5

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

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

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

\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} ...


2

Your approach is fine, though the sum should begin at $k=1$ (or $k=2$). I don't think the expression has a nice closed form. Noticing that the mean of the variable is $\lambda = 10$, and that the approximation $(1-1/n)^n \approx e^{-1}$ should work reasonably around $n\approx 10$, then the function $g(n)= n \, (1-1/n)^n$ is approximately linear around the ...


2

The number of virus attacks are dependent on the time that the PC is on, thus the distribution of the virus attacks is $$V|T\sim Po\left(\frac{t}{5}\right)$$ and thus $$f_{VT}(v,t)=\frac{e^{-t/5}\left(\frac{t}{5}\right)^v}{v!}\cdot\frac{5}{4t^2}$$ The probability that $V=2$ is $$\frac{1}{40}\int_1^5 e^{-t/5}dt=\frac{5e^{-1/5}-5e^{-1}}{40}\approx 5.64\%$$ ...


2

For $\phi \in [0,10]$ the log likelihood is: $$\log\left[ \frac{\exp(-\phi) \phi^2}{2!} \frac{\exp(-2\phi) (2\phi)^4}{4!} \right]=-3\phi+6 \log\phi +\log16-\log(2!)-\log(4!)$$ Setting the derivative to $0$ gives $-3+6/\phi=0$ or $\phi=2$.


2

If phone calls arrive at a switchboard at an average rate of $3$ per minute, and at another switchboard at an average rate of $4.2$ per minute, then at the two switchboards combined, they arrive at an average rate of $7.2$ per minute.


2

Number in a day is modeled by $\text{Poisson}\left(rt\right)$. Arrival times are modeled by $\text{Gamma}(n, r)$ with $n$ being the $n$th arrival.


2

While Matthew Pilling and tommik provide answers that show the mathematics of getting the answer, I will provide the intuition involved. We know that, on this specific day, Joe played 4 solos. This provides a minimum number of songs - specifically, there must have been at least 4 songs. Note that it is very possible for the Poisson distribution to produce 0 ...


2

The answer of @Matthew Pilling is perfect (+1). A Bayesian approach will lead to the same solution avoiding a lot of calculations: (constants are not considered until the end of the process) $$\mathbb{P}[X|J=4]\propto \mathbb{P}[X]\cdot \mathbb{P}[J=4|X]\propto\frac{5^x}{x!}\cdot \binom{x}{4}\left(\frac{1}{2}\right)^x\propto\frac{\left(\frac{5}{2}\right)^{x}}...


1

Define $X$ to be the payment by the company in one month. The number of accidents in one month is distributed $\text{Poisson}(.3)$. Define a variable $Y$ that is the number of accidents per month, then the event that $X=5000$ is the same as $Y=0$ and $X=10000$ is the same as $Y=0\text { or } 1$. So using the definition of expected value $$\begin{split}E(X)&...


1

For b, you want the conditional probability $$P(X=8|X\geq 6) = \frac{P(X=8 \cap X \geq 6)}{P(X\geq 6)}= \frac{P(X=8)}{P(X\geq 6)}$$ For c, you want to find the probability of zero calls between $12:30$ and $13:00$ on any day, and then use a binomial distribution. You have $7$ days, and you want any $4$ of them to have zero. Remember, the requirements for a ...


1

Let $X$ be the earnings, then by definition of expected value, $$\begin{split}\mathbb E(X)&=60 \mathbb P(X=60)+70\mathbb P(X=70) +80\mathbb P(X=80)\end{split}$$ Let $p=.0803$, the chance that there are $3$ or more defects on the roll. The inspector earns $\$60$ if he inspects 4, 5, or 6 total rolls, given by probability $p^4+4p^4(1-p)+{5\choose 2}p^4(1-...


1

To get an answer it is necessary to fix a certain $n$. So let's set $n=5$ as per Neyman Pearson's Lemma, the critical region is $$\mathbb{P}[Y\geq k]=0.05$$ where $Y\sim Po(5)$ It is easy to verify with a calculator (or manually in 5 minutes) that $$\mathbb{P}[Y\geq 10]=3.18\%$$ and $$\mathbb{P}[Y\geq 9]=6.81\%$$ It is evident that there's no way to have a ...


