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5 pts_ Suppose that Yi.-Kz. In form a random sample from a distribution for which the p.d.f: is given by f(zle) = e } 2 > 0where the unknown parameter 0 > 0, ...

Question

5 pts_ Suppose that Yi.-Kz. In form a random sample from a distribution for which the p.d.f: is given by f(zle) = e } 2 > 0where the unknown parameter 0 > 0, Suppose also that the improper prior distribution of 0 is given by T(0)Find the posterior distribution w(8lx).

5 pts_ Suppose that Yi.-Kz. In form a random sample from a distribution for which the p.d.f: is given by f(zle) = e } 2 > 0 where the unknown parameter 0 > 0, Suppose also that the improper prior distribution of 0 is given by T(0) Find the posterior distribution w(8lx).



Answers

Determine whether the distribution is a discrete probability distribution. If not, state why. $$\begin{array}{ll} \boldsymbol{x} & \boldsymbol{P}(\boldsymbol{x}) \\ \hline 0 & 0.1 \\ \hline 1 & 0.5 \\ \hline 2 & 0.05 \\ \hline 3 & 0.25 \\ \hline 4 & 0.1 \end{array}$$

For this exercise. We have a sample of size N drawn from a normal distribution with a mean zero and standard deviation is square root of one over towel one over towel is called the precision parameter. And for the prior distribution of towel, we are asked to assume it is a gamma distribution which is given by this formula. So this is our model distribution with unknown parameter. How and this is our prior distribution for the unknown parameter. The first we can calculate our joint distribution for our sample of size n. So these all come from this distribution. So the joint distribution or the joint pdf is the following products. So each of these is a normal distribution, except that instead of Sigma, we have the square root of one over towel substituted and so this can be simplified to the following. And now, to look at our posterior distribution, I'm going to say that it is proportional to the following and I'm going to ignore the denominator in the posterior distribution because it is a constant and so we can determine the distribution just from the numerator. So what I'm doing here is I'm going to multiply the prior distribution with this joint distribution. Now, if I just represent all of the constants in this expression with C and it looks like this and we can see that this is the form of a gamma distribution where this is alpha minus one and this is one over beta. And so for therefore, we can say that our posterior distribution is indeed a gamma distribution with the first parameter alpha plus and over two. And the second parameter is one over the following, and so that is the final answer.

This problem. We have been given a table, and we want to determine whether the distribution is in fact, a discrete probability distributions. If it's not, we need to determine why. As you remember, there are two things that a probability distribution must follow. And the first is that each probability is greater than equal zero, which means that the probabilities have to be non negative. The second is that the some of the probabilities most people want now start with the first condition there. Look at all of your probability. So look at everything that falls underneath The p of X value in that truck have a 0000 and then one. All of those are greater than or equal to zero. That's what means that holds true. Next, you need to check. Do they add up to one? So 2000 Yeah, plus zero plus one. Does that equal one? And yes, it does. And so it satisfies both of these two conditions. And so that tells us Yes, it is a discrete probability distribution function. Yeah,

In this exercise, we're We're told that X is a binomial distribution with parameters n and Unknown parameter P, and we are also given a prior distribution which is a beta with parameters alpha and beta. So if we draw a sample of size n, the number of successes will be based on a probability mass function given as follows. So this is just the PMF for a binomial distribution with parameters NNP Now for our prior distribution, it's a beta with parameters alpha and beta. And we also know that the end points are zero and one because P has to be between zero and one, it is the probability of success. So with endpoints zero and one, the pdf for P then simplifies to the following. And now to consider our our posterior posterior distribution, I'm only going to look at the numerator. So we will say that the posterior distribution is proportional to the following. We can learn the form of the posterior distribution, looking only at the numerator since what will be in the denominator is just a normalizing constant. Now there are various constants in this expression, so we can say that this expression is proportional to the following and we can see that this distribution is the form of a data distribution where this value is alpha minus one and this value is beta minus one. So, in other words, this is the Colonel. Have a beta distribution with parameters Alpha plus X and beta plus n minus X, So our posterior distribution is a beta with parameters alpha plus X and beta plus n minus X.

Yeah, this problem. We would like to determine if the distribution that we have been given as a discrete probability distributions now, before we before we actually talk about this specific problem, it is important to remember the two properties that makes the distribution discrete probability distributions. Mhm. The first is that each probability has to be paraded than or equal to zero. Probabilities cannot be negative, and that's what this first part tells us. The second part is that the sum of the probabilities must equal one. Whenever you add up the probabilities of everything, it must give us a value of one. So let's check it here and see if these two things are true. The first part is very easy to verify. We just looked through at that chart. All values are positive. That's what satisfies that first condition there. Now for the second one we want to know does p of zero plus p of one possibly of two lost p of three just before we want to know Does that equal one mhm now P of zero is 0.1 p of one is 0.5 pf two is 0.5 p of three is 0.25 NPF 40.1. And so when we add all these together, this does give us one. And so that checks out. And so it satisfies both of these two conditions here. And so that tells us, yes, it is a discrete probability distributions.


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