Given $f$ is a bounded continuous function on $\mathbf{R}$ and $\mu$ is a probability measure such that for all $x \in \mathbf{R}$ $$ f(x) = \int_\mathbf{R}f(x+y)\mu(dy) $$  Please help to show that $f(x+s) = f(x)$ for all $s$ in the support of $\mu$. (The support of $\mu$ is the smallest set $E$ s.t. $\mu(E)=1$.) The hint is that we can try to use martingales to show $E[f(x+X_0)-f(x)]^2=0$, where $X_0$ has distribution $\mu$.  So far I've constructed a martingale $M_n=f(x+S_n)$. Here $S_n = \sum_{j=1}^nX_j$ and $\{ X_j\}$ are i.i.d. with distribution $\mu$. But I've no idea on how to connect this martingale with the previous hint.  I came cross another post about the same question on the forum, but it ends with the martingale that I stated here. So I'm wondering if anyone can provide some further insights about how to finish the proof. Since this is a hw problem, I've been working on it for three days only to get to this point. I sincerely appreciate any inputs that people will give.  Thanks a ton!
124239
Assume f is a bounded continuous function on $\mathbf{R}$ and $X$ is a random variable with distribution $F$.  Assume for all $x \in \mathbf{R}$ that  $$ f(x) = \int_\mathbf{R}f(x+y)F(dy) $$  Please help conclude that $f(x+s) = f(x)$ where $s$ is any value in the support of $F$.  The hints that I have come across are to use Martingale theory and consider $\{ X_n\}$ to be i.i.d. with distribution $F$ and make a martingale with some function of $S_n = \sum_{j=1}^nX_j$.    Thanks!
4361
Let $(X , \mathscr{M}, \mu)$ be a measure space. Prove that if $f$ is measurable, then $|f|$ is measurable.    A function $f$ is measurable if $f^{-1}((a , \infty)) = \{x : f(x) &gt; a\} \in \mathscr{M}$ for every $a \in \mathbb{R}$.   Ok so this is what I have so far:  Consider two cases: $a &lt; 0$ and $a \geq 0$. If $a &lt; 0$, then we have $$|f|^{-1}((a , \infty)) = \{x : |f(x)| &gt; a \} = X \in \mathscr{M}$$ since $\mathscr{M}$ is a $\sigma$-algebra. I am confused on how to prove the case when $a \geq 0$. If $a \geq 0$, we have $$|f|^{-1}((a , \infty)) = \{x : |f(x)| &gt; a \} = ?$$ I am assuming we will have to use that fact that $f$ is measurable.  Any help will be appreciated!
18402
Let $f$ integrable and let $E_\alpha=\{x\mid |f(x)|&gt;\alpha\}$ for $\alpha\geq 0$. Show that $$\int\left|f(x)\right|dx=\int_0^\infty m(E_\alpha)d\alpha$$ where $m$ is the Lebesgue measure.   By hypothesis, $f$ is measurable, therefore there exists a sequence of simple function such that $\varphi_n\nearrow |f|.$ Moreover, $E_\alpha^n\nearrow E^\alpha$. By the continuity of the measure, $m(E_\alpha^n)\to m(E_\alpha)$ for all $\alpha\geq 0$. By the convergence monotone theorem, $$\int_0^\infty  m(E_\alpha)d\alpha=\lim_{n\to\infty }\int_0^\infty  m(E_\alpha^n)\underset{(*)}{=}\lim_{n\to\infty }\int\varphi_n(x)dx=\int\left|f(x)\right|dx.$$   I don't understand the equality $(*)$, why $m(E^n_\alpha)=\varphi_n(x)$ ?
32443
Exercise 5.3.1 in Durrett's "Probability Theory and Examples" states  Let $X_n$, $n\ge 0$, be a submartingale with $\sup X_n &lt; \infty$. Let $\xi_n=X_n-X_{n-1}$, and suppose $E(\sup \xi_n^+)&lt;\infty$. Show that $X_n$ converges almost surely.  Here are my thoughts: It looks like we need to use theorem 5.2.8 (Martingale convergence theorem), which states that $X_n$ converges almost surely if $\sup EX_n^+&lt;\infty$. I am trying to understand why we can just do the following (or at least I'm guessing we can't):  Suppose $X_n &gt; 0$ (edit: on some set with positive measure), then $0&lt; \sup X_n &lt; \infty$, and therefore $EX_n^+ \le \sup X_n &lt; \infty$. So we're done.  Otherwise, if $X_n \le 0$ a.e., then $\sup X_n^+ = 0$ a.e. Once again, we're done.  Am I missing something? Since I didn't use all the information, I probably am.  Thanks for the help.
88460
How do I prove the following? I don't know where to start.     If $X$ is a random variable with $E^{\mathbb P}[X] = \mathbb P(X&gt;0)=1$   and $ \mathbb Q$ is the  probability measure defined by $ \mathbb Q(A)=E^{\mathbb Q}[X1_A] $ then $E^{\mathbb Q}[Y]=E^{\mathbb P}[XY]$
116254
if $x_i$ are IID  continuous random variables, with $E[x_i]=μ$, is the following correct?   $E\left[\prod_{i=1}^n(1+⍺\space x_i)^i\right] = \prod_{i=1}^nE\left[(1+⍺\space x_i)^i\right] =  \prod_{i=1}^n(1+⍺\space E\left[x_i\right])^i = (1+⍺\space μ)^n$  thanks!
