4

The definition of the Geometric distribution you're using is one for which $X_1, \dots, X_n$ have the probability mass function $$p_\theta(k) = (1-\theta)^k\theta\text{, }\quad k = 0, 1, 2, \dots\text{.}$$ Now we have that $$f_{X_1, \dots, X_n}(x_1, \dots, x_n) =\prod_{i=1}^{n}p_\theta(x_i) = (1-\theta)^{\sum x_i}\theta^n$$ and as you stated, this is of the ...


3

The probability of success on each trial is $\theta,$ and $X_1$ is the number of failures before the first success, so $X_1\in \{0,1,2,3,\ldots\}.$ $Y= X_1+\cdots+X_n$ is a sufficient statistic for $\theta,$ i.e. the conditional probability distribution of $(X_1,\ldots,X_n)$ given $Y$ does not depend on $\theta.$ Now let $W= \begin{cases} 1 & \text{if } ...


3

By definition, $\ E\big(X|\sigma(Y)\big)\ $ is a $\ \sigma(Y)$-measurable function which satisfies the identity $$ \int_AE\big(X|\sigma(Y)\big)\,dP=E\big(XI_A\big)\\ $$ for all $\ A\in\sigma(Y)\ $. By the Radon-Nikodym theorem it is uniquely defined $\ P$-almost everywhere. Now $\ A\in\sigma(Y)\ $ if and only if $\ A=Y^{-1}(B)\ $ for some measurable $\ B\...


3

The joint density is $$ \text{constant} \times x^{a-1} e^{-\lambda x} \cdot y^{\beta-1} e^{-\lambda y} $$ Now put $c-y$ in place of $x$: \begin{align} & \text{constant}\times (c-y)^{a-1} y^{b-1} e^{-\lambda(c-y)} e^{-\lambda y} \\[8pt] = {} & \text{constant} \times \left( 1 - \frac y c \right)^{a-1} \left( \frac y c \right)^{b-1} \\[8pt] = {} & \...


3

Surprisingly, the expected number is infinity. Suppose that hights have a PDF $f_X$ and CDF $F_X$. Given that the height of the first person is $x_0$, the probability of each other person (assuming iid) to be taller is $1-F_X(x_0)$. The number of people that should be measured before we find someone taller is geometric, so $E(N\vert X=x_0)=\tfrac{1}{1-F_X(...


2

I see the problem is in modelling, so let me propose two models: Buses arrive every 10 minutes all day, starting at midnight 0:00. I finish my work somewhere between 5:00 and 5:10 pm and I wish to take the first bus. Let's say the time I finish work is uniformly distributed. I finish my work at 5:00 pm sharp. Buses arrive every 10 minutes, but their ...


2

First note that $F(x)=1-P(|X|\le x)=1-P(-x\le X\le x)=1-[G(x)-G(-x)+P(X=-x)]$ so that $dF(x)=dG(-x)-dG(x)-dP(X=-x)$ for $x>0$. In the expression you know, the first integral where the variable of integration $x$ is negative is problematic. Thus take $x=-m$ to get $$\begin{align*}\mathbb{E}\left[ X^{2}\boldsymbol{1}_{\{\mid X\mid > n^{1/3}\}} \right] &...


2

$A\sim\text{Binomial}\left(n, \frac 12\right)$. We know for binomial distributions, $E(A)=np=\frac n2$, and $Var=np(1-p)=\frac n4$. We know the formula for variance $Var(A)=E(A^2)-[E(A)]^2$, so $E(A^2)=\frac n4+\frac{n^2}{4}=\frac{n(n+1)}{4}$.


2

No need to find the distribution of $Y$. $EY=\frac 1{\sqrt {2\pi}} \int e^{x^{2}} e^{-x^{2}/2}dx=\infty$ which implies that $EY^{2}$ is also $\infty$.


1

In general you can integrate this type of function. Integrate $e^{-x^2}$ is much more difficult. However you can use that $$ \frac{d}{dx} e^{-x^2} = -2xe^{-x^2} $$ Your given function looks basically like $xe^{-x^2}$ so given above antiderivative and using substitution, the integral should be much easier to calculate.


