I checked whether this question was answered before but because of notation, it's hard to see. I am a reading a paper that defines the following two RVs $$ z \mid y \sim Binomial(\pi, y) \\ y \sim Poisson(\lambda) $$ then concludes (by integration and Bayes Rule) that $$  z \sim Poisson(\pi \cdot \lambda) \\ y - z \sim Poisson( (1 - \pi)\cdot \lambda) $$ I tried working it out on paper but since I am not a trained statistician, I am not sure where I am going wrong. If I want to derive $z \sim Poisson(\pi\lambda)$, then I use the conditional probability i.e. $$ p(z) = \int_y p(z, y) \,\, dy $$ where $p(z, y)$ is the joint probability. Expanding this out, I have $$p(z) = \int_y {y\choose z} \pi^z (1 - \pi)^{(y - z)} \frac{e^\lambda \lambda^y}{y!} \, \, dy $$ but I am guessing I have to get to the following equation somehow $$p(z) = \frac{e^{\pi\lambda} (\pi\lambda)^{z}}{z!}$$ but I was not able to manipulate the integral above to get this form. Not sure if that's even possible.
5485
I have been stuck with a problem for a couple of days regarding the distribution of outcomes from a two-stage process. Specifically, what is the distribution of the number of successes in a poisson process followed by a series of bernoulli trials, where the number of trials is determined by the result of the poisson process. If this is not clear, I’ll explain by way of example.  Suppose we have a hunter who is setting out to catch rabbits. To do this he sets up a trap. Rabbits pass over the trap at a rate of $r$ times per day (poisson distributed). However, the trap is not very good. If a rabbit walks over the trap, there is only a $p$ (probability) chance that it activates.  If the trap has no limit on the amount of rabbits it can catch, what does the distribution of rabbits caught per day look like?  My intuition is that it should follow a poisson distribution with mean $r.p$. However, I have been unable to prove this analytically. I figure that the probability of a given number of ‘successes’ should be calculable through the poisson distribution and the binomial distribution. Something like:  $ P(X=x) = \sum_{i=x}^{Inf} poisson_{pmf}(i | r) . binomial_{pmf}(x | i, p) $  We can then substitute in the respective mass functions into the above equation. However, beyond that I am getting stuck. Any help or a point in the right direction would be great appreciated.
30668
The classic example... $X \sim Po\left (\lambda\right ), Y \sim Po\left (\mu\right)$, X and Y are independent.  Show that the conditional distribution of X is binomially distributed. Or in other words, $P(X=k\mid X+Y = n) = P (\tilde{X} = k), \tilde{X} \sim B\left (n ,\frac{\lambda}{\lambda + \mu}\right )$.   I've so far managed to reach to this step, and have been stuck since. Just somehow gotta get a $\frac{1}{n!}$ in the denominator, that would then complete the proof..or at least I think..  $$P(X=k\mid X+Y=n) = \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{P(X+Y = n)}=  \frac{\frac{\lambda^{k}\mu^{n-k}}{k!(n-k)!}}{\sum_{i=1}^{n} \frac{\lambda^{i}\mu^{n-i}}{i!(n-i)!}}$$  Thanks for the help!
31422
Suppose that $X\sim N(0,1)$, $Y\sim N(0,1)$ with correlation $(X, Y) =ρ$ where $ρ ∈ (−1, 1)$.  Show the following,   Correlation $(Φ(X),Φ(Y))=\dfrac6π \arcsin\dfrac ρ2 $.  Here $Φ(X)$, and $Φ(Y)$ denote the CDF of Random variables $X$ and $Y$ respectively.  What I know so far:  $$\text{Cov}( Φ(X),Φ(Y)) =E(Φ(X)Φ(Y)) - E(Φ(X)) ×E(Φ(Y)) $$  We also know since  $X \sim N(0,1)$ and $Y\sim N(0,1)$;  $Φ(X)\sim\text{unif}(0,1)$, $Φ(Y)\sim\text{unif}(0,1)$. Hence $$E(Φ(X)) =E(Φ(Y)) =\frac12.$$  I am stuck in finding $E(Φ(X)Φ(Y))$ [I tried to find it using double expectation].  What I have written may be the wrong way of attempting the question. So I am thankful for any help.
