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4

No, your process does not make any sense. Even if the mean of $X_k$ is $1$, this does not mean that $$\frac {X_1 + \dots + X_N}{N} = 1$$ for some $N$. This is a huge misunderstanding; I'll try to briefly explain why is it so. Random variables are not numbers When talking about a random variable, you're talking about a function. When you write $X_1$, ...


4

Actually, the answer to this lies in quite a bit more advanced topic called rough path theory (beware: PDF). A rough path is a way of "enhancing" a $\alpha$-Hölder continuous path with some extra information. A rough path is an ordered pair, $\textbf{X}=(X, \mathbb{X})$ where $X\colon [0,T]\to V$ where $V$ is some Banach space (typically $\Bbb{R}$) and a ...


4

If $n$ is the number of tosses, then $T=n-H$, so $H-T=2H-n$. Since $n$ is large, $H$ has a close to normal distribution, mean $n/2$, and variance $(n)(1/2)(1/2)$. So $2H-n$ has close to normal distribution, mean $0$ and variance $n$. Let $W$ be a normal with mean $0$ and variance $n$. Let us find the mean of $|W|$. This is $$\int_{-\infty}^\infty \frac{|w|}...


3

tl;dr: look at the characteristic function, invoke independence of the summands to get a product of characteristic functions, compute the closed-form expression, and finally squint hard at the resulting expression to recognize a known characteristic function. (This is a good method whenever you have a r.v. defined as the sum of independent (possibly ...


3

Your last line should have been $$m_{X+Y}(t) = \int_{-\infty}^\infty e^{ts} f_{X+Y}(s) \mathop{ds}= \int_{-\infty}^\infty e^{ts} \int_{-\infty}^\infty f_X(s-y) f_Y(y) \mathop{dy} \mathop{ds}.$$ Making the change of variables $s=x+y$ gives you the answer.


3

For a set $\mathbf{A}$, $\mathbf{1}_\mathbf{A}$ often designates the characteristic function of $\mathbf{A}$, that is, the function defined by $\mathbf{1}_\mathbf{A}(x) = 1 \text { if } x \in \mathbf{A}$ and $\mathbf{1}_\mathbf{A}(x) = 0$ otherwise. When $\mathbf{A}$ is measurable, so is $\mathbf{1}_\mathbf{A}$ and $\int \mathbf{1}_\mathbf{A}d\mu = \mu(\...


3

The calculation is incompletely justified, and one should not write $\frac{X_1+\cdots+X_N}{N}=1$. Observe that $E(X_1+\cdots+X_N\mid N=k)=k$. This is because the $X_i$ and $N$ are independent, so for fixed $k$, we are just looking at $X_1+\cdots +X_k$, and by the linearity of expectation, the expectation of the sum is $(k)(1)$. By the Law of Total ...


3

Only a way to solve it correctly. In the answers of Ant and André it is explained what you did wrong. If $S_N:=X_1+\cdots+X_N$ then: $$\mathbb ES_N=\sum_{k=1}^{\infty}\mathbb E(S_N\mid N=k)\Pr(N=k)$$ Observe that $\mathbb E(S_N\mid N=k)=\mathbb ES_k=\mathbb EX_1+\cdots+\mathbb EX_k=k$. I leave the rest to you. Edit To explain why $\mathbb E(S_N\mid ...


3

According to Earliest Known Uses of Some of the Words of Mathematics, the term "characteristic function" in this sense (actually its French equivalent "fonction caractéristique") was first used by Poincaré in 1912 (except that with his notation, that function was what we now call the moment generating function). To me as an analyst, the terminology never ...


3

Yes, it does follow. Notice that $x \in \lim \inf_{i \to \infty} A_i$ iff there is some $j$ such that $x \in A_{i}$ for all $j \le i$. In this case we clearly have $\lim \inf_{i \to \infty} \chi_{A_i} (x) = 1$. Conversely, if $\lim \inf_{i \to \infty} \chi_{A_i}(x) = 1$, then there is no infinite subsequence $(A_{i_k})_{k}$ such that $x \not \in A_{i_k}$ for ...


