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4

The $X_{n}$ are independent and absolutely continuous. This gives us the liberty to start with the assumption that $X_{n}\neq X_{m}$ for each pair $\left(n,m\right)$ where $n\neq m$ without any loss of generality. Note that the event $E_n$ can be interpreted as: $X_n$ is the "greatest" of $X_1,\dots,X_n$. The $X_i$ (for $i\in\{1,\dots,n\}$) all have equal ...


3

Because a characteristic function (Fourier transform of a real function) is hermitian $\phi_X(-t)=\phi_X(t)^*$ (complex conjugate). Hence $\phi_X(t)\phi_X(-t) = |\phi_X(t)|^2 \ge 0$. But $\sin(t)/t$ is negative for some values of $t$. Hence...


2

Hint: Since $$\sum_{n \geq 1} \mathbb{P}(X_n \neq -1) < \infty$$ it follows from the Borel-Cantelli lemma that for almost all $\omega \in \Omega$ there exists $N = N(\omega)$ such that $$X_n(\omega)=-1 \qquad \text{for all} \, n \geq N.$$


2

Suppose that $$\limsup_{n \to \infty} \frac{|S_n(\omega)|}{n}<\infty$$ for some $\omega \in \Omega$. Then $$X_n = S_n-S_{n-1}$$ implies $$\begin{align*} \limsup_{n \to \infty} \frac{|X_n(\omega)|}{n} &\leq \limsup_{n \to \infty} \frac{|S_n(\omega)|}{n} + \limsup_{n \to \infty} \frac{|S_{n-1}(\omega)|}{n} \\ &= \limsup_{n \to \infty} ...


2

For $p\geqslant 1$, we have $\lVert X_n\rVert_p^p=2/n$, hence $X_n\to 0$ in $\mathbb L^p$. Since (in general) convergence in $\mathbb L^1$ implies convergences in probability, which implies in turn convergence in distribution, we can answer questions 2), 3) and 4). Note that for these question, we did not use independence. Question 1) can be solved using ...


1

Well, to the extent that you let the Dirac delta function act as a density, then yes. An RV with zero variance is almost surely constant, so its just a degenerate discrete random variable.


1

You have two issues: $\text{Var}(X)=4$ not $8$ so $\text{Var}(X_1+X_2+X_3+X_4+X_5)=20$ not $40$ $\text{Var}(W)=\dfrac{\text{Var}(5W)}{5^2}$ not $\dfrac{\text{Var}(5W)}{5}$


1

Writing $$\left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k = \left( \frac{1}{n} \sum_{k=1}^n X_k \right) \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1}$$ we get $$\mathbb{E} \left[ \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^k \right] = \frac{1}{n} \sum_{k=1}^n \mathbb{E} \left[ X_k \left( \frac{1}{n} \sum_{i=1}^n X_i \right)^{k-1} \right].$$ Since the ...


1

It is correct. Maybe some details can be added, for example just after the first displayed equation, we could add "and thus, taking the expectation, $\mathbb E\left[X^2\right] \geqslant \mathbb E\left[Y^2\right] $". And there is no need to work with the $\mathbb L^2$-norm as we only consider the square of expectations.


1

From my simulations, it is clear that the answer is going to depend strongly on $k$ and $n$. For all $k$, as $n$ increases, $\Pr[X_\alpha = X_{(n)}]$ increases: the probability that the given statistic is the maximum order statistic becomes increasingly likely. The effect of increasing $k$ is to make the rate at which this probability increases as a ...


1

We only need the fact that $X_n$ are uncorrelated, independence is not necessary. $E\left(\sum_{i=1}^n \dfrac{X_i}{n} - m\right)^2 = E\left(\sum_{i=1}^n \dfrac{X_i-m}{n}\right)^2 = \sum_{i=1}^n E\left(\dfrac{X_i-m}{n}\right)^2 \leq\sum_{i=1}^n \dfrac{K}{n^2} = \dfrac{K}{n} $ So we get $\sum_{i=1}^n \dfrac{X_i}{n}$ converges to $m$ in $L^2$, and we will use ...


1

The moment generating function of the random variable $\,\,\dfrac{1}{n}\sum\limits_{j=1}^{n}X_j\,\,$ is (for $\lambda<n$) $$ \mathbb{E}\left[e^{\lambda\frac{1}{n}\sum\limits_{j=1}^{n}X_j}\right]{}={}\left(1-\lambda/n\right)^{-n}e^{\theta\lambda}. $$ One way to justify that, as $n\to\infty$, this MGF tends to the MGF for the required degenerate ...


1

What we want to do is obtain a general expression for the (conditional) probability distribution of the random variable $X$ representing the minimum number drawn among $N = n$ numbers chosen from $\{1, 2, \ldots, 10\}$ without replacement. To this end, we make the following critical observation: If the minimum number is $X = x$, then the remaining $n-1$ ...


1

You can easily check your work with a computer algebra system. There are a number of such packages around, for applications such as Maple and Mathematica. For your example, the bivariate joint pmf of $(Y_1,Y_2)$ is given as $f(y_1,y_2)$: $$f=\frac{1}{18} \left(y_1+2 y_2\right); \quad \text{domain}[f] = \{ \{y_1, 1, 2\}, \{y_2, 1, 2\} \} \text{ && } ...


1

For a fixed $t$, taking logarithm gives $\log \phi_{Y_n}(t) = \sum_{k=1}^n \log \cos(kt \sqrt{\dfrac{3}{n^3}})$ Using $$1-\cos(x) = \dfrac{x^2}{2} - O(x^4), x\to 0$$ $$\log(1-x) = -x + O(x^2), x\to 0$$ we have $$1- \cos(kt \sqrt{\dfrac{3}{n^3}}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{9k^4t^4}{n^6}) = \dfrac{3k^2t^2}{2n^3} + O(\dfrac{k^4}{n^6})$$ and ...


1

A number of observations: (i) As functions of independent random variables, $\dfrac{X_n}{\sqrt{n}}\,\,$ is independent of $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\,$. (ii) $\,\,\dfrac{X_n}{\sqrt{n}}\sim\mathcal{N}\left(0, 1/n\right)\,\,$ and $\,\,\displaystyle \sum\limits_{m=1}^{n-1}{X_m^2}\sim\chi^2(n-1)\,$, since they are linear combinations of ...


1

You have to go through each case: $ \begin{cases} 1 & a+b<C \\ \frac{C^2}{2 a b} & (a\geq b\lor a\geq C)\land (a<b\lor b\geq C)\land C>0 \\ \frac{C-b}{a} & a+b=C \\ -\frac{a-2 C}{2 b} & a<b\land a<C\land b\geq C \\ -\frac{b-2 C}{2 a} & a>b\land b<C\land a\geq C \\ -\frac{a^2-2 a C+(b-C)^2}{2 a b} & ...



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