59224
Assuming $X\sim\mathcal{P}(\lambda)$ and $Y\sim\Gamma(w,1)$ prove that $P(X\ge w)=P(Y\le \lambda)$. How this fact is lead from the connection between the poisson and exponential distributions?   I don't know from where to start. poisson is defined only for discrete situations but the exponential is only for continious situation. How can I prove the fact ? EDIT: for gamma distribution I wrote that $f_Y(y)=\frac{y^{w-1}e^{-y}}{\Gamma(w)}$, but I have problem with integrating it. About Poisson: its function is $\displaystyle \sum _{w_i=0}^w P(X=w_i)$ which I don't know how to sum into a final expression. How can I continue?
34837
Being $X$ and $Y$ independent random variables distributed as $\mathrm{Exponential}(1)$ and let $T=X+Y$ Calculate $E(X|T)$.  This is my attempt at solving this:  Being $Z=\sum_{i=1}^{n}P_{i}$ where $P_{i}$~$\mathrm{Exp}(\lambda)$ then using it's MFG you have that $$M_{P_i}(t)=\frac{\lambda}{\lambda-t}$$ then $$M_{Z}(t)=M_{\sum{P_{i}}}(t)=(M_{P_{1}})...(M_{P_{n}})=(\frac{\lambda}{\lambda-t})^n$$ with this being the MFG of a $\mathrm{Gamma}$ distribution with parameters $(n,\lambda)$  With this result is easy to see that $T$~$\mathrm{Gamma}(2,1)$  Now $E[X|T]=\int xf_{X|T}(x|t)dx$  Then using Bayes' theorem we can make $$f_{X|T}(x|t)=\frac{f_{T|X}(t|x)f_{X}(x)}{f_{T}(t)}$$ then $$f_{T|X}(t|x)=Pr(T=t|X=x)$$ $$=Pr(X+Y=t|X=x)$$ $$=Pr(x+Y=t|X=x)$$ $$=Pr(Y=t-x)=f_{Y}(t-x)$$  Using this result we can now rearrenge $$f_{X|T}(x|t)=\frac{f_{Y}(t-x)f_{X}(x)}{f_{T}(t)}$$ $$=\frac{e^{-(t-x)}.e^{-x}}{\frac{te^{-t}}{\Gamma(2)}}$$ $$=\frac{1}{t}$$ then $$E[X|T]=\int (x)(\frac{1}{t})dx$$  And this where I got stuck because $1_{x\epsilon(0,\infty})$, I made this explanation step-by-step hoping that someone can se easily check where I went wrong. I would aprecciate a step-by-step answer, thanks.
79012
In "Data Analysis" by D. S. Sivia, there is a derivation of the Poisson distribution, from the binomial distribution.   They argue that the Poisson distribution is the limiting case of the binomial distribution when $M\rightarrow\infty$, where $M$ is the number of trials.  Question 1: How can that argument intuitively be understood?  Question 2: Why is the large-$M$ limit of $\frac{M!}{N!(M-N)!}$ equal to $\frac{M^{N}}{N!}$, where $N$ is the number of successes in $M$ trials? (This step is used in the derivation.)
41990
Let $X_1,X_2,...$ be a sequence of i.i.d Bernoulli random variables   with parameter $p$. Let $N$ be a Poisson random variable with with   parameter $\lambda$ which is independet of the $X_i.$      (a) Find the probability generating function of $Z=\sum_{i=1}^{N}X_i.$      (b) Use (a) to identify the distribution of $Z$.   The pgf of a sum of i.i.d random variables is the product of their respective pgf's so:  $$G_Z(s)=G_{X_1}(s)\cdot G_{X_2}(s)\cdot...\cdot G_{X_N}(s),$$  and sice the $X_{i}$ are Bernoulli distributed with parameter $p$ we have that  $$G_Z(s)=(1-p+ps)^N.$$  I'm not sure how to proceed from here. I can't express my $G_Z(s)$ in terms of a random variable, so what should I do with the $N$?