3

No, it's not true. Suppose $X_1$ has uniform distribution over $\{0,1,2\}$ and $X_2$ has uniform distribution over $\{0,2,3\}$ $E(X_1 | X_1+X_2=0) = 0$ $E(X_1 | X_1+X_2=1) = 1$ $E(X_1 | X_1+X_2=2) = 1$ $E(X_1 | X_1+X_2=3) = \frac{1}{2}$ $E(X_1 | X_1+X_2=4) = \frac{3}{2}$ $E(X_1 | X_1+X_2=5) = 2$


2

So if you define $X_n = \frac{1}{\sqrt{n}} \sum_{i=1}^n (V_i^2-1)$, then you want to show for each $\epsilon>0$ there is a constant $M$ such that $P[|X_n|>M]\leq \epsilon$ for all $n \in \{1, 2, 3, ...\}$. No. Try $\{V_i\}_{i=1}^{\infty}$ independent with $$V_i = \left\{ \begin{array}{ll} 0 &\mbox{ with prob $1 - \frac{1}{4^i}$} \\ 4^{i/2} &...


2

I am sure this will not be useful to you, but it is generally not true. Let $ X_1 $ take the values $ \{ 1, 10 \} $ with equal probability, and $ X_2 $ take values $ \{1,9\} $ with equal probability. Then what are $ E[X_1 \; | \; X_1 + X_2 = 11] $ and $ E[X_2 \; | \; X_1 + X_2 = 10] $?


2

If $X\sim \chi^2_n$ then $\Pr(X\ge 0) =1$. But $\Pr(X_1^2+X_2^2 - X_3^2 \ge 0 ) <1$, because the probability that $X_1^2+X_2^2$ is small and $X_3^2$ is large is not $0$.


2

It is not true in general that $\mathbb{E}[e^{itX}]=e^{it\mathbb{E}[X]}$. For instance, suppose that $X=1$ with probability $\frac{1}{2}$ and $X=-1$ with probability $\frac{1}{2}$. Then $\mathbb{E}[X]=0$, hence $e^{it\mathbb{E}[X]}=1$. On the other hand, $$ \mathbb{E}[e^{itX}]=\frac{1}{2}\Big(e^{it}+e^{-it}\Big)=\cos(t) $$


2

These questions make little sense from the classical (Kolmogorov's axiomatic) probability theory point of view. For example, let $\xi$ be $U[0,1]$, and $\eta = \xi$ if $\xi$ is irrational and $1/\pi$ otherwise. Then these variables have the same distribution, yet $\eta$ never takes rational values. So the answers to both questions would be $0$. However, ...


2

Integrate $\int_0^\infty x^4 e^{-x^2/2}\; dx$ by parts using $u = x^3$, $dv = x e^{-x^2/2}\; dx$. Or change variables with $x = \sqrt{t}$ and use properties of the Gamma function.


2

$$\mathbb{E}[Z^4]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz=\frac{2}{\sqrt{2\pi}}\int_{0}^{\infty}z^4e^{\large-\frac{z^2}{2}}dz$$ set $\frac{z^2}{2}=u$, we have $z\,dz=du$ and $z^3=2\sqrt{2}u\sqrt{u}$, thus $$\mathbb{E}[Z^4]=\frac{4}{\sqrt{\pi}}\int_{0}^{\infty}u\sqrt{u}\,e^{-u}du=\frac{4}{\sqrt{\pi}}\Gamma\left(\frac{5}{2}\...


2

If you have seen the Moment Generating Function, we can solve this by using it. We have $$ \mathbf{E}(e^{tZ}) = e^{\frac{t^2}{2}}.$$ By equating the coefficient of $t^4$, we have $$\frac1{24}\mathbf{E}(Z^4) = \frac18.$$ This gives $\mathbf{E}(Z^4)=3$.


2

It seems that you know how to handle the case where the convergence in probability is replaced by almost sure convergence. Let's do the general case. As David Mitra suggests, the key point is to extract an almost everywhere convergent subsequence. Suppose that we do not have the convergence in $\mathbb L^1$. Then there exists a positive $\delta$ and an ...


