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Suppose $X$ is a Poisson random variable with mean $lambda$. The parameter $lambda$ is itself a random variable whose distribution is exponential with mean 1. Show ...

Question

Suppose $X$ is a Poisson random variable with mean $lambda$. The parameter $lambda$ is itself a random variable whose distribution is exponential with mean 1. Show that $P[X=n]=left(frac{1}{2}ight)^{n+1}$.

Suppose $X$ is a Poisson random variable with mean $lambda$. The parameter $lambda$ is itself a random variable whose distribution is exponential with mean 1. Show that $P[X=n]=left(frac{1}{2} ight)^{n+1}$.



Answers

Suppose $X$ is a Poisson random variable with mean $lambda$. The parameter $lambda$ is itself a random variable whose distribution is exponential with mean 1. Show that $P[X=n]=left(frac{1}{2} ight)^{n+1}$.

In this question, we are asked to show that the expectation on a binomial random variable is equal to end times p the four discrete random variable. The expectation is equal to the summation overall x of X times, the probability mass function and for a binomial random variable this is equal to So this is a binomial random variable with and trials and probability of success is p, and this is summed over X equals zero up to N. So if we consider the first term in the summation for X equals zero, the first term will equal zero. So we can also rewrite this as follows. No, I'm just writing out the formula for the probability mass function now in this combination operator. If we were to write that out in full, the numerator would have in factorial times and minus which would be n times n minus one times n minus two and so on. So we can factor out in end like this. We could just say it and and we could do the same with X so that in the bottom we would have x factorial. It would be X times X minus one and so on so we can rewrite that like that. And so now we can get rid of these excesses and we can move an end and a p weaken factor in end times p out to the left side. So this is where we're taking the p from. So we'll have a TV exponents X minus one instead of X. And now we will let why equal X minus one. So now the summation is from why equals zero two y equals and minus one. And we have n minus one. Choose why? Times P to the exponents. Why? Times one minus p to the exponents and minus one minus. Why, I just put a little bracket here to help us. And now let's make one more substitution. Let's let in equal and plus one. But let's only do the substitution for everything in the summation. So right here. So we're basically gonna leave this and alone Just leave it as in. So now we have an times p has the summation from why equals zero to em of n choose way times PD exponents. Why? Times one minus p to the exponents and minus. Why now? I haven't changed anything about this equation since thesis last line. So this quote, if this equality still holds, So all we're doing is representing in with M plus one. But they are equal by definition, right here. So this is still all valid. We can, of course, say that it is still equal now. One thing we can note here is that this is the crucial step is that this is the summation of all probability masses for the binomial random variable. So it's it's really just end times p. And to some, why equals zero to m of the probability masses forgetting why successes out of em trials with the probability of success of P. And when you sum all the probability masses of a discreet random variable, it is by definition, equal to one. So we can simply say that this is equal to N. P. And that completes showing that the expectation of a binomial random variable is equal to N times P

Were given independent random variables. X one through X n where X I has a negative binomial distribution with parameters are I N. P. Where asked These moment generating functions show that the some of X one through xn has a negative financial distribution as well. And to find the parameters of this distribution well, we have the moment generating function of random variable X I is p times e to the t over one minus one minus p times to the tee, all to the r I Yeah. Now, since we're assuming the X one through X in these air, all independent, it follows that woman generating function. Uh, x one some through X n This is going to be the product. The moment generating functions for each X I So we have p times e to the t over one minus one minus p times e to the t all to the r I. Well, our one first multiply it all the way through with p times eaten t over one minus one minus p times e to the t to the r n. And because the expressions inside the parentheses are all the same, using power rules We can write this as p Times e to the t over one minus one minus p times E to the t all to the r one summed through R M. So this is the moment generating function of a negative binomial riel variable in particular with parameters some of the r i and P. Yeah, so by uniqueness of moment generating functions, it follows that he's some from x 12 X n is negative by normally distributed with parameters some of the awry from I equals one to end and P.

