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4

Let $(X_j)_{j \in \mathbb{N}}$ be a sequence of independent random variables such that $\mathbb{P}(X_j = 1) = \mathbb{P}(X_j =-1) = \frac{1}{2}$. The event $\{X_j = 1\}$ models that the $j$-th child is a boy, $\{X_j = -1\}$ models the event that it is a girl. Then $$S_n := \sum_{j=1}^n X_j$$ describes how much more boys are born than girls until the $n$-th ...


3

[This answers the original edition of the question, which did not assume $X_1,X_2,X_3,\ldots$ are independent.] No. Suppose $R\sim\mathrm{Uniform}(0,1)$ and $(Y_1,Y_2,Y_3,\ldots)\mid R\sim\mathrm{i.i.d. Bernoulli}(R)$. Then the strong law of large numbers implies that $$ \Pr\left( \lim_{n\to\infty} \frac{Y_1+\cdots+Y_n} n= R \mid R\right) = 1. $$ So ...


3

The following fact is well known (and strongly related to the Riesz-Fisher theorem): If $X_n$ tends to $X$ in $L^1$, we can find a subsequence $X_{\phi(n)}$ such that $X_{\phi(n)}$ tends to $X$ both in $L^1$ and almost surely. Applying Fatou's lemma to this subsequence yields $$ E(|X|^2) = E(\liminf_{n\to\infty} |X_{\phi(n)}|^2) \leq ...


3

Let $Y = E(X \mid G)$. Note that $$\int X^+ - X^-\, dP = \int X\, dP = \int Y\, dP = \int_{Y>0} Y\, dP + \int_{Y<0} Y \, dP = \int_{Y>0} X\, dP + \int_{Y<0} X \, dP $$ More over since $X$ and $Y$ have the same distribution $E [|X|] = E[|Y|]$. This implies $$\int X^+ + X^-\, dP = \int |X|\, dP = \int |Y|\, dP = \int_{Y>0} Y\, dP - ...


3

Start from definitions. You are trying to find $E[X|\sigma(X)]$. This by definition must satisfy $\int_A E[X|\sigma(X)]dP = \int_A XdP$ for all $A\in\sigma(X)$, and be $\sigma(X)$ measurable. Try $E[X|\sigma(X)]=X$ and verify it satisfies all definitions. Since $X(\omega)=x$ is equivalent to $X^{-1}(x)\in\sigma(P)$, it follows that $E[X|X=x]=x$.


3

To flesh out Did's comment: let $P(X=1)=P(X=-1)=1/2$. Let $Y=-X$ and note that $X,Y$ are not independent. Then define $f(x,y)=x+y$. $0=E[X+Y|Y=1]\neq E[X+1]=1.$


3

Let me give you first a non-rigorous "proof" of the statement; afterwards I'll explain how to justify it. If we define $$f_n(x) := n x 1_{[0,1/n]}(x) + 1_{[1/n,\infty)}(x),$$ then $f_n \uparrow 1_{(0,\infty)}$. Obviously, $f_n$ is not differentiable at $x=0$ at $x=1/n$, but for the moment we forget about this (that's the non-rigorous part) and apply ...


2

Hint: $$1_{\{|X|>a\}} \leq \left| \frac{X}{a} \right|^4 1_{\{|X|>a\}} \leq \left| \frac{X}{a} \right|^4. $$ Now integrate both sides.


2

Define an array as a sequence of random variables on a probability space $(\Omega, \mathcal{F},P)$. Introduce doubly infinite arrays of random variables $X_{n,j},~\mathcal{F}_{n,j}$ for $j,n\geq 1$, and set sub $\sigma -$algebras of a $\sigma -$algebra. Adapting the array to the filtration, then $\{X_{n,j}\}$ is a MDA if the relation ...


2

Sorry, your enunciation of the "Good Set Principle" in measure theory is not correct. Let us look in details: Let $P(x)$ means that "$x$ has property $P$". Let $\mathcal{G}:= \{ X \ | \ X \in \Sigma, P(X) \ \}$. Then it is true that $$(\ast) \hspace{1cm} \forall B \ ( B \in \Sigma \Longrightarrow P(B) ) $$ is equivalent to $$(\star) \hspace{1cm} ...


