Show that the mean X̅ of a random sample of size n from a...

Show that the mean $\bar{X}$ of a random sample of size $n$ from a distribution having pdf $f(x ; \theta)=(1 / \theta) e^{-(x / \theta)}, 0<x<\infty, 0<\theta<\infty$, zero elsewhere, is an unbiased estimator of $\theta$ and has variance $\theta^{2} / n$.

Transcript

00:01 Hi, i'm david and i'm here to help the answer your question.

00:03 Now let me bring up your question here.

00:06 In this question here, we're going to discuss about the sample mean.

00:11 And here we're given the bdf of the random variable x of the form fx given the titor, equal to the 1 of the teta, e to the power minus x over teta, for the x squared than 0 and then the teta will be smaller than infinity.

00:30 Bigger than 0.

00:32 Now this will be the density for the exponential with the mean it will equal to the theta therefore we know that the e on the x it will equal to the theta and now we define the same amount x par by formula equal to 1 over n so much than the i guess from 1 up to the n xi and the question first asked to find to show that it is unbiased estimator of the titor.

01:03 So, means then what you find the e on the x bar.

01:06 To find the e in the x bar will be e of the 1 over n, summation of the xi, i goes from 1 to n.

01:14 Here, one of the n will be the constant i can bring outside the expected value.

01:18 Then we have the summation on the e on the xi, i goes from 1 up to the n.

01:25 And then we know that.

01:28 It just equal to the summation, the xi just equal to the same thing, so just equal to the theta, i goes from 1 up to n, so get equal to 1 n times.

01:40 This summation equal to the n theta, so we can cancel the n and then get equal to the theta...

Question

Show that the mean $\bar{X}$ of a random sample of size $n$ from a distribution having pdf $f(x ; \theta)=(1 / \theta) e^{-(x / \theta)}, 0<x<\infty, 0<\theta<\infty$, zero elsewhere, is an unbiased estimator of $\theta$ and has variance $\theta^{2} / n$.

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