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6

The moment generating function of a Poisson random variable $X$ with parameter $\lambda$ is $$ \mathbb{E}\left[e^{tX}\right]=\sum_{n=0}^\infty \frac{e^{-\lambda}e^{nt}\lambda^n}{n!}=e^{\lambda(e^t-1)} $$ which converges for all $t$, and in fact is the restriction to $\mathbb{R}$ of an entire function. In particular, the power series for the moment generating ...


4

Somewhat more generally, let $X$ be any random variable whose moment generating function $M(z) = \mathbb E[e^{tX}]$ is analytic in a neighbourhood of $0$. This says that the series $\sum_{j=0}^\infty \mathbb E[X^j] t^j/j! $ has positive radius of convergence, i.e. $|\mathbb E[X^j]| \le C D^j j!$ for some constants $C, D$. Then we have uniqueness in the ...


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|}...


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

You can make a table. The names of the rows are the outcomes of the first dice (d1). The names of the columns are the outcomes of the second dice. In total we have 36 possible outcomes. And the cells marked with $\color{red}x$ are the outcomes where the second dice has a greater outcome than the first dice. $ \begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{...


3

The probability of a tied result on a fair die: $\mathsf P(A=B) ~=~ \tfrac 1 6$ By symmetry: $\mathsf P(A>B)~=~\mathsf P(B>A)$. By total probability: $~\color{blue}{\mathsf P(B>A)}~$ $\color{blue}{=~ \tfrac 12(1-\mathsf P(A=B)) ~\\=~ \dfrac 5{12}}$ Note: if you wanted a tie or greater: $~\mathsf P(B\geq A) ~$$=~ \mathsf P(A=B)+\mathsf P(B>A)...


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

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

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


3

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} &...


3

$X_i-\mu$ is a standard normal random variable, hence $$ \mathbb{E}[|X_i-\mu|]=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}|x|e^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}\int_0^{\infty}xe^{-\frac{x^2}{2}}\;dx=\sqrt{\frac{2}{\pi}}$$ Therefore the sum is equal to $$\frac{\sqrt{\pi}}{2n}\cdot n\sqrt{\frac{2}{\pi}}=\frac{1}{\sqrt{2}}$$ for all $n$. Even without ...


3

The straight forward, but somewhat tedious solution is to simply list the outcomes: First throw 1 then throw 1, no score First throw 1 then throw 2, score First throw 1 then throw 3, score etc. Then it's just a matter of counting scores vs no scores (every outcome is equally probable). I get $15$ outcomes that scores and $21$ that doesn't (there's $36$ ...


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

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 $P(X \in [-1,0)) = P(X \in [0,1))$.


2

Not in general. $$\Pr\left(X_{1}+X_{2}>x\right)=\int\Pr\left(X_{1}+X_{2}>x\mid X_{2}=y\right)dF\left(y\right)=\int\Pr\left(X_{1}>x-y\right)dF\left(y\right)$$ where independence was used. Since the rv's are nonnegative this equals: $$\int_{0}^{x}\Pr\left(X_{1}>x-y\right)dF\left(y\right)+\int_{x}^{\infty}\Pr\left(X_{1}>x-y\right)dF\left(y\...


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

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

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

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

It's a general fact that if $\{A_n\}$ is a sequence of measurable sets, then $\bigcup_{n=1}^{\infty}A_n$ is also measurable, with $$ \mu\Big(\bigcup_{n=1}^{\infty}A_n\Big)\leq \sum_{n=1}^{\infty}\mu(A_n)$$ This property is sometimes called countable subadditivity. One way to prove it is as follows: let $B_1=A_1$, and for $n>1$ let $B_n=A_n\setminus (\...


1

If, for all $T$, $P(|X_T-X_0|\geq 2c)=1$, then there is nothing to prove, the process is bounded so it is integrable. If for some $T$ the above probability $\rho$ is less than $1$, then \begin{align}P(\tau_1=\infty)\leq P(\tau_1\geq NT)&\leq P(|X_T-X_0|\leq2c,|X_{2T}-X_T|\leq 2c,\cdots,|X_{NT}-X_{(N-1)T}|\leq 2c) \\&=\prod_{i=1}^N P(|X_{iT}-X_{(i-1)...


1

I assume $m$ refers to probability $$A \subset B \to P(A) \le P(B)$$ In our case we have $$P(B) \le P(\bigcup_j A_j)$$ We also have $$P(\bigcup_j A_j) \le \sum_j P(A_j)$$ by countable subadditivity.



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