2

We start with the Pearson differential equation: $$f'(x)=-\frac{\left(a_1 x+a_0\right) }{b_2 x^2+b_1 x+b_0}f(x),$$ Define $g(x)=b_2 x^2+b_1 x+b_0$ (using a trick in Diaconis et al.(1991)). Consider (f g)'(x), also written $(f(x) g(x))'=f'(x) g(x) +f(x) g'(x)$. We have $$(f g)'(x)=(-a_0 + b_1 - a_1 x + 2 b_2 x) f(x)$$ We assume that the distribution has ...


2

Stauffer, Dietrich; Aharony, Anthony (1994), Introduction to Percolation Theory (2nd ed.), CRC Press, ISBN 978-0-7484-0253-3 I used this for my thesis and it is pretty good.


2

Choose some $A_1\in\mathcal{A}$ such that $A_1\neq \emptyset,X$. Then at least one of $A_1,A_1^c$ contains infinitely many elements of $\mathcal{A}$. Indeed, suppose that both $A_1$ and $A_1^c$ contain only finitely many elements of $\mathcal{A}$. For any $E\in \mathcal{A}$ we can write $$ E=(E\cap A_1)\cup(E\cap A_1^c) $$ with $E\cap A_1,E\cap A_1^c\in\...


2

The hidden assertion here is that either $A_1$ or $A_1^c$ must have an infinite number of measurable subsets, so wlog assume it's the latter. The justification of this assertion is: Any set $B\in\cal A$ can be written $B=(B\cap A_1)\cup (B\cap A_1^c)$, a union of a measurable subset of $A_1$ with a measurable subset of $A_1^c$. How many possible $B$'s are ...


2

If $A_1$ is neither $\emptyset$ nor $X$, then so is $A_1^C$. If both have only a finitely many measurable subsets, then, since every subset of $X$ can be written as the union of a subset of $A_1$ and a subset of $A_1^C$, $\mathcal{A}$ would be finite. If $A_1^C$ has infinitely many measurable subsets, we are done, otherwise we can by relabeling replace $A_1$ ...


2

I will try to start from the simplest case possible and then build up to your situation, in order to hopefully develop some intuition for the notion of convolution. Convolution essentially generalizes the process of calculating the coefficients of the product of two polynomials. See for example here: Multiplying polynomial coefficients. This also comes up ...


2

This is overkill since I use the non trivial result that if the sequences $ x_n, x$ are in $l_1$, then $x_n \to x$ in norm iff $x_n(k) \to x(k)$ for all $k$. It is straightforward to verify that $\mu \emptyset = 0$ and $\mu A \ge 0$ for any $A \in \cal A$. Suppose $A_k \in \cal A$ are disjoint, let $x_n(k) = \mu_n A_k$, $x(k) = \mu A_k$, then we have $...


2

Since $\overline{X}_n=\frac74$ and $\mathrm{Var}\left(X_n\right)=\frac3{16n}$, we have by Chebyshev's Inequality, $$ P\left(\left|X_n-\tfrac74\right|\ge\lambda\right)\le\frac3{16n\lambda^2} $$ Plugging in $\lambda=\frac1{20}$ yields $$ P\left(X_n\ge1.8\right)\le\frac{75}{n} $$ Therefore, $$ P\left(X_n\le1.8\right)\ge1-\frac{75}{n} $$


2

Shortly after posting the question, I came across exactly what I was looking for, in a blog post by Tyrell McAllister. The post contains a pdf of the explanation, with a link to a comment where he explains why his version corresponds to Aumann's. Since link-only answers are discouraged on SE, I post here my own even shorter version, further simplifying ...


2

Assuming $|[X]|$ means absolute value of floor of $X$, that is $1$ if and only if $X \in [-1,0) \cup [1,2)$. So $$P(X<0||[X]|=1) = \dfrac{P(X \in [-1,0])}{P(X \in [-1,0)) + P(X \in [1,2))}$$ Also useful will be symmetry, so $P(X \in [-1,0)) = P(X \in [0,1))$.



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