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Q3: Assuming the time intervals between any two consecutive 11M minibuses are independent and are uniformly distributed in [0, T] for some T>0. We want to estimate T from n samples of the arrival time intervals between consecutive minibuses: X1, X2, ..., Xn. (i) (5 points) Find an unbiased estimator of T (you need to prove it is unbiased) based on the sample mean. (ii) (5 points) Find the maximum likelihood estimator hat(T)MLE of T. (Hint: You do not need to take derivatives for this maximization). (iii) (5 points) Is hat(T)MLE in (ii) unbiased? Support your result by calculating its bias. (Hint: First, find the cdf of hat(T)MLE and note that "max(X, Y) ≤ a" is always equivalent to "X ≤ a and Y ≤ a"). (iv) (5 points) Compare the MSE of the estimator you find in (i) and (ii).

          Q3: Assuming the time intervals between any two consecutive 11M minibuses are independent and are uniformly distributed in [0, T] for some T>0. We want to estimate T from n samples of the arrival time intervals between consecutive minibuses: X1, X2, ..., Xn.
(i) (5 points) Find an unbiased estimator of T (you need to prove it is unbiased) based on the sample mean.
(ii) (5 points) Find the maximum likelihood estimator hat(T)MLE of T. (Hint: You do not need to take derivatives for this maximization).
(iii) (5 points) Is hat(T)MLE in (ii) unbiased? Support your result by calculating its bias. (Hint: First, find the cdf of hat(T)MLE and note that "max(X, Y) ≤ a" is always equivalent to "X ≤ a and Y ≤ a").
(iv) (5 points) Compare the MSE of the estimator you find in (i) and (ii).

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q3 assuming the time intervals between any two consecutive 11m minibuses are independent and are uniformly distributed in 0 t for some t0 we want to estimate t from n samples of the arrival 68074

Added by Jared G.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Madhur L

07/11/2023

Step 1

Let Y be the random variable representing the time interval between consecutive minibuses. Since Y is uniformly distributed in [0, T], its expected value is E(Y) = T/2. The sample mean is defined as X̄ = (X1 + X2 + ... + Xn) / n. To estimate T, we can use the Show more…

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Q3: Assuming the time intervals between any two consecutive 11M minibuses are independent and are uniformly distributed in [0, T] for some T>0. We want to estimate T from n samples of the arrival time intervals between consecutive minibuses: X1, X2, ..., Xn. (i) (5 points) Find an unbiased estimator of T (you need to prove it is unbiased) based on the sample mean. (ii) (5 points) Find the maximum likelihood estimator hat(T)MLE of T. (Hint: You do not need to take derivatives for this maximization). (iii) (5 points) Is hat(T)MLE in (ii) unbiased? Support your result by calculating its bias. (Hint: First, find the cdf of hat(T)MLE and note that "max(X, Y) ≤ a" is always equivalent to "X ≤ a and Y ≤ a"). (iv) (5 points) Compare the MSE of the estimator you find in (i) and (ii).

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Transcript

00:01 Hello student, in the given question, in part 1st, we have to show that g of x is equals to 2 upon n multiply by summation ranging from i equals to 1 to n into xi is an unbiased estimator of theta.

00:19 So, we need to show that expectation of g of x is equals to theta.

00:26 So, since xi is the uniformly distributed, xi is the uniformly distributed over 0, theta, we have expectation of xi is equals to theta by 2.

00:46 Therefore, we have the expectation of g of x is equals to expectation of 2 upon n multiply by summation over 1 to n xi which is equals to 2 by n multiply by summation ranging from i equals to 1 up to n of xi expectation of i sorry here not xi here is expectation of xi which is equals to which is equals to 2 by n multiply by n multiply by theta by 2 which is equals to theta.

01:38 Hence, g of x is equals to g of x is an unbiased estimator of theta is an unbiased estimator of theta.

01:56 Now, in second part, we have in second part, we have given a given t is equals to small t, we know that one of the variable is equal to t.

02:13 Let assume that without loss of the generality that x1 is equals to t, then the remaining n minus 1 observation must be uniformly distributed over 0, t.

02:27 Since, they are independent and identically distributed.

02:30 In part b, we have expectation of g of x given t is equals to small t which is equals to expectation of 2 by n multiply by summation over i equals to 1 to n of xi given t is equals to small t which is equals to 2 by n is constant outside multiply by t plus summation over i equals to 1 up to n expectation of xi which is equals to 2 by n by t plus n minus 1 by 2 multiply by t which is equals to n plus 1 given n into t...

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