# Tag Info

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The short answer is that the Poisson approximation is faster and easier to compute and reason about, and among other things tells you approximately how big the exact answer is. Here's a simple example: suppose you're trying to get something to happen in a video game that is rare; maybe it happens 1% of the time you do something, independently. You'd like to ...

5

You can use the equality: $$\mathbb EX=\mathbb E[X\mid X\geq1]P(X\geq1)+\mathbb E[X\mid X=0]P(X=0)$$leading to:$$1=\mathbb E[X\mid X\geq1](1-P(X=0))=\mathbb E[X\mid X\geq1](1-e^{-1})$$ hence to:$$\mathbb E[X\mid X\geq1]=\frac{1}{1-e^{-1}}=\frac{e}{e-1}$$

4

Multiply both sides by $\frac{e^5(n+1)!}{5^n}$ simplify and get $$5=n+1\rightarrow n=4$$ how do you know you must multiply both sides by... I do not know. It is just a conclusion. To realize that you can rewrite you equation in the following way: $$\frac{e^{-5}\cdot 5^n}{n!}=\frac{e^{-5}\cdot 5^n\cdot 5}{(n+1)\cdot n!}$$ As you can see, both sides contain $... 4 Let$c=P(X=0)$. From the expression you wrote, you can infer by induction that $$P(X=k)=c\gamma^k/k!$$ for all integers$k \ge 0$. To find the value of$c$, sum this equation over all$k \ge 0$and recall the Taylor series for$e^\gamma$. 4 Let$X$denote the total number of games played, and let$J$denote the number of games Joe played. The conditional distribution of$X$given$J=4$, namely$X|J=4$, is supported on$\{4,5,\ldots\}$and has pmf $$P(X=k|J=4)=\frac{P(J=4|X=k)P(X=k)}{\sum_{k=4}^{\infty}P(J=4|X=k)P(X=k)}$$ which is non zero whenever$k\geq 4$. Using the facts that$X\sim \text{...

3

The CDF methods you showed is for continuous distribution. With discrete laws the pdf (more precisely the pmf, probability mass function) is not the derivative of the CDF. With a discrete distribution, it is enough to directly calculate the pmf: $$\mathbb{P}[Y=y]=\mathbb{P}[X^2+3=y]=\mathbb{P}[X=\sqrt{y-3}]=\frac{e^{-\lambda}\lambda^{ \sqrt{y-3}}}{ (\sqrt{y-... 3 Describing Radioactive Decay Suppose we have an extremely large number of atoms, n, each atom having a very small chance of radioactive decay in a second. Suppose that the large number and small individual probability end up giving a measurable expected number of decay events, v, in a second. For all intents and purposes, the radioactive decay of one ... 3 A hierarchical model is not the same as the product of two random variables. Here is a simple example of a product: Let$$\Pr[X = x] = \theta^x (1-\theta)^{1-x}, \quad x \in \{0,1\}, \\ \Pr[Y = y] = p(1-p)^{y-1}, \quad y \in \{1, 2, 3, \ldots\}.$$So X is Bernoulli with parameter \theta and Y is geometric with parameter p. Now XY is a new random ... 3 One of the common reasons to use a Poisson distribution is when you don't really have a good guess at what n is exactly (though you're sure it's large) - to be precise, a Poisson distribution can be described as follows: The Poisson distribution of rate \lambda is the limit of the binomial distributions with n trials and an expectation of \lambda ... 3 The Poisson approximation only gives an approximate answer. To get an exact answer you must use a B(25,0.05) distribution. The probability can be easily found by$$1-\sum_{k=0}^4\binom{25}{k}(0.05)^k(0.95)^{25-k}$$3 The fact that X_1 and X_2 have a symmetric joint distribution implies that they are identically distributed because$$ P(X_1 = i) = \sum_{j \ge 0} P(X_1 = i, X_2 = j) = \sum_{j \ge 0} P(X_2 = i, X_1 = j) = P(X_2 = i). $$If you see the joint distribution of X_1 and X_2 as a non-negative function on \mathbb N^2, then the distribution of X_1 is ... 3 Observe that$$E[e^{Xt}]=\sum_{k=0}^{\infty} e^{kt}P(X=k) = \sum_{k=0}^{\infty} e^{kt}e^{-\lambda } \frac{\lambda ^k}{k!} = e^{\lambda(e^t-1)} \quad (*). $$Also$$\frac{d^3}{d^3t}E[e^{Xt}]|_{t=0} = \frac{d^3}{d^3t}(1+E[X]+\frac{1}{2!}t^2E[X^2]+\frac{1}{3!}t^3E[X^3]+ O(t^4X^4))|_{t=0}= E[X^3] \quad (**).$$Combining (*) and (**) results$$\frac{d^3}{d^3t}E[...