We're giving independent exponential, random variables X and Y, with common parameter of Lambda. In part, they were asked to use convolution to show the X Plus y has a gambling distribution and to find the parameters of that distribution well, we have the probability density function of X because it is an exponential distribution is Lambda Times E to the negative Lambda X, where X is greater than zero. And likewise, this is also the probability density function of why, um did you the negative lambda Why or why is great in the zero So the probability density function of the they're able w which will take to the X plus why we determined using convolution probability density function This is the convolution probability density function for X with the probability density function for why, which is the integral from negative infinity to infinity Uh, f x of x times f y of instead of why we have w minus x which is equal toe. Why, according to our random variables, the X you mean in substituting we get It's a girl from negative infinity to infinity of you really were only considering X and y great in zero so X and Y are both created and zero. This implies that X is greater than zero. The B minus X is also greater than zero. So we have the excess lesson W and therefore the X lies between zero and W. So really only integrating instead from like infinity to infinity. From zero to W. This is thespian son, which both of these air non zero and we get lambda eat to the negative Lambda X Chinese Linda E to the negative Lambda W minus X dx. This can be written as the integral from your w of Lambda squared. And then we have the the negative Lambda X Times to the Lambda X becomes 20 which is one then times e to the negative Lambert W DX. And so we can clearly just factor at the Lambda Square to each of the negative Lambda W and integrates to get Lambda squared w each of the negative lambda W for W greater than zero. Now we see that this is a game, a distribution with parameters Alfa Equalling to in the parameter beta Equalling one over Lambda, next in Part B were asked to use the previous part or the previous exercise. I mean so exercise 64 to obtain the same result. I recall that Exercise 64 tells us that if X and y are independent gamma random variables with the same scale parameter beta, then the moment then that the sum of X and Y is also gamma distributed with the shape parameter which is the sum of the shape parameters for X and Y and the same scale parameter beta. So if X is exponentially distributed with parameter lambda course, this distribution is the same as a gamma distribution with parameters. Alfa equals one and beta equals one over Lambda. Of course, we also have that the random variable why is also exponentially distributed with parameter lambda. And of course, this distribution again is the same as a gamma distribution with parameters one and one over Lambda Well, by the previous exercise we have that's X plus Why? Well, this is also came a distribution and it has parameters. Well, the some of the alphas. So one plus one is two and the same scale perimeter data which in this case is one over lambda. When we see that this is the same result as we found in part a finally in part C rest to generalize part B so that if x one through xn are independent, exponential, random variables all becoming parameter lambda up to find the distribution of their some. Now we want to establish this using Meg moment generating functions. We have the moment generating function of the some X one some all the way through X m. Well, because these air independent, exponential random variables, this is the same as women generating function of X one times movement hearing function next to all the way up through moment generating function of Exxon, which for in exponential random variable, this is going to be the parameter lambda over the perimeter Lambda minus teeth. So we have lambda over Lambda minus t times Mhm lambda over Lambda minus T. And we have this multiplication and times. So this is the same as Lambda Overland the minus T to the end, which we can actually rearrange by flipping this as one over when the minus t Rolanda to the end, which is one minus t overland to the end. Yeah. In other words, one minus one over Lambda Times T Uh huh. make it even more explicit. Yeah, all to the end of power. We see that this is the moment generating function of the gamma distribution with perimeters and and one over Lambda. So it follows by the uniqueness of moment generating functions. That's the random variable X one summed all the way through. Xn is a gamma distribution with parameters and one over lambda. Let me see how this is.

