Daniel Xiang
• Member for 6 years
• Last seen more than a week ago
• Chicago, IL, United States

The space of bounded linear operators on $X$, denoted $\mathcal{L}(X)$, is a Banach space, since $X$ is complete. The norm associated with $\mathcal{L}(X)$ is the operator norm, i.e. \begin{align*} \|...

$\{X_n\}$ is a sequence of measurable functions, i.e. $\{f_n\}$ in the theorem statement. Since we suppose $X_n \to X$ pointwise, we also have $X_n \to X$ a.e. Since $P(\Omega) < \infty$, by the ...

If $\{E_n\}$ is a countable collection in $\mathcal{L}$, then define the sets \begin{align*} F_1 &\doteq E_1 \\ F_n &\doteq E_n \backslash (E_1 \cup E_2 \cup \dots \cup E_{n-1}) \end{align*} ...

(a) Let $x \in \bigcup_{n=1}^{\infty} \bigcap_{k=n}^{\infty} A_k$. Then for some $n$, we have that $x \in \bigcap_{k = n}^{\infty}A_k$. Thus we also have $x \in A_k$ for all $k \geq n$. This means ...

The average value $A$ of a function $p$ over the interval $(a,b)$ is given by \begin{align*} A = \frac{1}{b-a}\int_a^b p(x)dx \end{align*} What are your $a$ and $b$ in this situation?

Fix $\epsilon > 0$. Now choose $N$ so large that $\sum_{k=N}^{\infty} \frac{1}{2^k} < \epsilon$. For any $m,n > N$, we have \begin{align*} |a_m - a_n| \leq \sum_{k=n+1}^m |a_k - a_{k-1}| < ...

Your proof looks great! For the continuous case, your probability is actually a density, so you could write \begin{align*} \int_{\mathbb{R}} f(y \mid x) dy = \int_{\mathbb{R}} \frac{f(y,x)}{f(x)} dy = ...

Here is a martingale argument. Note that by continuity from below, \begin{align*} \{T_1< \infty\} = \bigcup_{n=1}^{\infty}\{T_1 < T_{-n}\} \Rightarrow P(T_1 < \infty) = \lim_n P(T_1 < T_{-...

Assuming you are wondering how to systematically get all $x$ for which the original expression is true, you can take logs of both sides (base 10 log) to obtain \begin{align*} 8 = \log x^{12-4\log x} = ...

Here is the trick. Let $(Y_1,\dots,Y_n)$ be an independent copy. Then \begin{align*} 2\text{Var}(\max X_i) = E(\max X_i - \max Y_i)^2 = \int_0^\infty P((\max X_i - \max Y_i)^2>t)dt. \end{align*} If ...

Well if you just want $E(w_1 Z_1 + w_2 Z_2)$, you don't need to find the distribution of $Z_2$. Instead just use the tower property to obtain \begin{align*} E(w_1 Z_1 + w_2 Z_2) &= w_1 E Z_1 + w_2 ...

\begin{align*} E_k T_A = E_k \left(T_A \sum_{\ell \in S} \textbf{1}_{\{Z_1 = \ell\}}\right) = \sum_{\ell \in S}E_k T_A \textbf{1}_{\{Z_1 = \ell\}}, \end{align*} where $E_k$ is shorthand for ...

First note that since the Binomial r.v. is obtained through a sum of independent Bernoulli random variables, we can write $X_n$ as \begin{align*} X_n = \sum_{i=1}^{n^2} Y_i, \end{align*} where $Y_i \... View answer 2 votes Since$f$is integrable, Fubini's theorem implies that the integral over$[0,1]\times [0,1]$of$fexists and can be computed iteratively via \begin{align*} \int_0^1 \int_0^1 f(x,y) dxdy &= \... View answer Accepted answer 2 votes If this weren't the case, i.e. \begin{align*} \limsup_{k\to\infty} x_{n_k} > \limsup_{n\to\infty} x_n \doteq c, \end{align*} then for anyN$, we can find an$m(N) > N$such that$x_{n_{m(N)}} &...

You would write your joint density $f_{XY}$ as \begin{align*} f_{XY}(x,y) = g(x)h(y) \textbf{1}_{\{|x|\leq 1, |y| \leq x^2\}}, \end{align*} which can no longer be split up into a product of functions ...

No. Let $f_n(x) = x^n$ on $[0,1]$. Then since continuous functions are uniformly continuous on compact sets, the $f_n$ are uniformly continuous, and converging pointwise to $\textbf{1}_{\{1\}}$, but ...

if the sum is zero, then each term is zero, which implies that $P(|X_n|/n > 1) = 0$ for any $n$. So $P(|X_n|/n \leq 1) = 1$. Since the statement holds for all $n$, the countable intersection of ...

Following Jalex Stark's hint, let \begin{align*} v \doteq (1/1,1/2,1/3,\dots), \end{align*} so that $v \notin \ell^1$ and $v \in \ell^2$. Now define the bounded operator $A : \ell^2 \to \ell^2$ as, \...

Following saz's hint, we have that since the $X_i$ are nonnegative, we can interchange the order of the expectation and summation, in which case, \begin{align*} E \sum_{n \geq 1} \prod_{i=1}^n X_i = \...

Integrate by parts. \begin{align*} \int_{-\infty}^{\infty} x^2 e^{-ax^2} dx &= \int_{-\infty}^{\infty} x d(\frac{-e^{-ax^2}}{2a})dx \\ &= -\int_{-\infty}^{\infty} \frac{-e^{-ax^2}}{2a} dx + [\...

Take $A = e_1^{\intercal}$ and $B = e_2$. Then $\|AB\|_{\infty} = \|e_1^{\intercal}e_2\|_{\infty} = 0$, but clearly $\|A\|_{\infty} = \|B\|_{\infty} = 1$.

Consider the first sequence. Clearly $\{2\} \subset \cap_1^{\infty} C_k$ since $2-\frac{1}{k} < 2 \leq 2$ for all $k$. Now suppose $x \in \cap_1^{\infty} C_k$. Then $x \in C_k$ for all $k$, so ... View answer Accepted answer 2 votes There are some errors in your calculations. We have \begin{align*} L(x,\theta) &= \prod_1^n \theta x_i^{\theta -1 } \\ \Rightarrow \ln(L(x,\theta)) &= n \ln \theta + \sum_1^n \ln(x_i^{\theta ... View answer Accepted answer 2 votes By Martin's argument,T$is continuous at 0. Thus if$x_n \to x$, then$x - x_n \to 0$, so$T(x - x_n) \to T(0) = 0$, so$T(x) - T(x_n) \to 0$, which implies$T(x_n) \to T(x)$. To show boundedness ... View answer Accepted answer 2 votes Since you aren't given that$f$is measurable, you can't say that$\{x \in E : f(x) > \alpha\}$is measurable for any$\alpha \in \mathbb{R}$. Instead, recall that continuous functions are ... View answer 2 votes Let$x \in [0,1]$. If$x$rational, then$f(x) = g(x)$is given. If$x$is irrational, let$\{q_n\}$be a sequence of rationals converging to$x. By continuity, \begin{align*} f(q_n) \to f(x) \\ g(... View answer Accepted answer 2 votes The number of customersX$can be expressed as a sum of the customers that arrive before the bank employee checks the line after$M$minutes. In other words, we can write$Xas \begin{align*} X = \... View answer 2 votes LetT_i$be the time it takes to reach node$i$for$i = 1, ..., 5$. Then the total time$T\$ it takes for one to reach node 6 from node 0 can be written as a sum \begin{align*} T = T_1 + \dots + T_5 ...