3

Given the data, we have $$\Pr[X \le 1] = 1 - \Pr[X > 1] = 1 - 0.476 = 0.524, \\ \Pr[X = 2] = \Pr[X > 1] - \Pr[X > 2] = 0.476 - 0.168 = 0.308, \\ \Pr[X = 3] = \Pr[X > 2] - \Pr[X > 3] = 0.168 - 0.069 = 0.099, \\ \Pr[X = 4] = \Pr[X > 3] - \Pr[X > 4] = 0.069 - 0.036 = 0.033, \\ \Pr[X > 4] = 0.036.$$ We can multiply these probabilities by ... 3 \begin{align} & \Pr(S_T\le s) = \operatorname E(\Pr(S_T\le s\mid T)) \\[8pt] = {} & \sum_{t=1}^\infty \Pr(S_t\le s\mid T=t) \Pr(T=t) \\[8pt] = {} & \sum_{t=1}^\infty \int_0^s \frac 1 {\Gamma(t)} (\lambda u)^{t-1} e^{-\lambda u} (\lambda\, du) \cdot (1-p)^{t-1} p \\[8pt] = {} & \int_0^s \sum_{t=1}^\infty \frac 1 {(t-1)!}(\lambda u(1-p))^{t-1} ... 3 I'd like to obtain: number of Electric Vehicles (EVs) which arrive to a charging station during one day; and their Time-of-Arrivals (ToA); Unfortunately the total number of arrivals in a day is not deterministic quantity thus you cannot "know" it but you can have a probability information about how many they are. It is a random variable, ... 3 Given that the random vector\textbf X$has certain number of entries$n$, it follows the multinomial distribution parameters$n, \textbf p$. This is because$X$is created by incrementing each coordinate with the vector of probabilities$\textbf p$,$n$times. Then the joint distribution is found by: $$\begin{split}f(x_1,...,x_m,z)&=f(x_1,...,x_m|z)\Pr(... 3 For part (a), merge the first and third process. Each arrival to the merged process arrives to process i \in \{1,3\} with probability \frac{\lambda_i}{\lambda_1 + \lambda_3} and so$$\Pr[N_1(\tau) = k] = \left(\frac{\lambda_1}{\lambda_1 + \lambda_3}\right)^k \left(\frac{\lambda_3}{\lambda_1 + \lambda_3}\right).$$For part (b), merge all three processes. ... 3 It is not necessary to use any kind of bounds. The average number of fish you catch per day is 10, so \lambda=10. The distribution of the number of salmon you catch per day is then S\sim\text{Poisson}(10\times .3)=\text{Poisson}(3). The probability that S\ge5=1-(.0498+.1494+.2240+.2240+.1680)=.1848. For part a, we could try P(S=s \&T=t) (... 2 Let d\mu(t)=\lambda e^{-\lambda t}dt$$P(Z=n|X)=e^{-X}\frac {X^{n}} {n!}.$$So$$P(Z=n, X \leq x)= {\int_0^{x} {e^{-t} t^{n}}/n!} d\mu(t).$$This gives$$P( X\leq x|Z=n)=\frac {{\int_0^{x} {e^{-t} t^{n}}/n!} d\mu(t)t} {{\int_0^{\infty} {e^{-t} t^{n}}/n!} d\mu(t)}.$$Finally$$E(X|Z=n)=\frac {{\int_0^{\infty } t{e^{-t} t^{n}}/n!} d\mu(t)} {{\int_0^{\infty} ... 2 This might be a useful approach to consider, based on my reading of the information: The probability of a page containing a misprint is given as 1.25%. I interpret this to imply$p=0.0125$. To determing the probability of having exactly 3 misprinted pages out of 200, using the binomial distribution, one then has: $$P(k=3)=\binom{200}{3} p^{3}(1-p)^{\left(... 2 It is \frac {\sum\limits_{k=1}^{\infty} ke^{-1} \frac1 {k!}} {1-P(X=0)} and P(X=0)=e^{-1}. Can you compute the sum? The answer is \frac e {e-1}. 