117331
Suppose that $X$ and $Y$ are chosen randomly and independently according to the uniform distribution from $(0,1)$. Define $$ Z=\frac{\max(X,Y)}{\min(X,Y)}.$$ Compute the probability distribution function of $Z$. Can anyone give me some hints on how to proceed? I can only note that $\mathbb{P}[Z\geq 1]=1$ and $$F_Z(t)= \mathbb{P}[Z \leq t]=\mathbb{P}[X\leq Y, Y\leq tX]+\mathbb{P}[Y \leq X, x\leq tY]$$
104952
The definition of zero-truncated Poisson (ZTP) distribution is:   \begin{align} g(k;\lambda)= P(X = k | X &gt; 0) &amp;= \frac{f(k;\lambda)}{1-f(0;\lambda)}  \\[8pt]                                &amp;= \frac{\lambda^ke^{-k}}{k!(1-e^{-\lambda})}  \\[8pt]                                &amp;= \frac{\lambda^k}{(e^\lambda - 1)k!} \end{align}  The definition of conditional probability is:  $$ P(A|B)=\frac{P(A\cap B)}{P(B)} $$  Now, $A = f(k;\lambda)$ and $B = 1-f(0;\lambda)$, but how is $A\cap B = f(k;\lambda)$?  Source:
66471
I'm trying to obtain the Jeffreys' prior for a negative binomial distribution. I can't see where I go wrong, so if someone could help point that out that would be appreciated.  Okay, so the situation is this: I am to compare the prior distributions obtained using a binomial and a negative binomial, where (in both cases) there are $n$ trials and $m$ successes. I get the right answer for the binomial case, but not for the negative binomial.   Let's call the Jeffreys' prior $\pi_J(\theta)$. Then,  $$ \pi_J(\theta)\propto [I(\theta)]^{1/2}. $$  Under the regularity conditions (fulfilled as we are dealing with the exponential family),  $$ I(\theta)=-E\left(\frac{\partial^2 \log L(\theta|x)}{\partial \theta^2}\right) $$ where for the negative binomial $n$ is $x$ in the above expression (the total number of successes $m$ is fixed, $n$ is not). The distribution --I think-- is  $$ p(m|\theta)\propto\theta^m(1-\theta)^{n-m} $$ since $\theta$ is defined as the probability of success and $m$ is the number of successes. This is also the likelihood, since $m$ is a scalar and not a vector. Hence,  $$ L(\theta|n)\propto\theta^m(1-\theta)^{n-m}\\ \log L(\theta|n)=m\log\theta +(n-m)\log (1-\theta)\\ \frac{\partial\log L(\theta|n)}{\partial \theta}=\frac{m}{\theta}-\frac{n-m}{1-\theta}\\ \frac{\partial^2\log L(\theta|n)}{\partial \theta^2}=-\frac{m}{\theta^2}-\frac{n-m}{(1-\theta)^2} $$ so the Fisher information is  $$ I(\theta)=-E\left(\frac{\partial^2\log L(\theta|n)}{\partial \theta^2}\right)=\frac{m}{\theta^2}+\frac{E(n)-m}{(1-\theta)^2}=\frac{m}{\theta^2}+\frac{\frac{m\theta}{1-\theta}-m}{(1-\theta)^2}\\ =\frac{m(1-\theta)^2+\frac{m\theta^3}{(1-\theta)}-m\theta^2}{\theta^2(1-\theta)^2}=\frac{m(1-2\theta)+\frac{m\theta^3}{(1-\theta)}}{\theta^2(1-\theta)^2}\\ =\frac{m(1-2\theta)(1-\theta)+m\theta^3}{\theta^2(1-\theta)^3}=\frac{m(1-3\theta+2\theta^2+\theta^3)}{\theta^2(1-\theta)^3}\\ \propto\frac{1-3\theta+2\theta^2+\theta^3}{\theta^2(1-\theta)^3} $$  This, however, does not give me the correct answer. The correct answer is  $$ \pi_J(\theta)\propto \frac{1}{\theta(1-\theta)^{1/2}} $$ which means that the Information I get should be   $$ I(\theta)=\frac{1}{\theta^2(1-\theta)} $$ since the prior should be proportional to the square root of the information.   Can anyone find any mistakes? I wouldn't be surprised if I screwed something up with the set up of the distribution (successes vs failures with their respective probabilities, etc).   I used the expected value from  and I know the correct answer from .
