Question

3. (a) Let X?, X?, ..., X? denote independent and identically distributed random variables from a distribution with pdf given by f(x) = (1/?²) xe??/?, for x > 0, where ? > 0 is an unknown parameter. (i) Find the maximum likelihood estimator, ?? for ?. (ii) Determine whether ?? is an unbiased estimator. (iii) What is the maximum likelihood estimate of ? if a random sample of size 10 yields the sample values of 126, 120, 141, 135, 123, 134, 132, 125, 129 and 138? (b) Let X?, X?, ..., X? denote independent and uniformly distributed random variables on the interval [0, ?]. Find (i) the pdf of X???, the kth order statistic, where k is an integer between 1 to n. (ii) E[X???]. [Hint: ??¹ x??¹(1 ? x)??¹ dx = ?(?)?(?)/?(?+?) where ?(?) is a gamma function and, ? and ? are unknown parameters.] 

          3. (a) Let X?, X?, ..., X? denote independent and identically distributed random variables from a distribution with pdf given by
f(x) = (1/?²) xe??/?, for x > 0,
where ? > 0 is an unknown parameter.
(i) Find the maximum likelihood estimator, ?? for ?.
(ii) Determine whether ?? is an unbiased estimator.
(iii) What is the maximum likelihood estimate of ? if a random sample of size 10 yields the sample values of 126, 120, 141, 135, 123, 134, 132, 125, 129 and 138?

(b) Let X?, X?, ..., X? denote independent and uniformly distributed random variables on the interval [0, ?]. Find
(i) the pdf of X???, the kth order statistic, where k is an integer between 1 to n.
(ii) E[X???].
[Hint: ??¹ x??¹(1 ? x)??¹ dx = ?(?)?(?)/?(?+?) where ?(?) is a gamma function and, ? and ? are unknown parameters.]
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3. (a) Let X?, X?, ..., X? denote independent and identically distributed random variables from a distribution with pdf given by f(x) = (1/?²) xe??/?, for x > 0, where ? > 0 is an unknown parameter. (i) Find the maximum likelihood estimator, ?? for ?. (ii) Determine whether ?? is an unbiased estimator. (iii) What is the maximum likelihood estimate of ? if a random sample of size 10 yields the sample values of 126, 120, 141, 135, 123, 134, 132, 125, 129 and 138? (b) Let X?, X?, ..., X? denote independent and uniformly distributed random variables on the interval [0, ?]. Find (i) the pdf of X???, the kth order statistic, where k is an integer between 1 to n. (ii) E[X???]. [Hint: ??¹ x??¹(1 ? x)??¹ dx = ?(?)?(?)/?(?+?) where ?(?) is a gamma function and, ? and ? are unknown parameters.]

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Elementary Statistics a Step by Step Approach
Allan G. Bluman 9th Edition
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3. (a) Let X₁, X₂, ..., Xₙ denote independent and identically distributed random variables from a distribution with pdf given by f(x) = (1/̠²)xe⁻ˣ/β, for x > 0, where β > 0 is an unknown parameter. (i) Find the maximum likelihood estimator, β̂ for β. (ii) Determine whether β̂ is an unbiased estimator. (iii) What is the maximum likelihood estimate of β if a random sample of size 10 yields the sample values of 126, 120, 141, 135, 123, 134, 132, 125, 129 and 138? (b) Let X₁, X₂, ..., Xₙ denote independent and uniformly distributed random variables on the interval [0, β]. Find (i) the pdf of X₍ₖ₎, the kth order statistic, where k is an integer between 1 to n. (ii) E[X₍ₖ₎]. [Hint: ∫₀¹ xᵃ⁻¹(1 − x)ᵇ⁻¹ dx = Γ(α)Γ(β)/Γ(α+β) where Γ(⋅) is a gamma function and, α and β are unknown parameters.]
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3. (a) Let X₁, X₂, ..., Xₙ denote independent and identically distributed random variables from a distribution with pdf given by f(x) = (1/̠²)xe⁻ˣ/β, for x > 0, where β > 0 is an unknown parameter. (i) Find the maximum likelihood estimator, β̂ for β. (ii) Determine whether β̂ is an unbiased estimator. (iii) What is the maximum likelihood estimate of β if a random sample of size 10 yields the sample values of 126, 120, 141, 135, 123, 134, 132, 125, 129 and 138? (b) Let X₁, X₂, ..., Xₙ denote independent and uniformly distributed random variables on the interval [0, β]. Find (i) the pdf of X₍ₖ₎, the kth order statistic, where k is an integer between 1 to n. (ii) E[X₍ₖ₎]. [Hint: ∫₀¹ xᵃ⁻¹(1 − x)ᵇ⁻¹ dx = Γ(α)Γ(β)/Γ(α+β) where Γ(⋅) is a gamma function and, α and β are unknown parameters.]