2

I assume $X_1,X_2$ are independent. As Saty suggested, it's an easier transformation of variables if you set $$Y_1 = X_1+X_2\\ Y_2 = \dfrac{X_1}{X_1+X_2}.$$ Then $X_1 = Y_1Y_2$ and $X_2 = Y_1-Y_1Y_2$ and the Jacobian is $$ J = \begin{vmatrix} y_2 & 1-y_2 \\ y_1 & -y_1 \\ \end{vmatrix} = -y_1. $$ Then, ...


2

By rescaling the bins, you are changing the number of internal states, which shifts the origin of the entropy scale. That's actually the exact same problem classical thermodynamics has that only gets resolved in quantum theory, where the number of microscopic states is limited by the quantization of the phase space. When the absolute entropy scale is set, ...


2

Set $$M_t := f(X_t) - \int_0^t Af(X_s) \, ds$$ and write $P_t f(x) := \mathbb{E}^x f(X_t)$ for the semigroup associated with $(X_t)_{t \geq 0}$. Using the definition of the generator, it is not difficult to see that $$\frac{d}{dt} (\mathbb{E}^x f(X_t)) = \frac{d}{dt} P_t f(x) = P_t Af(x).$$ By the fundamental theorem of calculus, this implies $$P_t ...


2

The word "chance" is very loosely used in the question. Let's examine the case for 5 riders. If it is the expected value that is sought, E[x] = np, so for 5 riders, $5\cdot\frac15 = 1$ If it is the probability that is sought, P(none of the 5 have a puncture) = $(\frac{4}{5})^5 = \frac{1024}{3125}$ Thus P(at least one puncture occurs) = 1 - ...


2

Let $\Omega = \{a,b\}$. Define $X,Y:\Omega\to\mathbb R$ by $$X(\omega) = \begin{cases}0,& \omega=a\\ 1,&\omega=b. \end{cases} $$ and $$Y(\omega) = \begin{cases}1,& \omega=a\\ 0,&\omega=b. \end{cases} $$ Then $\sigma(X)=\sigma(Y) = 2^\Omega$, but $XY$ is identically zero so $\sigma(XY)=\{\varnothing,\Omega\}$.


2

Suppose $\lim_{t\rightarrow-\infty} \frac{f_Y(t)}{f_X(t)}$ exists. We get: $$ \frac{F_Y(t)}{F_X(t)} \leq 1 \: \: \: \: \forall t $$ and so if $\lim_{t\rightarrow-\infty}\frac{F_Y(t)}{F_X(t)}$ exists, it must be less than or equal to 1. But it gives a case of $0/0$, so using L'Hopital's rule we get: $$1 \geq \lim_{t\rightarrow-\infty} \frac{F_Y(t)}{F_X(t)} ...


2

I'll take the statement "All players are equally skilled." to mean that every player has a $50\%$ chance to win against every other player. $7$ players have been eliminated by $p_5$; the probability that $p_1$ was not among them is $24/31$. If $p_1$ has not been eliminated, the identity of the winner of $p_5$'s branch of the tournament is irrelevant to her; ...


2

Suppose $X$ and $Z$ are independent, and they both have symmetric densities, so for all $t \in \mathbb{R}$: $$ f_X(t) = f_X(-t) \: \: , \: \: f_Z(t)=f_Z(-t) $$ Suppose $Y=X+Z$. Claim 0: The vector $(X,Z)$ has the same distribution as the vector $(-X,-Z)$. Claim 1: The vector $(Z,Y)$ has the same distribution as the vector $(-Z,-Y)$. Claim 2: ...


2

Binomial Confidence Intervals. Perhaps the most common practical use of the normal approximation to the binomial is to use data to find a confidence interval for the binomial success probability $\theta =$ P(Success) on any one trial. If $X$ is the number of Successes in $n$ trials, then $\theta$ is estimated as $\hat \theta = X/n.$ Assuming $\hat \theta$ ...


2

The result is false in this generality. Fix some $z\in S$ and define $O\mu=\mu_c+\mu_d(S)\delta_z$, where $\mu_d$ and $\mu_c=\mu-\mu_d$ are the discrete and continuous parts of $\mu$, respectively. Then $O$ satisfies your conditions, but cannot possibly be given by a kernel, if there is a non-discrete probability measure $\nu$ on $S$. Why not? If ...


