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A random sample X1,x2, X,, is drawn from distribution with the pdf f(s;8) = 4S(1 2}49-1 for 0 < $<l,$ >0. The method of moments estimator of $I+11nonc of t...

Question

A random sample X1,x2, X,, is drawn from distribution with the pdf f(s;8) = 4S(1 2}49-1 for 0 < $<l,$ >0. The method of moments estimator of $I+11nonc of the answers provided herc4(1-2)

A random sample X1,x2, X,, is drawn from distribution with the pdf f(s;8) = 4S(1 2}49-1 for 0 < $<l,$ >0. The method of moments estimator of $ I+1 1 nonc of the answers provided herc 4(1-2)



Answers

Refer to Exercise $3.86 .$ The maximum likelihood estimator for $p$ is $1 / Y$ (note that $Y$ is the geometric random variable, not a particular value of it). Derive $E(1 / Y) .[\text { Hint: If }|r|<1$, $\left.\sum_{i=1}^{\infty} r^{i} / i=-\ln (1-r) .\right]$.

In this question, we are told that X is a binomial random variable parameter and is 15 and peace have. So that's this column. And were asked to compute the probability distribution from the cumulative distributions. So that's basically taking the difference of each of the values. So we're going to have X and the affects From zero all the way up to 15. So there's zero 123 or Yeah, 567, 8, 9 and 10 11, 12, 13, 1415. We have our values of X. And we are going to fill in our values of P of X. So for X0, that's just zero for excess one. It's just .05 for access to its Like point or 37 -1005. Which is this value of here? When X is three we take point or one 76 -1037. The basis point or 139. When X is four, can we take the difference with zero Finite 290176 .0416. Okay. And then when X is five, we take .1509 minus wind, zero, finite too, .0917 6.3036 .1509 5 to 7 XS seven. We have .5- the previous value .1964, excessive heat 6964. So that's again .1964. This is nine. We have .8491 -6964. That's 1 5- seven. Can we have 9408 minus the previous fragile four, And then Sure Or minus 940 Hit point for 416 and then .996, -18-4 .4139 For 13, 999, 5 -996, 3 00, 14, it looks five And then for 15 it's just zero. So we have our accumulative distribution converted to the probability.


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