72596
Given $n$ independent geometric random variables $X_n$, each with probability parameter $p$ (and thus expectation $E\left(X_n\right) = \frac{1}{p}$), what is $$E_n = E\left(\max_{i \in 1 .. n}X_n\right)$$    If we instead look at a continuous-time analogue, e.g. exponential random variables $Y_n$ with rate parameter $\lambda$, this is simple: $$E\left(\max_{i \in 1 .. n}Y_n\right) = \sum_{i=1}^n\frac{1}{i\lambda}$$  (I think this is right... that's the time for the first plus the time for the second plus ... plus the time for the last.)  However, I can't find something similarly nice for the discrete-time case.    What I have done is to construct a Markov chain modelling the number of the $X_n$ that haven't yet "hit". (i.e. at each time interval, perform a binomial trial on the number of $X_n$ remaining to see which "hit", and then move to the number that didn't "hit".) This gives $$E_n = 1 + \sum_{i=0}^n \left(\begin{matrix}n\\i\end{matrix}\right)p^{n-i}(1-p)^iE_i$$ which gives the correct answer, but is a nightmare of recursion to calculate. I'm hoping for something in a shorter form.
94952
I'm reading some course notes on Bayesian statistics and in one of the slides entitled 'model evidence' it writes: $$p(y|m)=\int{p(y,\theta|m)d\theta}=\int{p(y|\theta,m)p(\theta | m)d\theta}$$ &quot;Because we have marginalised over $\theta$ the evidence is also known as the marginal likelihood.&quot; I have two issues here.  I don't understand how does $p(y,\theta|m)$ become $p(y|\theta,m)p(\theta | m)$. Is this derived from the multiplication rule of dependent events $P(A,B)=P(A|B)P(B)=P(B|A)P(A)$? If so - I don't see how. What is this linked to &quot;we have marginalized over theta&quot;? What does $m$ really stand for? I know what it's just supposed to stand for but it's just beyond my comprehension. How is it related to model parameters?
50264
This is a problem in KaiLai Chung's A Course in Probability Theory.     Given a nonnegative random variable $X$ defined on $\Omega$, if $\mathbb{E}(X^2)=1$ and $\mathbb{E}(X)\geq a &gt;0$, prove that $$\mathbb{P}(X\geq \lambda a)\geq (a-\lambda a)^2$$   for $0\leq\lambda \leq 1$.   Let $A=\{x\in \Omega:X(x)\geq \lambda a\}$, we get $$\int_A (X-\lambda a)\geq a-\int_A\lambda a -\int_{A^c}X$$ and $$\int_A (X^2-\lambda^2 a^2)=1-\int_A\lambda^2a^2-\int_{A^c}X^2$$ I want to contrast $\int_A (X-\lambda a)$ and $\int_A (X^2-\lambda^2 a^2)$, but I don't know how to do it, could anyone gives me some hints?