92992
Let $S=\{f:\mathbb{R} \to  \mathbb{R}\}$ that satisfies: $\forall f\in S$, $f'$ exists and $f'$ is continuous and $f(0)=f(1)=0$.  Please prove that $\exists \delta :\forall f\in S$ s.t.  $\int_{0}^{1} (f(x))^2dx\leq \delta \int_{0}^{1} (f'(x))^2dx$. Thanks in advance.
5851
Folland, "Real Analysis", Chapter 5.3, Exercise 36:     Let $\mathcal{X}$ be a separable Banach space and let $\mu$ be   counting measure on $\mathbf{N}$.  Suppose that   $\left\{x_n\right\}_1^\infty$ is a countable dense subset of the unit   ball of $\mathcal{X}$, and define $T \colon L^1(\mu) \to \mathcal{X}$   by $Tf = \sum_1^\infty f(n) x_n$.  (a)  $T$ is bounded.  (b)  $T$ is   surjective.   I have proved (a).  I would like help on (b).  Here are my ideas so far.    Say we want to show $y \in \mathcal{X}$ is in the image of $T$.  Reduce to the case $||y|| = 1$ and let $x_{n_m} \to y$.  My thought is to produce for each $N$ a function $f_N \in L^1(\mu)$ such that $||y - \sum_n^\infty f(n)x_n|| &lt; 1/N$ and also such that the $f_N$ converge in $L^1(\mu)$ to some $f \in L^1(\mu)$.  Then use the continuity of $T$ to conclude $T(f) = y$.    The way I have been setting up my inequalities is as follows. Pick $M_N$ such that for all $m &gt; M_N$, $||y - x_{n_m}|| &lt; 1/N$.  Then I want to define $f_N$ so that $$\sum_n^\infty f_N(n) = \sum_{m &gt; M_N}^\infty f_N(m) = 1, $$ hence $$||y - T(f_N)|| = ||\sum_{m &gt; M_N}^\infty f_N(m) y + \sum_n^\infty f_N(n)x_n|| \leq \sum_{m &gt; M_N}^\infty f_N(m) || y - x_{n_m} || &lt; \frac{1}{N}. $$  Presumably, if the $f_N$ are appropriately chosen, then I can find a dominating function and apply the Dominated Convergence Theorem to finish up.    However, getting all these hypotheses to hold simultaneously has been difficult.  In essence, the difficulty seems to be in requiring that the $f_N$ be "normalized" as well as possess a dominating function.  If this method is feasible, at least it seems to require a bit of analysis to show the desired series all converge.  Perhaps I am missing something cleaner.  If so, a simpler solution would be much appreciated.    -Thanks.
17237
Let $X$ be a random variable and let $g,f$ be real-valued, nondecreasing, and bounded.   Show that $\mathbb{E}[f(X)g(X)]\geq \mathbb{E}[f(X)]\mathbb{E}[g(X)]$  Having a hard time seeing where to start for some reasons, any hints?  I managed to write:  $\mathbb{E}[f(X)g(X)] = \frac{1}{4}\mathbb{E}[(f(X)+g(X))^2 - (f(X)-g(X))^2]$  But I'm not sure if this is in the right direction or not.
53104
Let $X$ be a (Lebesgue-)measurable set of $\mathbb{R}^n$ and $f:X \to \mathbb{R}^n$ continuous function that preserves measurability ($A$ meausurable $\implies f(A)$ measurable).  Prove: for all $A \subset X$,$\space$$\lambda(A)=0 \implies \lambda(f(A)) = 0$ .  I'm totally stuck. Initially I made some progress but now I'm at a point where it feels like the statement shouldn't be true at all.  Additionally i'd like to know if the statement works for general topological measure spaces (or even just metric measure spaces). That is:   Let $X$ be a (Lebesgue-)measurable set of $Y$, Topological (or metric) measure space and $f:X \to Y$ continuous function that preserves measurability etc...
