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Help with estimators Uni Economics

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Could anyone help with Q2? Vaguely understand estimators but been trying to do this question for hours with no success. Help with Q3 and 4 would also be appreciated haha

thanks
Original post by Shazzy69

Could anyone help with Q2? Vaguely understand estimators but been trying to do this question for hours with no success. Help with Q3 and 4 would also be appreciated haha

thanks


Let μ^\hat{\mu} be one of the estimators you are given. Then μ^\hat{\mu} is an unbiased estimator for μX\mu_X if and only if E(μ^)=μXE(\hat{\mu}) = \mu_X .

I suggest you try them in the order (iii), (ii), (i). I'll show you (iii) and then see if you can continue.

E(μ^3)=E((1/n)1nXi)=(1/n)E(1nXi)=(1/n)nE(X)=E(X)=μXE(\hat{\mu}_3) = E((1/n) \sum_{1}^{n} X_{i}) = (1/n) E(\sum_{1}^{n} X_{i}) = (1/n) n E(X) = E(X) = \mu_X

The second last step is due to the fact that the XiX_i are iid. So this estimator is unbiased. Now see if you can do similar for (ii). Remember that the bias of the estimator is the difference between its expected value and μX\mu_X .

Part (i) is slightly trickier, but think about

(1/n)1n(XiμX+μX)2)(1/n) \sum_{1}^{n} (X_i - \mu_X + \mu_X)^2)

and connect one of the terms in its expansion with σX2\sigma^{2}_{X}
Reply 2
Original post by Gregorius
Let μ^\hat{\mu} be one of the estimators you are given. Then μ^\hat{\mu} is an unbiased estimator for μX\mu_X if and only if E(μ^)=μXE(\hat{\mu}) = \mu_X .

I suggest you try them in the order (iii), (ii), (i). I'll show you (iii) and then see if you can continue.

E(μ^3)=E((1/n)1nXi)=(1/n)E(1nXi)=(1/n)nE(X)=E(X)=μXE(\hat{\mu}_3) = E((1/n) \sum_{1}^{n} X_{i}) = (1/n) E(\sum_{1}^{n} X_{i}) = (1/n) n E(X) = E(X) = \mu_X

The second last step is due to the fact that the XiX_i are iid. So this estimator is unbiased. Now see if you can do similar for (ii). Remember that the bias of the estimator is the difference between its expected value and μX\mu_X .

Part (i) is slightly trickier, but think about

(1/n)1n(XiμX+μX)2)(1/n) \sum_{1}^{n} (X_i - \mu_X + \mu_X)^2)

and connect one of the terms in its expansion with σX2\sigma^{2}_{X}


Thanks so much for your help, for part ii) I've got n/n-2(E(X)) - Is that correct and if so is that what this biased is equal to?

For part i) i'm still quite confused about how you expanded it and how to link in σX2\sigma^{2}_{X} Sorry to be a pain
Original post by Shazzy69
Thanks so much for your help, for part ii) I've got n/n-2(E(X)) - Is that correct and if so is that what this biased is equal to?


If you mean E(μ^2)=nn2E(X)=nn2μX E(\hat{\mu}_2) = \frac{n}{n-2}E(X) = \frac{n}{n-2} \mu_{X} then you are correct. For the bias, simply subtract off μX\mu_X from this.

For part i) i'm still quite confused about how you expanded it and how to link in σX2\sigma^{2}_{X} Sorry to be a pain


Here are two hints:

(i) (xa+a)2=(xa)2+a2+2a(xa) (x - a + a)^2 = (x-a)^2 + a^2 + 2 a (x-a)

(ii) What is (1/n)1n(XiμX)2 (1/n) \sum_{1}^{n} (X_i - \mu_X)^2 ? You should recognize it.
Reply 4
Original post by Gregorius
If you mean E(μ^2)=nn2E(X)=nn2μX E(\hat{\mu}_2) = \frac{n}{n-2}E(X) = \frac{n}{n-2} \mu_{X} then you are correct. For the bias, simply subtract off μX\mu_X from this.



Here are two hints:

(i) (xa+a)2=(xa)2+a2+2a(xa) (x - a + a)^2 = (x-a)^2 + a^2 + 2 a (x-a)

(ii) What is (1/n)1n(XiμX)2 (1/n) \sum_{1}^{n} (X_i - \mu_X)^2 ? You should recognize it.


Thank you I now understand that, but how do I get from (1/n)1n(Xi)2 (1/n) \sum_{1}^{n} (X_i)^2 to (1/n)1n(XiμX+μX)2) (1/n) \sum_{1}^{n} (X_i - \mu_X + \mu_X)^2) Am I missing some sort of rule/formula?
(edited 8 years ago)
Original post by Shazzy69
Thank you I now understand that, but how do I get from (1/n)1n(Xi)2 (1/n) \sum_{1}^{n} (X_i)^2 to (1/n)1n(XiμX+μX)2) (1/n) \sum_{1}^{n} (X_i - \mu_X + \mu_X)^2) Am I missing some sort of rule/formula?


They are identical. An old trick in manipulating these types of formulae.

x2=(xa+a)2 x^2 = (x - a + a)^2
Reply 6
I see, could I have some help with question 3 as well please?
Original post by Shazzy69
could I have some help with question 3 as well please?


I'm getting a distinct sense of deja vu here. See the following thread:

http://www.thestudentroom.co.uk/showthread.php?t=3753463

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