18106
Let $X$ ~ $Bin(n,p)$ and $Y$ ~ $Bin(m,p)$ be two independent random variables. Find the distribution of $Z=X-Y$.   see also   I figured this out:  $$ P(Z=z)=\cases{\sum_{i=o}^{min(m,n)} Bin(k+i,n,p)*Bin(i,m,p), &amp;if $z\ge0$;\cr \sum_{i=0}^{min(m,n)} Bin(i,n,p) * Bin(i-z, m, p),&amp;otherwise. \cr}$$  I also validated it by Monte Carlo simulation. For $n=30$, $m=20$ and $p=0.5$, I get the following distribution, where the circles are the analytical probabilities and the line connects the MC estimates.    Because that looked to me pretty much like a binomial distribution, I gave it a try and figured out that its actually a binomial, just shifted by m to the left. This can be simply written as $P(Y=y) = Bin(y+m, m+n, p)$. Hence, given equal success probabilities, the sum of two independent binomially distributed random variables is binomial, but also their difference, just shifted to the left.  This question here  is actually the same as mine, but received no answer and only the comment that there would be no simple formula. But the above formula looks pretty simple to me.   Is it correct that for the case of equal success probabilities, the above equations actually describes the distribution of $Z=X-Y$? I read in a book that $Z$ could not be binomial distributed because it had negative support. Is it right to call it a shifted binomial?
40707
A car hire firm has three cars, which it hires out on a daily basis. Number of cars demanded per day follows a poisson distribution with mean 2.1  a) Find the probability that all cars are in use on any one day.   b) Find the probability that all cars are in use on exactly three days of a five day week.  What I tried:  For the a) part: normal poisson formula with mean = 2.1 and x = 3   $$\frac{(e-2.1)(2.1^3)}{6}$$  Got the answer as 0.1890. The book's answer is 0.350   For the b) part: did poisson to binomial. Found out p = 0.7. Then used 5C3. But I think my method here is wrong.
104418
$X$ and $Y$ are iid r.v.s where $f(x)=\lambda e^{-\lambda x}$, $x&gt;0$. We are given $Z=X/Y$ and asked to find the CDF and density for $Z$.  I tried doing this using the multivariate transformation formula, where   $f_{W_1,W_2}(w_1,w_2)=f_{X_1,X_2}(x_1,x_2)|J|$, but I ended up getting that $f_Z(z)=\frac{1}{(1+z)^2}$, which results in a negative CDF. I was hoping to see a different method of doing this. Any help appreciated.  Edit: Here is my work for obtaining my answer:  $Z=X/Y$; we then let $W=Y$ and $X=ZW$. The Jacobian is then calculated to be $w$. That means we have:  \begin{align} f_Z(z) &amp; =\int_0^\infty f_X(x)f_Y(y)w\,dw \\[10pt] &amp; =\int_0^\infty \lambda e^{-\lambda x}\lambda e^{-\lambda y}w \, dw \\[10pt] &amp; =\lambda^2 \int_0^\infty e^{-(z+1)\lambda w}w\,dw \\[10pt] &amp; =\frac 1 {(z+1)^2} \end{align}
96676
Suppose $X$ and $Y$ are independent random variables on $\mathbb{R}$ having pdf $f(t)=\frac{1}{\pi} \frac{1}{1+t^2}$. Define $Z=\frac{X+Y}{3}$ determine the pdf of $Z$.   So the pdf of $(X,Y)$ is $f(x,y)=\frac{1}{\pi^2 (1+x^2)(1+y^2)}$.   We shall find the CDF of $Z$ and then differentiate it.  $F(z)=P(Z \leq z) = \int_{-\infty}^{\infty} \int_{-\infty}^{3z-x} \ \frac{1}{\pi^2 (1+x^2)(1+y^2)} \  dy \ dx= \int_{-\infty}^{\infty} \frac{1}{\pi^2}\left(\frac{\pi}{2(1+x^2)}- \frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.  At this stage I am having trouble evaluating $\int_{-\infty}^{\infty} \left(\frac{\tan^{-1}(x-3z)}{1+x^2}\right) \ dx$.  How do I do this? Or otherwise is there different more elegant way to get the required pdf?