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2

I think you want $X_n$ in your sums instead of $X_0$. Anyway: Case 1: Yes, monotone convergence is sufficient to prove it. Case 2: If $Y$ takes on one sign, or more generally is bounded in one direction or another, then monotone convergence will give you this. In general this could fail. The most general but still practical way to check this that I can ...

2

$$\text{Boy, High income} =4\\ \text{girl, High income} = 6\\ \text{Boy, low income} = 6\\ \text{Girl, low income} = x$$ $$P(\text{Male - M}) = \frac{10}{16+ x}\\ P(\text{High income - H}) = \frac{10}{16+ x}\\ P(\text{Male High income}- MH) =\frac{4}{16+x}$$ Independence means that $P(M \text{and} H) = P(M) P(H)$ $$\frac{10}{16+x}\frac{10}{16+x} = ... 2 Just do some in-your-head approximations, no table-lookups: 95\,\% is about 2\sigma, here \sigma=\sqrt{npq}=5, so anything between 40 and 60 heads will not raise your suspicion at this confidence level 2 Yes, the conditional distribution of the count of emperor penguins among the given total of 90 penguins will be binomial with parameters n=90, p=0.10. The fact that the penguins arrive in a Poisson process is irrelevant to the question.$$\mathsf P(N_e=10\mid N_e+N_p=90) = \binom{90}{10} 0.10^{10}0.90^{80}$$However, the notations for random ... 2 Your answer is ok except that the parts about s and t seems confused. Only one time period is mentioned, and how much time it is doesn't matter because we are given the number of arrivals during that time. (If we were asked "What is the probability that the number of arrivals will be 90?" that would be another matter.) No ratio of times is involved. ... 2 P(F\mid E) = \dfrac{P(F\cap E)}{P(E)}=\dfrac{0.4}{0.8} = \dfrac{1}{2} 2 Perhaps the most commonly used example is the natural filtration of Brownian motion, i.e. \mathcal{F}_t=\sigma(W_s:s \in [0,t]). This sort of thing is usually how filtrations are created. 1 You can look at$$\mathcal B_r=\Big\{A\subseteq [0,1]\times[0,1]\ \bigg|\ A\cap [0,r]\times[0,1]\text{ is Borel, }\land (A\cap(r,1]\times[0,1]\in\{(r,1]\times[0,1],\varnothing\})\Big\}$$For r\in[0,1]. Where Borel means the standard Borel sets of [0,1]\times[0,1]. Then for r\leq s we have \cal B_r\subseteq B_s, since whenever A\in\cal B_r, we ... 1 You can do the same thing as you do in the discrete-time case by just thinking of the X_n as being indexed by the rationals. Explicitly, choose a bijection f:\mathbb{N}\to\mathbb{Q}, and let \mathcal{F}_r be the \sigma-algebra generated by the X_n such that f(n)<r. 1 Define the event D as "Declared faulty" and the event F as that a computer is faulty. By law of total probability$$ P(D) = P(D\mid F)P(F) + P(D \mid F^c)P(F^c) $$We know that they give the correct answer with probability y so P(D \mid F)=y and P(D\mid F^c)=1-y. Also we know P(F)=x and P(F^c)=1-x so it remains to put everything together. 1 This model is well know as "double-well" potential. The closed form solution to \mathbb{E}(X(t)) is not known. See Iacus, ch. 1.13.8. Physically, you would expect that the global solution exists and is unique, since the drift term pushes the particle back to area around the origin every time X(t)^2 > \tfrac{8}{9}. We can proceed with simulation, ... 1 The marginal (unconditional) distribution of X under this hierarchical model is$$\begin{align*} \Pr[X = k] &= \sum_{n = k}^\infty \Pr[X = k \mid N = n]\Pr[N = n] \\ &= \sum_{n=k}^\infty \binom{n}{k} p^k (1-p)^{n-k} e^{-\lambda} \frac{\lambda^n}{n!} \\ &= \frac{(p\lambda)^k e^{-\lambda}}{k!} \sum_{n=k}^\infty \frac{((1-p)\lambda)^{n-k}}{(n-k)!} ...

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Your question is not well posed. When you specify $X \sim \text{Binomial}(n,p)$ you are telling us what $n$ is. It is a fixed constant, not a random variable. In that sense, $E[n] = E[n|\text{anything}] = n$.

