6

You are trying to prove that $$\frac{n(n-1)...(n-k+1)}{n^k}\to 1$$ as $n\to \infty$. Note that the product above has a finite number (k) of terms, and so when finding the limit of the above product, you may look at the product of the limits: $$\lim_{n\to\infty}\frac{n(n-1)...(n-k+1)}{\underbrace{n\cdot n\cdot ...\cdot n}_{k}}\\=\bigg(\lim_{n\to\infty}\frac{n}...


3

Differentiate! $$a_k=\Big(\frac{d}{d\lambda}\Big)^k\sum\cdots\big|_{\lambda=0}=0$$


3

Intuitively, for approximating $\operatorname{Binomial}(n,p)\approx\operatorname{Poisson}(\lambda=np)$, the smaller $p$ is, the closer the variance $npq=\lambda(1-p)$ to $\lambda$, so you expect better approximation. In the proof, you use $$ \frac{n(n-1)\dots(n-k+1)}{n^k}\left(1-\frac{\lambda}{n}\right)^{n-k}\approx e^{-\lambda} $$ to show $$ \binom{n}{k}p^...


3

In the proof, you need $np\to \lambda$. But if $n$ grows large and $np\to\lambda$, then we must have $p\to 0$, or else $np\to\infty$.


2

For simplicity of notation let $X=N(A)$, $Y=N(B)$ and $Z=X+Y=N(A\cup B)=N(A)+N(B)$, let $a = EX = \lambda|A|$ and $b=EY$. The probability that $(X,Y)=(k,l)$ can be found by conditioning on $Z$ as follows: $$\begin{align*} P((X,Y)=(k,l)) &= \sum_{n\ge0}P(Z=n)\times P((X,Y)=(k,l)|Z=n)\\ &=P(Z=k+l)\times P((X,Y)=(k,l)|Z=k+l)\tag{*}\\ &=\frac{(a+b)^...


2

For part $b)$ You need $P(X\ge1)=1-P(X=0)=1-\dfrac{e^{-4}\times4^0}{0!}=0.9817$ And your part $c)$ is correct.


2

$\mathbb{E}\left[t^{X}\mid\lambda=u\right]=e^{-u\left(1-t\right)}$ so that $\mathbb{E}\left[t^{X}\mid\lambda\right]=e^{-\lambda\left(1-t\right)}$ and: $$G_{X}\left(t\right)=\mathbb{E}t^{X}=\mathbb{E}\left[\mathbb{E}\left[t^{X}\mid\lambda\right]\right]=\mathbb{E}e^{-\lambda\left(1-t\right)}=\frac{1}{2}\int_{0}^{2}e^{-\lambda\left(1-t\right)}d\lambda=\begin{...


1

The hierarchical model is: $$N \mid M \sim \operatorname{Poisson}(\lambda M), \quad M \sim \operatorname{NegBinomial}(r,p)$$ where $\lambda, r, p$ are fixed parameters, and $M$ is parametrized such that $M \in \{0, 1, 2, \ldots \}$. Then the PMF of the unconditional distribution of $N$ is $$\Pr[N = n] = \sum_{m=0}^\infty \Pr[N = n \mid M = m]\Pr[M = m] = \...


1

The inequality holds without the limit at least for large enough $n$ and it is quite easy to prove. Let $a=1-(1-s)\frac {\lambda_n} n$ and $b=1-(1-s)\frac {\lambda} n$. For large $n$ and $s \in [0,1]$ note that $a,b \in [0,1]$. Hence $|a^{n}-b^{n}|=|a-b| |a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1}| \leq n|a-b|$ and this gives the inequality.


1

I'm assuming $\lambda=\frac{21}{2}$ is the average number of blackouts in a week. Then, the average number of blackouts in one day is just $\frac{\lambda}{7}=\frac{3}{2}$, since it is a Poisson process, and thus the average number of events is proportional to the observation interval. So if you let $X$ be the number of blackouts in a day, then $$P(X=k)= e^{...


1

Since the interarrival times are exponential random variables with rate $\lambda$, the probability for any given one of them to exceed $T$ is $e^{-\lambda T}$, and they are independent. So the number of arrivals $N$ needed for this to happen is a geometric random variable with parameter $p = e^{-\lambda T}$. Its mean is thus $1/p = e^{\lambda T}$.


1

If you want to calculate $P\left(X_n=j\right)$, you need to calculate the amount of permutations $\tau_n:\{1,...,n\}\longrightarrow \{1,...,n\}$ with j fixed points. If we were to construct such a permutation we'd have $binomial(n,j)=\frac{n!}{\left(n-j\right)! j!}$ possibilities to choose j fixed points. Now we have n-j ponints left, to construct ...


1

Note that you've aggregated the Poisson intensity across the $16$-year observation period, rather than treated the data as a sample of size $n = 16$ of annual rates. Since the research question is whether the true annual rate is below a certain hypothesized rate, it seems more natural to work with annual rates, which is also how the data are provided. In ...


1

Under the conditions described, as I understand them: \begin{eqnarray} \mathbb{P}\left(X=k, Y=j\right) &=& \frac{\lambda^{k+j}}{\left(k+j\right)!}e^{- \lambda}{k+j \choose k}p^k\left(1-p\right)^j\\ &=& \frac{\left( \lambda p\right)^k}{k!}e^{-\lambda p}\frac{\left( \lambda \left(1-p\right)\right)^{\,j}}{j!}e^{-\lambda \left(1-p\right)}\ . \...


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