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Let Xi,Xz, variables isXn "4 Exponential(8). The probability density function for these randomf(r18)e-5 , 1 > 0; 8 > 0.Find the maximum likelihood estima...

Question

Let Xi,Xz, variables isXn "4 Exponential(8). The probability density function for these randomf(r18)e-5 , 1 > 0; 8 > 0.Find the maximum likelihood estimator for 8.b. Compare the MLE to the estimator 8 (Xi + Xn)/2. Which do you recommend? Explain. You can use the following facts: E(X;) = 8 and Var(X;) = 82 , for anyConstruct a 100(1 _ a)% C.I. for B using the pivotal quantity (2nBxLe) /8. You can use the fact that (2nBuLe) /8 xn=Is 1/8 MLE unbiased for 1/82 Prove youT claim. You ca

Let Xi,Xz, variables is Xn "4 Exponential(8). The probability density function for these random f(r18) e-5 , 1 > 0; 8 > 0. Find the maximum likelihood estimator for 8. b. Compare the MLE to the estimator 8 (Xi + Xn)/2. Which do you recommend? Explain. You can use the following facts: E(X;) = 8 and Var(X;) = 82 , for any Construct a 100(1 _ a)% C.I. for B using the pivotal quantity (2nBxLe) /8. You can use the fact that (2nBuLe) /8 xn= Is 1/8 MLE unbiased for 1/82 Prove youT claim. You can use the facts Xi-IXX; Gamma(n, 8) and E [(Ei-1X;)*] (8*T(n + k)) /T(n) when ~n and Ci-IX; Gamma(n, B).



Answers

Let $Y_{1}, Y_{2}, \ldots, Y_{n}$ denote a random sample from the density function given by $$f(y | \theta)=\left\{\begin{array}{ll}\left(\frac{1}{\theta}\right) r y^{r-1} e^{-y^{\prime} / \theta}, & \theta>0, y>0 \\0, & \text { elsewhere }\end{array}\right.$$ where $r$ is a known positive constant. a. Find a sufficient statistic for $\theta$ b. Find the MLE of $\theta$ c. Is the estimator in part (b) an MVUE for $\theta$ ?

In problem 15. We want to identify this probability density function over this interview, then find the mean variance and the standard division. Without integration we can rewrite F of X to be in the form, one divided by it multiplied by E to the power of minus one, divided by eight, multiplied by X. This is in the form A. But deployed by E is about uh minus X and we have X starting at zero to infinity. This is X. This is a form of the exponential probability density function, then F of X is example nature. Then we can find the mean for the exponential distribution equals one divided by a here equals one divided by it, then it's one divided by one divided by eight equals And we can get the variants equals one divided by a squared equals one divided by one divided by eight. All squared equals 64. And finally we can get sigma which equals square root of the variance of X equals square root of 64 equals it.

For part one. In order to verify that this is a probability density function, we're going to integrate over its entire domain, which is going to be from zero up to infinity. And we could take care of this improper and a girl right away by just rewriting this with the limit. So let's go ahead and put in front here. The limit as a dummy variable like beat approaches infinity, and then we don't run into an issue of trying to substitute zero or infinity directly in here. So let's go ahead and take the anti derivative of this. Well, first off, we're going to have to use a use substitution and let's just quickly do that here. So, for instance, um, if the U is negative t over seven, do you is going to be negative 1/7 and then we could multiply that back over to the other side and get negative seven. Do you? Because DT and so in the end of this will have another factor of negative seven out front if we had a great desk. And so then the integral of either the EU is either you plug back in and we're gonna end up getting, um, negative. The seventh cancel e to the negative t over seven from zero to be. And of course, we still have the limit as be approaches zero there as well. So let's keep writing this. This would be e negative e So the negative b over seven minus a negative e to the zero. The two minuses would be a plus. Then the first term, if we let me go off to infinity, will be negative. Infinity e to the negative Infinity is all zero e to the zero is one. Therefore, the probability density function does equal one which we wanted to. So that's good. The next part is finding the probability that the value is between zero and four and we'll go ahead and integrate this and so again, same and a girl 1/7 e to the negative t over seven. Well, we could just jump right ahead to the anti derivative which we just found, which was negative e to the negative t over seven from 0 to 4 four. And then if we replace this year, we're going to get negative. Eat. So the negative four over seven and then minus minus will be a plus e to the zero. And so it's basically going to be eating. Zero, of course. Is one subtract something. So I'm gonna put this into the calculator here just to verify what this is. So we have negative e to the negative for seventh, and then all of that, we could just say, plus one that will get us approximately. Well, that is our answer or exact answer. But of course, it helps toe have a dust more. So this is about 43% 0.435 to 8 would be the probability that the event is between X zero and access for. So that's part B. And then lastly, we get thio the expected value where we are integrating again over the entire domain. So from zero to infinity. So we're gonna have to implement that idea of, um, the limit as he goes off to infinity there. So I'm just gonna go ahead and do that here, So let's go ahead and rewrite that a little bit. So we've got the moment as B goes off to the family and then we go from zero to be of X. Times are function, which is the the negative teeth over seven 1/7. Sorry, I forgot that part. And then DT for this one, we're actually gonna have to use integration by parts, so this will be a bit longer here, Um, because of already having done part a and verifying that Let's go ahead and get rid of that. And let's continue over here. So if we do integration by parts, we're gonna have our derivative column and are in a girl column. The derivative column is going to be extra votive of excess. One derivative of one is zero, and a girl column is 1/7 Mhm. So the negative t over seven. Now, instead of multiplying by negative 1/7 we're gonna be dividing by negative 1/7 which is equivalent Thio each step of the way, multiplying by a factor of negative seven here. So the first one is going to just be negative e to the negative t over seven. And that makes sense, because that was the integral that we had gotten in the previous step down here anyway, so that's good. And then times another factor of negative seven would be positive. Seven e to the negative t over seven, and we're going to multiply these terms with an additional positive factor, these terms with a negative factor. So let's go ahead and write out now the Senate girl would turn into so it would become. We still have the limit as B goes off to infinity, so we'll still keep that there. Bring that down. I'm just gonna not right it to save some space here. Hope them will have negative x times e to the negative t over seven and then that minus seven, be to the negative t over seven from, uh, zero could be. So if we plug being, it would be. And then we let me go off to infinity. That's what end up being e to the negative infinity. The second term would also be minus in to the negative infinity. Both of those end up becoming zero. So we can just completely ignore the first two terms and then really just focus on the last two terms here, and so that will be plugging zero in the x zero. So then it will just be minus seven and then e to the zero that eating zero is a one minus minus turns the expected value into seven. So we find out that the expected value is seven, and that's it.


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