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The length of the longer piece is $\max\{X,1-X\}$. So your expectation would be $\int_0^1 \max\{x,1-x\}\,dx$.

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Let $X$ represent the number of forks that enter the sink in a year. $$X \sim \text{Poi}(2400)$$ $Z$ represent the number of smashed forks in a year. $$X \sim \text{Poi}( 2400 p)$$ $Y$ represent the mass of smashed forks in a year. $$Y = E_1 + \ldots + E_Z$$ where $E_i$ is are independent exponential random variable of mean $10$ and variance ...

1

Hint: You need to compute $$\displaystyle\int_{-\infty}^\infty e^{tx}\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\,dx=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}+tx}\,dx$$ Note that $-\dfrac{x^2}{2}+tx=-\dfrac{1}{2}(x^2-2tx)$. Now complete the square.

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Let $f\left(x\right)$ denote the probability density of $X$. Note that $X$ can take values from $-\infty$ to $\infty$; that is the range of the integral (not $0$ to $1$). The expectation is $$E\left[e^{tX}\right]=\int_{-\infty}^{\infty}f\left(x\right)e^{tx}dx=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-x^{2}/2}e^{tx}dx.$$

1

Denote by $X_1$ (respectively $X_2$) the random variable representing the number of kills of player $1$ (respectively player $2$). We assume that $X_1$ and $X_2$ are independent from each other, i.e. they are playing in different games. Then $$\mathbb P\left(X_1\le X_2\right) =\sum_{k=0}^{10}\sum_{i=k}^{10}\mathbb P\left(X_1=k\right)\mathbb ... 1 We have$$P(A \cap B \cap C)=P(A \cap (B \cap C)) = P(A)P(B \cap C)=P(A)P(B)P(C)$$You've calculated the other probabilities correctly. 1 Denote by$$\tau_j := \inf\{n \geq 1; X_n = j\}$$the hitting time and define, iteratively,$$\tau_j^k := \inf\{n > \tau_j^{k-1}; X_n = j\}$$for k \geq 2. Hints: It suffices to show that P_i(X_n = j i.o.)=0 for all j \geq 1. Show that$$P_j(\tau_j = \infty) \geq P_j(\tau_0<\infty) = f_{j0}.$$Conclude that$$p:= P_j(\tau_j< ...

1

We start by describing the distributions of $X$ and $Y$ separately. Write $X=\sum_{i=1}^N X_i,$ where $X_i$ is the time to collect the $i$-th coupon. The $X_i$'s are independent geometrically distributed: that is, $X_i\sim G(p_i)$ where $p_i=(N+1-i)/N$. We have $\mathbf{E}X_i=1/p_i$ and $Var X_i=(1-p_i)/p_i^2$, so $$\mathbf{E}X = \sum_{i=1}^N \frac{1}{p_i} ... 1 There are n-1 edges available to person # i. There is an independent probability p of any of these being connected. This description indicates it is a binomial distribution.$$\begin{align} \mathsf E(C_i) & = (n-1)\,p \\[2ex] \mathsf {Var}(C_i) & = (n-1)\,p\,(1-p) \end{align} The interdependence of friend counts between ...

1

hint: $P((A\cup B\cup C)^c)=1-P(A\cup B\cup C)=1-P(A)-P(B\cup C)+P(A\cap (B\cup C))=1-P(A)-P(B)-P(C)+P(B\cap C)+P((A\cap B)\cup (A\cap C))=1-P(A)-P(B)-P(C)+P(A\cap B)+P(B\cap C)+P(A\cap C)-P(A\cap B\cap C)=1-\dfrac{4}{5}-\dfrac{2}{3}-\dfrac{3}{7}+\dfrac{8}{15}+\dfrac{2}{7}+\dfrac{12}{15}-\dfrac{8}{35}=....$

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Hint: $p(A)=4/5, p(B)=2/3, p(C)=3/7, p(A\cap B)=2/7, p(B\cap C)=2/7, p(A\cap C)=12/15, p(A\cap B\cap C)=8/35$ Now find $p(A\cup B\cup C)$ then, use $1-(A\cup B\cup C)$ to get the required result. Note that: $p(A\cup B\cup C)=p(A)+p(B)+p(C)-p(A\cap B)-p(B\cap C)-p(A\cup C)+p(A\cap B\cap C)$

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