2 Your description of how you simulated the process is ambiguous. I'm suspecting you've drawn a different gamma-distributed \ \lambda\ and then a \ \lambda-negative exponentially distributed time interval for every successive time stamp. If that is so, then you wouldn't have been properly simulating the point process Ross describes in his section 5.4.3... 2 The set of points within 4 meters of him has area \pi\cdot(\text{4 meters})^2. Thus the expected number of trees within that disk is \dfrac{\pi\cdot 4^2}{300}\approx 0.1675516. That number of trees has a Poisson distribution. You're looking for the probability that that number is 0. 2 I hope that I correctly grasped your question. And to answer the question, let us consider a concrete example. Let N=(N_t)_{t\geq0} and \tilde{N}=(\tilde{N}_t)_{t\geq0} be independent Poisson processes and consider$$X_t=N_{\tilde{N}_t}, \qquad \text{i.e.,}\qquad X_t(\omega)=N_{\tilde{N}_t(\omega)}(\omega).$$Then each X_t has a mixed Poisson ... 2 A partial answer, showing how someone who does this kind of thing regularly gets started on the process I hope that this "here's how I'm thinking through this" answer is of some value to you. Here's a sketch of an answer to a very specific question that's at least closely related to yours. Consider an long strip in the xy-plane, oriented along ... 2 Your answer cannot depend on y. If you condition on Y as you have done here, then your probability must be computed over all outcomes of Y. So in particular,$$\Pr[X < Y] = \sum_{y=0}^\infty \Pr[X < Y \mid Y = y]\Pr[Y = y]$$by the law of total probability. Then$$\Pr[X < Y \mid Y = y] = \Pr[X < y] = 1 - e^{-\lambda y}$$because F_X(x) = ... 2 You want to solve for h(\cdot) where for every \lambda>0,$$E[h(S)]=\sum_{k=0}^\infty h(k)\frac{e^{-n\lambda}(n\lambda)^k}{k!}=c\,e^{-3\lambda}\lambda^2$$for some positive constant c. That is,$$ \sum_{k=0}^\infty \frac{h(k)n^k}{k!}\lambda^k=c\,e^{(n-3)\lambda}\lambda^2 =c\sum_{j=0}^\infty \frac{(n-3)^j}{j!}\lambda^{j+2} =\sum_{k=2}^\infty \frac{c(... 2 Using this approach, for a smooth function$g$and a random variable$X$with mean and variance$\mu,\sigma^2$we can approximate: $$E[g(X)] \approx g(\mu) + \frac12 g''(\mu) \sigma^2$$ $$E[g^2(X)] \approx g^2(\mu) + g''(\mu)\sigma^2 + [g'(\mu)]^2 \sigma^2$$ $$Var(g(X)) \approx [g'(\mu)]^2\sigma^2$$ (which might be regarded as a linear approximation ... 2 Predict with what accuracy? If you want an approximately 95% prediction set, then maybe$x\in [1,9].$In R: qpois(c(.025,.975), 5) [1] 1 10 sum(dpois(1:9, 5)) [1] 0.961434 x = 0:15; pdf = dpois(x, 5) plot(x, pdf, type="h", lwd=3, ylab="PDF", main="POIS(5)") abline(v = c(.5, 9.5), col="red", lty="dotted&... 2 Just use the linearity of the expectation, plus the facts that$Var(X)=E[X^2]-E[X]^2$and, for the Poisson distribution,$Var(X)=E[X]=\lambda\$: $$E[(X+1)^2]=E[X^2]+2E[X]+1=Var(X)+E[X]^2+2E[X]+1=\lambda^2+3\lambda+1$$

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