Let $(X,Y)$ denote a two-dimensional random vector with an absolutely continuous distribution with density function $$p(x,y) = \frac{1}{y}\exp(-y), \qquad 0 < x < y < \infty.$$ Find $E(X^r\mid Y)$ for $r = 1,2,\ldots$ My solution: $$p_y(y) = \exp(-y)$$ $$p_{x\mid y}(x\mid y) = \frac{p(x,y)}{p_y(y)} = \frac{1}{y}$$ $$E(X\mid Y) = \int_0^y x p_{x\mid y}(x\mid y) \, dx […]

Let $X$ be a set, $F$ a $\sigma$-field of subsets of $X$, and $\mu$ a probability measure on $X$. Given a random variable $f:X\rightarrow\mathbb{R}$, define $$\chi_f(t)=\int_Xe^{itf}d\mu$$ Show that $\chi_f$ is continuous and $|\chi_f(t)|\leq 1$. I’m not sure how to start proving either statement. The function $\chi_f$ seems quite intangible, as it is an integral over […]

Assume we have a Hamiltonian system on $(\mathbb{R}^{2n},\omega)$ with Hamiltonian $H = H(q,p)$. In a paper I read, it says, without clarification, that the natural Liouville measure $\mu$ obtained by the volume form $\Omega = \omega \wedge \cdots \wedge \omega $ restricts on the energy level surfaces $N = \left\{(q,p) \in \mathbb{R}^{2n}: H(q,p) = c\right\}$ […]

This question arose when someone (and surely not the least!) commented that something like $\left(X\mid Y=y\right)$ , i.e. $X$ under condition $Y=y$, where $X$ and $Y$ are real-valued random variables and $P\left\{ Y=y\right\}>0 $, is not a well defined random variable. To see if he is right I need the definition of real-valued random variable. […]

I am trying to prove the following exercise (exercise 3, chapter 7 of Rudins Book “Real and Complex Analysis”): Suppose that $ E $ is a measurable set of real numbers with arbitrarily small periods. Explicitly, this means that there are positive numbers $ p_i $, converging to $ 0 $ as $ i\rightarrow \infty […]

A while ago it was made clear that a nowhere dense subset $P \subset [0;1]$ whose Lebesgue measure $\mu(P) = \mu([0;1]) = 1$ doesn’t exist. But is it possible in principle to define a nowhere dense subset $P \subset X$ so that its Lebesgue measure $\mu(P) = \mu(X) < \infty$ for some metric space? Or […]

Let $m$ be a probability measure on $Z \subseteq \mathbb{R}^p$, so that $m(Z)=1$. Consider a locally bounded $f: X \times Y \times Z \rightarrow \mathbb{R}_{\geq 0}$, with $X \subseteq \mathbb{R}^n$, $Y \subseteq \mathbb{R}^m$ compact, such that $f(\cdot, \cdot, z)$ is continuous, and $f(x,y,\cdot)$ is measurable. Also, $f$ is uniformly ($\forall (x,y,z) \in X \times Y […]

I am trying to show that $\int_{\mathbb{R}_{\ge 1}} 1/x^2 < \infty$. (1) By definition, $\int_{\mathbb{R}_{\ge 1}} 1/x^2 = \underset{0 \le \phi \le 1/x^2}{sup} \int_{\mathbb{R}_{\ge 1}} \phi$ for $\phi$ a simple function. (2) If we let $\psi_n$ be a simple function s.t. $\psi_n = \Sigma_{k=2}^n 1/k^2 \chi_{[k,k-1]}$ then since the p-series $\Sigma_{n=2}^\infty 1/n^2 < \infty$ we […]

Let $f \in L^1(\mathbb{R})$. Find $$ \lim_{n \rightarrow \infty} \int_{-\infty}^\infty f(x)\sin(nx) dx \,. $$ LDCT is a no go, as well as MCT and FL, which are really the only integration techniques we’ve developed thus far in the course. Integration by parts isn’t applicable either since we just have $f \in L^1$. I’ve tried splitting […]

Sry if my question is stupid, but I just wondered if is there is like a corresponding counterpart to the Lebesgue measure on $\mathbb{R}^n$ for (some?) metric spaces $(E,d)$? Since the natural way to measure distances in $(E,d)$ is given by $d$, shouldn’t there be a “natural” way to measure sets as well? Sure I’m […]

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