1

The answers are the same, notice $1-e^{-\lambda_1t}+\frac{\lambda_1}{\lambda_1+\lambda_2}e^{-\lambda_1 t}=1-\frac{\lambda_2}{\lambda_1+\lambda_2}e^{-\lambda_1t}$. For your calculation the $\lambda_2$ in the second term was somehow dropped $$\begin{split}\Pr(T_1<T_2+t)&=\int_0^\infty\Pr(T_1<T_2+t|T_2=t_2)f_{T_2}(t_2)dt_2\\ &=\int_0^\infty\left(1-...


1

Just note that for $a,b\in \Bbb Z$ $$ \begin{align*} F_{X,W}(a,b)&=\Pr [X\leqslant a, X+Y\leqslant b]\\ &=\sum_{\{(t,s)\in \mathbb{Z}^2:t\leqslant a,t+s\leqslant b\}}f_{X,Y}(t,s)\\ &=\sum_{(t,s)\in\mathbb{Z}^2}\mathbf{1}_{(-\infty ,a]}(t)\mathbf{1}_{(-\infty ,b]}(t+s)f_{X,Y }(t,s)\\ &=\sum_{(t,s)\in\mathbb{Z}^2}\mathbf{1}_{(-\infty ,a]}(t)\...


1

If by "rate $\lambda$" you mean a process $X_t$ such that $\mathbb E[X_t] = \lambda t$, then this is just linearity of expectation: $$\mathbb E[2 X_t + 3 Y_t] = 2 \mathbb E[X_t] + 3 \mathbb E[Y_t] = (2\lambda + 3 \mu) t$$ On the other hand, $2 X_t + 3 Y_t$ is certainly not a Poisson process (e.g. this never takes the value $1$), while if $X_t$ and $...


1

Simple answer is that for both problems, you have a span of two hours, so the calculation is identical for both of them. On average 7 customers arrive in 1 hour so 14 customers arrive in 2 hours. Using $X\sim\text{Poisson}(14)$ to be the number of customer arrivals in a 2 hour time interval, we have $$\Pr(X=2)=\frac{e^{-14}(14)^2}{2!}$$ In terms of a Poisson ...


1

You can try doing it directly. Let $f(\lambda)$ be the log-likelihood after ignoring the additive terms that do not involve $\lambda$, then (please check my calculation) $$ f(\lambda) = -n\lambda+\ln{\lambda}\sum {x_i} +n\ln{\lambda} -\lambda\sum{y_i}. $$ Then differentiating with respect to $\lambda$ and equating it to zero get $$ \hat{\lambda}= \frac{1+\...


1

Because the application of the poisson distribution is an approximation only. The quality of a tablet is binomial distributed as $X\sim \textrm{Bin}(500,0.001)$. Then $$P(X=0)=\binom{500}{0}\cdot 0.01^0\cdot 0.99^{500}=0.00657...=0.6750\%$$ But your approximation is not so bad.


1

First, the mean of $X_1-Y_1$ is $\lambda-\lambda^{-1}$. Thus, the MM estimator of $\lambda$ is $$ \hat{\lambda}_n=\frac{1}{2}\left(m_n+\sqrt{m_n^2+4}\right), $$ where $m_n:=n^{-1}\sum_{i=1}^{n}(X_i-Y_i)$. For the asymptotic distribution of $\hat{\lambda}_n$, note that $$ \sqrt{n}\left(m_n-(\lambda-\lambda^{-1})\right)\xrightarrow{d}N\!\left(0,\lambda+\lambda^...


1

A Poisson Distribution gives the probability of a number of events in an interval generated by a Poisson process. The Poisson distribution is defined by the rate parameter, $\lambda$, which is the expected number of events in the interval: expected here means "on average". It's all about the average number of events in the given time interval. Now ...


1

You have an error in your exponential. Its $e^{-\lambda t}$. Then you use conditional probabilities and denote the poisson process by $(N_t)_{t\geq 0}$: $$P(X_t = k)=\sum_{n=0}^{\infty} P(X_t = k| N_t = n)P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n P(N_t = n)=\sum_{n=0}^{\infty} \left(\dfrac{1}{6}\right)^n \dfrac{e^{-\lambda t}(\lambda t)^n}{n!...


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