plane. We want to find $$\int_D \frac{x^2+y^2}{2\pi}e^{-(x^2+y^2)/2}\,dy\,dx,$$ where $D$ is the unit disk. Switch to polar coordinates. We want to find ...


0

I think there is a typo in the definition of the order on $X$, I think you mean $\forall x \in \mathbb Z$? I'm going to assume that you consider monotone functions $f: X \to \mathbb{R}_{\geq 0}$ such that $f$ is measurable. Potentially you could get rid of the measurability assumption by assuming that the $\sigma$-algebra contains all sets ...


1

The inequality $$|e^{\imath \, tu}-1 \leq |t| \cdot |u| \tag{1}$$ follows easily from the fact that $$e^{\imath \, tu}-1 = \frac{1}{\imath} \int_0^{tu} e^{\imath x} \, dx.$$ From $(1)$ we see that $$\begin{align*} \int |e^{\imath \, t u}-1| \, \nu(du) \leq \int \min\{|t|\cdot |u|,2\} \nu(du) = |t| \int_{-\frac{2}{t}}^{\frac{2}{t}} |u| \, \nu(du) +2 ...


0

I don't really know what the etiquette is on answering your own question (whether it's frowned upon to answer your own question), but I'll post my solution/work anyway... So, we have $\displaystyle g(\theta) = \frac{1}{n}\sum\left(\frac{X_i}{X_i+\theta Y_i} - \frac{1}{2} \right) = \frac{1}{n} \sum \left( U_i - \frac{1}{2} \right)$. We verified that ...


1

It may seem a naive answer, but have you tried this calculation using the fact that the logarithm of a product is the sum of the logarithms? So that the logarithm of the probability you are looking for is the sum of the logarithms of the 360 probabilities?


0

As commenters said, the equality of $\liminf$ and $\limsup$ indeed implies the existence of limit, with the caveat about infinite limits: having $\lim=+\infty$ is usually considered diverging to $\infty$. Is this the best approach when $X$ is not known? Depends. If you are using tools that naturally fit the language of $\liminf$ and $\limsup$ (Fatou's ...


1

Forgetting some irrelevant factorial and exponential terms, you seem to be trying to factor the ratio $$R=\frac{\theta^{u-v}\lambda^v}{(\theta+\lambda)^u}.$$ You are not very specific about the complete expression you get for $R$, but indeed, $$R=\left(\frac{\lambda}{\theta}\right)^v\left(\frac{\theta}{\theta+\lambda}\right)^u.$$ And $R$ is also, as Casella ...



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