For this question were given a possum process with a rate of Lambda. And we're also told that on the interval from zero to end, we had any events. So from right so say this is time n when you in this time that we have had and events that is given to us and so were asked between zero and one What is the probability that we have X events so x of the and events. So since it is a porcelain process, the number of events that occur in the given time is denoted as x of T. So what we're looking for is the probability that the number of events that occur between zero and one so in one unit of time, is equal to X. Given that given that we had given that the number of events that occurred between zero and N is equal Thio in. So this is our conditional probability given that between zero and n we had any events taking place. What is the probability that between zero and one x of those events take place in that time period? And so, by our probability laws, this is the probability that number of events taking place by one time unit secretly X intersected with the event that any events take place within the time from zero to, um divided by probability of any events taking place between zero and end time. So another way to look at that is to if we have actual events taking place between zero and one, we also then need to have had and minus X events taking place between Time one and M. So we can rewrite this probability in the numerator as probability that by time one we have X events times the probability in the duration and minus one. So that's from time. One until time and we have n minus X events. And we can simply multiply these two probabilities together since non overlapping time intervals are independent for opposing process. And we have to keep our denominator, which is the probability that we have any events by time N. And so we can work this out t e to the negative Lambda Times one. So that's just even negative. Lambda Times landed times one to the exponents X over X factorial times E to the negative lambda and now the time duration is and minus one, so we can have landed times and minus one. All of this to the exponents and minus X divided by end minus X factorial. And the denominator is still probability that by time and we have had any events And so let's right this out in more detail. So that's e to the negative round to end times lambda in to the exponents 10 divided by in factorial. So this is equal to the following. So this should be in brackets here like that. So all I've really done in this part is collected like terms from from this more cumbersome equation here. And so this gets a bit simpler. We can cancel these out. Lambda two exponents n is seen right here, and this comes out to you and minus one to the exponents and minus x times and factorial divided by X factorial times n minus X factorial times end to the exponents and so this is the answer. But you might also want to note that it is also equal to and shoes X times one over m to the exponents X times and minus one over em to the exponents and minus X, which is the binomial distribution for n Bernoulli trials with probability of success, one over n and probability of failure and minus one over in. Which makes sense because if you think of the duration, say from zero to end and some sub interval, which is 10 to 1, we're told that we have had any events between Time zero and N and for a plus on process. If we know that we've had any events, those events are equal, equally likely to have occurred anywhere along this time interval, so they're uniformly distributed. And for a uniformed distribution, the likelihood of any individual event falling between zero and one is equal to one divided by the total range of time. So that's n so basically, the probability of falling in this interval here is one divided by the total range of time, which is n. So that's our probability of success. And so Q is one minus that which is equal to and minus one over in, which also makes sense because this duration here is and minus one, and so the probability of an individual event falling in that time duration is and minus one divided by the total time, which is m, which is what we have here for the probability of failure. So another way to have answered this question would be to make this rationalization and then just presented as a binomial distribution. So now, for Part B, we're once again told that in the time interval from zero to end, we have any events that have occurred. So we're asked, what is the limiting conditional distribution of the number of events and the time interval from 0 to 1 as an approaches infinity. So remember, the probability X by one is equal to X is equal to what we had found in part a. So this part right here So we can rewrite this a little bit differently, though we can rewrite this as and minus one over em to the exponents and minus X times one over n to the exponents X times in factorial over Expect Auriol times in minus X factorial So all I have done here is I've just taken this end to the exponents m and rewritten it as and if exponents and minus x times into the exponents X So these air in the denominator over here and over here. And so the only mistake I think I've made so faras I should have expressed this as the conditional probability. So this would be probability number of events by time one equals X, given that the number of events by time and is equal to, um and so this is the conditional probability and we want the limiting conditional probability as n approaches infinity. So if we take the limit as an approaches infinity so we can just rearrange this a little bit. So we have and minus one over in the exponents and minus X times in factorial over and it exponents x times and minus X factorial times one over x factorial. So that's just some minor rearrangement of the previous expression. And so now if we take the limit when an approaches infinity, this component comes out to one and this component comes out to one and this component comes out to one over x factorial. So it's not too hard to see for this one. I mean, if we have, if Anna's approaching infinity, we basically have infinity over infinity so that that fraction is one for this one. As an approaches infinity in factorial is like ended exponents n and to the exploding exes like end of exploring X and end minus X factorial is like end to the exponents and minus X. So this product would give you and the exponents end, which is what we have in the numerator as well. So that comes out to one. And so the final answer comes out to one over x factorial.


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