2

Possibly you're thinking of the Poisson distribution. Suppose you have a one-in-$1{,}000{,}000$ chance of success on each trial, and there are $3{,}600{,}000$ trials. The expected number of successes is then $3.6$. If we ask for the probability that there are exactly $5$ successes, we get $$ \frac{3.6^5 e^{-3.6}}{5!} = \frac{3.6^5 e^{-3.6}}{120} \approx ...


2

For an alphabet with $Q$ symbols the answer is $$Q^N - N![z^N] \sum_{q=1}^Q {Q\choose q} \left(\sum_{p=1}^{k-1} \frac{z^p}{p!}\right)^q.$$ This is based on the observation that it is easier to count strings where no symbol appears $k$ or more times. Observe that a string of length $N$ in an alphabet of $Q$ symbols is a partition into sets of the ...


2

Let $\mathcal A,\mathcal B,\mathcal C$ be three independent $\sigma$-algebras and let $\mathcal{D}$ be the $\sigma$-algebra generated by $\mathcal{B}$ and $\mathcal{C}$. We want to show that $\mathcal{A}$ and $\mathcal{D}$ are independent. Indeed: if $\mathcal A,\mathcal B,\mathcal C$ are the natural $\sigma$-algebras of $X,Y,Z$ the product $YZ$ is a ...


1

In some sense, there is an $x$ on the right-hand side because the equality $$\mathbb{E}_x(Y \circ \theta_s \mid \mathcal{F}_s^+)(\omega) = \mathbb{E}_{B_s(\omega)}(Y)$$ holds only $\mathbb{P}_x$-almost surely. For a function $f: \mathbb{R} \to \mathbb{R}$ the Markov property reads $$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s} f(B_{t}), ...


1

It seems like you may be referring to the Law of Large Numbers. The Wikipedia page linked to gives a good explanation, from what I can see. See also: the relevant part of this SE answer for more information on the theorem, and why it is important, this SE question for another good, layman oriented explanation of the theorem, the top-rated answer to this SE ...


1

This is merely a slight extension of Nate's comment, but it should answer your question: On $(\Omega,\mathcal{A},\mathbf{P})$ we have a sequence $(X_i)_{i\in \mathbb{N}}$ of independent RVs $X_i:(\Omega,\mathcal{A})\to(\Omega_i,\mathcal{A}_i)$ and the corresponding sequence $\sigma(X_i)_{i\in \mathbb{N}}\subset\mathcal{A}$ of independent $\sigma$-algebras, ...


1

Hint: If $f(a)$ tends form a positive number to zero faster than $P(Y>a)$, then $P(Y>a) \geq f(a)$ will hold in general, because $f(a)$ is generally closer to $0$.


1

$$E|X|=E(|X-X_n+X_n|)\le E(|X-X_n|)+E|X_n|\ \forall n$$ Since $X_n\stackrel{1}{\to} X$, $\forall \epsilon>0$, for large enough $n$, $E|X-X_n|\le \epsilon$ Hence $$E|X|\le \epsilon +\sqrt{M}$$ for all $\epsilon>0\implies E|X|\le \sqrt{M} $. In general, if $E|X_n^p|\le M$ and $X_n\stackrel{p}{\to}X$, $p\ge 1$, then $E|X^p|\le M$. This can be proved in ...


1

The crucial point is the level of significance of the test, if you consider $\epsilon = 2^{-1}$ then you are rejecting the null conjecture (It is a random array) in many cases. For some reason you are bound to think that many $0$ may occur. So you will reject randomness at the event the first $m$ digits are zero. You could also think that it is not very ...


1

Here, \begin{align*} \tilde{f} = \frac{f}{\mathbb{E}_{\mathbb{P}}(f \mid \mathcal{Y})}. \end{align*} That is, \begin{align*} \mathbb{E}_{\mathbb{Q}}(g \mid \mathcal{Y}) = \frac{\mathbb{E}_{\mathbb{P}}(f g\mid \mathcal{Y}}{\mathbb{E}_{\mathbb{P}}(f \mid \mathcal{Y})}. \end{align*} This is the "abstract Bayes formula".



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