1

Another way of expressing it is that $X_i=1$ if and only if car $i$ is the first car in a group. Then by linearity of expectation, $\sum E(X_i)$ is the expected number of cars that are the first in a group, and clearly, this is the expected number of groups. If any car ahead of car $i$ is slower than car $i$, then car $i$ will be stuck behind some slower ...


1

A) What's the probability that 2 different vertices have a path of length 2 between them. You need to count the possible two-edged paths between two particular vertices (ie: how many other vertices are there in the graph) and determine the probability that at least one among these paths is made. B) What is the expected value of the number of paths of ...


1

For each subsequence $\sigma$ of $\omega$ let $X_\sigma$ be a random variable that is $1$ if the subsequence is increasing and $0$ otherwise. Then $$\Bbb E(N)=\sum_\sigma\Bbb E(X_\sigma)\,.$$ There are $k!$ possible orders of the $k$ integers in a $k$-term subsequence of $\omega$, exactly one of which is an ascending sequence. Thus, the probability that any ...


1

Another aprroach is to use the law of total expectation thus $$\mathbb{E}[X^2Y]=\mathbb{E}[\mathbb{E}(X^2Y|Y)]=\mathbb{E}[y\mathbb{E}(X^2|Y)]$$ And the quantity $$\mathbb{E}(X^2|Y)$$ is known because it is known how to factorize $$f(x,y)=f(y)f(x|y)$$ into two gaussian densities In your case you have $$f_{XY}(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}\text{exp}\left\...


1

Hints: There is a standard trick for such computations: We can find $a$ such that $Z=aX+Y$ and $X$ are independent. Since these variables are jointly normal you only need to choose $a$ such that the covariance is $0$ and this requires $a=-\frac {EXY} {EX^{2}}$. Now your computations become easy. For example $EX^{2}Y=E X^{2}(Z-aX)=E[X^{2}Z]-aEX^{3}=(EX^{2}) (...


1

If $\epsilon$ is small so the number of points is large you can ignore the curvature and consider packing $d$ dimensional balls into $d$ dimensional space. For dimensions up to $8$ the densest lattice packings are known. For higher dimensions the densest packing is unknown and is suspected to often be nonlattice. You can compute the $d$ volume of the $d$ ...


1

I think I found a way to resolve this. Do you agree? Let $Y = \mid X \mid$. Thus, if $G$ is a distribution function of $Y$ then $G(y)=1-F(y)$ and $dG(y) = -dF(y)$. Therefore \begin{equation} \mathbb{E}\left[X^2 \boldsymbol{1}_{\{\mid X \mid > n^{1/3}\}}\right]= \mathbb{E}\left[Y^2 \boldsymbol{1}_{ \{Y > n^{1/3}\}}\right]=\int_{n^{1/3}}^{\infty}y^2 dG(...


1

We have if $c>0$, $|y|\ge c \iff y \ge c \lor y \le -c$ $$P[|X-E[X]| \ge a]=P[X-E[X] \le -a] + P[X-E[X] \ge a]$$ Hence, $$P[|X-E[X]| \ge a]\ge P[X-E[X] \ge a]$$ Remark: Even if $X$ is positive, $X-E[X]$ can also take negative value.


1

\begin{align} & \operatorname{var}\left( \frac{X_1+\cdots+X_n} n \right) \\[8pt] = {} & \frac 1 {n^2} \operatorname{var}(X_1+\cdots +X_n) \\[8pt] = {} & \frac 1 {n^2} \left( \operatorname{var}(X_1)+\cdots+\operatorname{var}(X_n) \right) \\[8pt] & \text{and so on.} \\[10pt] \operatorname E\left( \overline X \right) = {} & \operatorname E\...


1

I am supposed to discuss your work, before presenting my answer. Unfortunately, while I can solve the problem, my formal knowledge of Probability is inadequate to examine your work. On the re-roll, Player 2's expectation is as follows: $(1/2)$ the time, he will roll $> 10.$ If he does so, his average gain is $(15.5)$. $(1/20)$ of the time, his roll will ...


1

Sorry to come 6 years late to the party, but perhaps this will be useful for someone looking at a similar problem. Here is one naive approach to derive at least some set of bounds. Let's take a discrete uniform distribution on the set $S = \{a, 0.3, b\}$ with $a \le 0.3 \le b$, which ensures that any random variable $X$ drawn from this distribution will ...


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