54185
Let $f_n \in L^1$ such that $f_n\rightrightarrows f$. Then I want to conclude that If $\mu(X)&lt;\infty,$ then $f\in L^1$ and $\int f_n\to\int f$, so I want to make use of the Dominated convergence theorem, the thing is that I don't know which would be my dominating function.  I was trying to figure out if the measure could play that role, but after thinking a little bit I have realize that the measure doesn't have to be bigger than the absolute value of the functions of the sequence.  Another thing I was thinking of, is to apply Fatou's lemma, but I can't ensure that my functions are positive.  So, Can someone help me to prove this theorem please ?  Thanks a lot in advance :).    NOTE: I think this theorem is taken from Folland's Real analysis, modern techniques, and their aplications 2nd Edition :), This is to have all the definitions so we can talk the same language as in the book (this is, same notation and theorems or machinery)
102394
Prove $X = (X_n)_{n \geq 0}$ is a martingale w/rt $\mathscr{F}$ where X is given by:  $X_0 = 1$  and for $n \geq 1$  $X_{n+1} = 2X_n$ w/ prob 1/2  $X_{n+1} = 0$ w/ prob 1/2  and $\mathscr{F_n} = \mathscr{F_n}^{X} \doteq \sigma(X_0, X_1, ..., X_n)$.  I think that $X_{n} = 2^{n} \prod_{i=1}^{n} 1_{A_i} \forall n \geq 0$ where $A_1 = \{ \omega \in \Omega | X_{2}(\omega) = 2 X_1(\omega) \} \in \mathscr{F}$ as follows:  $X_{1} = 2X_0* 1_{A_1} + 0*1_{A_1^c}$  $X_{2} = 2X_1* 1_{A_2} + 0*1_{A_2^c}$  $=2(2X_0* 1_{A_1} + 0*1_{A_1^c})*1_{A_2} + 0*1_{A_2^c}$  Is that right? If not, why? If so, here is my attempt:  (leaving out adaptability and integrability stuff)  We must show that $E[X_{n+1}|\mathscr{F_n}] \equiv E[2^{n+1} \prod_{i=1}^{n+1} 1_{A_i}|\mathscr{F_n}] = 2^n \prod_{i=1}^{n} 1_{A_i}$? Is that right?  help please?
50265
$F$ is a distribution function .Take any $a \in \mathbb{R}$.  Then evaluate : $\displaystyle\int_{\mathbb{R}} (F(x+a)-F(x)) dx$   As it is a general distribution function , I cannot employ densities to solve this problem . Any insights or a complete solution will be very appreciated .
101432
I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times.  However, I can't see how to proceed.  I went down what I thought were the typical paths and just reached dead ends.  Does someone maybe have a hint?  I think that with the appropriate nudge I should be able to solve this question.  If not, I'll edit and ask if someone knows how to solve the whole thing.  Here's the question:  Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$.  Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$.
104930
I'm working in the following exercise:  Suppose $(X, \mathcal A, \mu)$ is a finite measure space and suppose $\mathcal F$ is the set of all $\mathcal A$-measurable functions $f: X \rightarrow \mathbb R$. For $f, g \in \mathcal F$, let $$d(f, g)=\int_X\frac{|f-g|}{1+|f-g|}d\mu.$$  Show that:   $d(f, g)=0$ if and only if $f=g$ almost everywhere. $d(f, g)=d(g, f)$ $d(f, g)\leq d(f, h)+d(h, g)$ If $f_n$ is a sequence in $\mathcal F$ and if $f \in \mathcal F$ then $d(f_n, f)\rightarrow 0$ if and only if for every $\delta&gt;0$ the following holds: $$\lim_{n\rightarrow\infty}\mu(\{x \in X: |f_n(x)-f(x)|\geq\delta\})=0.$$ If $f_n$ is a Cauchy sequence on $\mathcal F$ then there exists $f \in \mathcal F$ such that $d(f_n, f)\rightarrow 0$.   i.e., I must show that this thing is a complete pseudometric space.  I have already done 1. 2. 3. 4., but I'm stuck on 5. I don't know which $f$ should the $f_n$ converge to.
86882
Suppose that $\mu$ is a measure on the Borel $\sigma$-algebra on $[0,1]$ and for every $f$ that is real-valued and continuously differentiable we have  $$ \left|\int f'(x)~\mu(\text{d}x)\right|  \leqslant \left(\int_0^1 f^2(x) ~\text{d}x\right) ^{1/2}. $$  (1) Show that $\mu$ is absolutely continuous with respect to Lebesgue measure on [0,1].  (2) If $g$ is the Radon-Nikodym derivative of $\mu$ with respect to Lebesgue measure, prove that there exists a constant $c&gt;0$ such that $$ \left|g(x)-g(y)\right|\le c\left|x-y\right|^{1/2},~~~~~~x,y\in[0,1]. $$  Attempt  The inequality given reminds me of Jensen's inequality but I've no idea how to use it on (1).
16510
Let $X_{1},\ldots,X_{n}\stackrel{\text{ i.i.d }}{\sim}N(\mu,\mu^{2})$.  $T=\left(\sum_{i=1}^{n}X_{i},\sum_{i=1}^{n}X_{i}^{2}\right)$ is a sufficient statistic for $\mu$. Also $T$ is minimal sufficient.  However, the question is: Prove that $T$ is not complete, hint: use $f(x,y)=x-2y^{2}/(n+1)$.  To prove that $T$ is not complete, I must find an example of a Borel function, $f$, for which $E_{\theta}(f(T))=0$ does not imply $f(T)=0$. However, I do not know how to interpret the 'hint', since I only can conclude that $f(T)=0$, whenever using $x=\sum_{i=1}^{n}X_{i}$ and $y=\sum_{i=1}^{n}X_{i}^{2}$ and calculating $E_{\theta}(f(T))=0$.