00:01
Okay, given a, let's see, n samples x1, blah blah to x -n.
00:08
I mean there are iid samples from distribution x with mean value mu and vireens, sigma square.
00:20
And suppose n is strictly greater than 3 and define two estimators b1, is defined to be just the mean value of all samples i from 1 to n.
00:35
And mu2, mu2 half is defined to be equal to 2 times x1, the first one minus the second one plus the third one over 2.
00:47
First, we're required to check whether those two estimators are unbiased or not.
00:56
I mean just we just need to take the expectation of those two estimators.
01:02
First, by the linearity of the expectation, we have 1 over n times expectation of xi, and i from 1 to n.
01:15
And again, as they are iid samples, so i mean xi and x capital x have the same, have the same distribution, that means they have the same mean value and vibrant.
01:34
So here we have new.
01:38
From by this summation, we get n times me, because each terms, because all terms are equal and out this division we have mu.
01:47
So it is unbiased.
01:51
Now what about mu? again, user linearity 1 over 2 times expectation of 2 times x1, minus expectation of x2 plus expectation of x3.
02:10
Okay, user leninity plug to out we have 2 mu, minus new plus new.
02:17
Those two terms will be cute and again, this is also unbiased.
02:24
So both new one and new two are unbiased.
02:36
Okay, then we are required to determine which one is better.
02:41
So the most common, i mean, the most common ways to determine which one is better is to consider its virus.
02:53
I mean, the one with the last virus is, always considered to be the better one.
03:03
Okay, as by the definition of the virus, we see the variance of the random variable y is equal the expectation of y squared minus expectation of y to the power t.
03:20
And here as muo -1 and mu -t have the same mean value, so the second term is equal.
03:32
Okay, so that means we all need to consider the the first term, i mean the second moment for me 1 and new 2.
03:49
And we just consider this guy and this guy.
04:00
The one with less second moment is the better one.
04:06
Okay, let's compute that.
04:10
For the first turn, square them, we have 1 square here times expectation of summation xx to the power 2.
04:22
I from 1 to n.
04:25
And use a square formula, we have 1 over n, every first beginning times the expectation...