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Transcript

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00:01 So in this question, first of all, we're told that the samples xi from i equals 1 to n are independent and identically distributed according to f of x is 1 over beta squared x e to minus x over beta for x greater than 0.
00:24 So first of all, we want to find the maximum likelihood estimator for beta.
00:29 So what we're going to do is we're going to find the likelihood.
00:31 So the likelihood of beta given our sample is just the product over the density functions for each of those samples.
00:44 So that's 1 over beta to the 2n.
00:47 So the product from i equals 1 to n xi times e to the minus 1 over beta times the sum from r equals 1 to n of xi.
01:00 So the log likelihood is just the log of the likelihood, as it's so.
01:09 So that's minus 2n log beta, plus the sum from i equals 1 to n, log xi, minus 1 over beta times the sum from i equals 1 to n of xi.
01:24 And the maximum likelihood estimator maximizes that log likelihood.
01:29 So dl by d beta evaluated at the mle is going to be equal to 0.
01:35 And that's minus 2n over the mle, plus 1 over the mle squared, times 1.
01:41 The sum from i equals 1 to n of x i.
01:46 So now we can move this to the other side and multiply by beta squared.
01:51 So beta hat times 2n is equal to the sum from i equals 1 to n x i.
02:00 So beta hat, the maximum likelihood estimator, is just 1 over 2n times the sum from i equals 1 to n of x i.
02:10 So it's a half of the average.
02:18 So yeah, this is our maximum likelihood estimator.
02:24 So now let's determine whether it's an unbiased estimator.
02:28 Well, first of all, we need to find the expected value of x.
02:32 So the expected value of x is the integral from 0 to infinity, 1 over beta squared, x squared, e to minus x over beta, d x.
02:43 So now let's integrate by parts.
02:45 Let's do u equals x squared over beta squared so that u dashed equals sorry let's do x squared over beta so that u dashed is 2x over beta and then let's have v dashed equals 1 over beta e to the minus x over beta so that v is minus e to the minus x over beta so then our integral is going to be you take the integrated part minus x squared over beta e to the minus x over beta evaluator between zero and infinity so when x is z is 0, this part's going to be 0.
03:21 And when x goes to infinity, this part's going to go to 0.
03:24 So that means that this whole bracket will be 0.
03:27 So we can ignore it.
03:28 And then we take off the differentiated part.
03:30 So that's plus 2 over beta integral from 0 to infinity, x e to the minus x over beta, dx.
03:41 So this is 0.
03:44 So then let's integrate by parts again.
03:49 Let's integrate by parts again, so that's right that u equals 2x, u -dash equals 2, v equals 1 over beta, e to the minus x over beta, that's v -dash, so v is e to the minus x over beta, and we get the expected value of x is going to be 2x, e to the minus x over beta, between zero and infinity, which again, for the same reasons is going to be zero, because the x is zero at the zero end, and the e to the minus x over beta is 0 at the infinity end.
04:22 And then we have to, and that's got a minus sign that v.
04:27 And then we have to take off, so that means adding 2 integral from 0 to infinity, e to the minus x over beta, dx.
04:36 So that gives us minus 2 beta, e to the minus x over beta between 0 and infinity.
04:43 In infinity we get nothing at 0, the exponential gives us 1, so we get 2 beta as our expected value of x.
04:52 So now we can calculate the expected value of 0.
04:54 Beta.
04:55 The expected value of beta hat is the expected value of this quantity here, a half 1 over n, sum from i equals 1 to n of x i, which is 1 over 2n, sum from i equals 1 to n, of the expected value of x i.
05:20 But all these x i have the same expected value, which is 2 beta.
05:24 So we get beta over n, sum from i equals 1 to n of just 1, and this gives us n, n, so this sun gives us n.
05:34 So we get that the expected value of beta hat is beta, therefore beta hat is unbiased.
05:47 Right.
05:48 So now let's have a look at the mle of beta.
05:51 If we have x -i is 126, 120, 121, 141, 135, 123, 134, 132, 132, 123, 134, 132, 1225, 122, 122, 125, 129 and 138.
06:09 So this is a sample of size 10.
06:12 Then beta hat is a half of the mean of that sample.
06:16 So let's calculate that.
06:17 So 126 plus 120 plus 141 plus 135 plus 123 plus 134 plus 132 plus 132 plus 134 plus 132 plus 125 plus 129 plus 138 and then divide that by 20 and we get beta hat is 65 .15.
06:34 Right, so now part b we have another sample xi from i equals 1 to n, but now x is uniform on the interval 0 to beta.
06:55 So let's get the pdf of the kth order statistic.
07:00 So x k is such that x i from i.
07:12 Equals 1 to k minus 1 is less than or equal to x k is less than or equal to x i for i equals k plus 1 to n so what we can see is that if x k is going to be less than some value so x k less than or equal to x then that means that all of the x i from i equals 1 to k have to be less than or equal to x.
07:51 So that means that yeah so x i from i equals 1 to k are all less than or equal to x.
08:07 So if we say that n, uh, if we say that n x is the number of the x i, which are less than or equal to x, and the probability that x k is less than or equal to x is equal to the probability that nx is greater than or equal to k.
08:34 So what is this probability? well, nx is going to be binomially distributed with n trials and the probability of success as the probability that any of the x is are less than k.
08:50 So probability x less than or equal to k.
08:54 But what's this probability? well, the probability that x is less than or equal to k is equal to k over beta for k between zero and beta.
09:10 Sorry, this should be x not k.
09:19 So that's our cumulative function here.
09:25 So then we can say that the probability that x k less than or equal to x is the probability that nx is greater than or equal to k, when nx is binomely distributed n x over beta.
09:45 So the probability that nx is greater than or equal to k, well let's say the probability that nx is equal to, let's call it j, is going to be n choose j times x over beta to the j times x over beta to the j times 1 minus x over beta.
10:10 So that'll be beta minus x over beta to the n minus j...
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