The Student Room Group

unbiased estimators

hi ive just started learning estimators today (self-teaching), so i might need quite a bit of help

i have just started this question

''A coin is tossed n times and the number of heads, H is counted. The probability of a head is p. Show that an unbiased estimator of p is H/n''

I found the sampling distribution, P(H=h), because of binomial distribution and said:

P(H=h)= nCh * (p^h) * (1-p) ^(n-h)

and i then said that E(estimator)=SIGMA { h/n*P(H=h) } from h=o to h=n.

I was given the result SIGMA r^2* (nCr) *(p^r)* (1-p)^{n-r} and so used this to evaluate my expectation

and got E(estimator)=[(p^2)/h]*(n-1)^2 +(p/h)

if it is unbiased the answer should be p so where have i gone wrong?

cheers
Reply 1
You've made it far too difficult than it actually is :tongue:.

Okay so this is taken from a binomial distribution, with H~B(n,p).

E(H/n) is the same as 1/n E(H).

You know E(H) is np, therefore that's the same as 1/n (np), giving you p.
Reply 2
Original post by CTArsenal
You've made it far too difficult than it actually is :tongue:.

Okay so this is taken from a binomial distribution, with H~B(n,p).

E(H/n) is the same as 1/n E(H).

You know E(H) is np, therefore that's the same as 1/n (np), giving you p.


oh wow thank you, i feel so stupid. thank you very much
Reply 3
Original post by CTArsenal
You've made it far too difficult than it actually is :tongue:.

Okay so this is taken from a binomial distribution, with H~B(n,p).

E(H/n) is the same as 1/n E(H).

You know E(H) is np, therefore that's the same as 1/n (np), giving you p.


the next part says

show that V=nZ(1-nZ) is a biased estimator and what multiple of Z would be an unbiased estimator where Z is the estimatior used in the first part :Z=H/n

i tried finding Var V but it got messy and theres very high powers in the expectation. do you know what i am supposed to do here?

and this time it does say that the result i mentioned in the op might be needed (which i did use to simplify my expression but i had h^4 etc in there aswell :/.
Reply 4
Original post by newblood
the next part says

show that V=nZ(1-nZ) is a biased estimator and what multiple of Z would be an unbiased estimator where Z is the estimatior used in the first part :Z=H/n

i tried finding Var V but it got messy and theres very high powers in the expectation. do you know what i am supposed to do here?

and this time it does say that the result i mentioned in the op might be needed (which i did use to simplify my expression but i had h^4 etc in there aswell :/.


I think you're getting confused between consistent estimators and unbiased estimators. To find out the bias you have to find out the expectation and subtract it by p (in this case); you can't prove if an estimator is unbiased by using the variance as an indicator.

As Z = H/n, V = H(1-H).
E(V) = E(H(1-H))
E(H - H^2) = E(H) - E(H^2)
E(H) = np
You know from S1 that Var(H) = E(H^2) - (E(H))^2 , therefore E(H^2) = Var(H) + (E(H))^2.
I'm sure you can do the rest from here :smile:

The bias will be the answer you get minus p. Finding the multiple that will make it unbiased is pretty straightforward from there.

Posted from TSR Mobile
Reply 5
Original post by CTArsenal
I think you're getting confused between consistent estimators and unbiased estimators. To find out the bias you have to find out the expectation and subtract it by p (in this case); you can't prove if an estimator is unbiased by using the variance as an indicator.

As Z = H/n, V = H(1-H).
E(V) = E(H(1-H))
E(H - H^2) = E(H) - E(H^2)
E(H) = np
You know from S1 that Var(H) = E(H^2) - (E(H))^2 , therefore E(H^2) = Var(H) + (E(H))^2.
I'm sure you can do the rest from here :smile:

The bias will be the answer you get minus p. Finding the multiple that will make it unbiased is pretty straightforward from there.

Posted from TSR Mobile


The question is referring to an estimator of the variance. The second part atleast
Reply 6
Original post by newblood
The question is referring to an estimator of the variance. The second part atleast


Take a picture of the question and post it up because you've made it sound a little ambiguous
Reply 7
Sure cheers
Reply 8
Original post by CTArsenal
Take a picture of the question and post it up because you've made it sound a little ambiguous


ive uploaded it btw
Reply 9
Original post by newblood
ive uploaded it btw


Okay,

So V = H(1-H/n)
V = H - (H^2)/n
E(V) = E(H) - E(H^2/n)
E(V) = E(H) - 1/n E(H^2)

You know that Var(H) = E(H^2) - [E(H)]^2
Therefore E(H^2) = Var(H) + [E(H)]^2
As a result, E(V) = E(H) - 1/n[Var(H) + (E(H))^2]

From there you should know what to do, as the expectation of V is just np(1-p) because you're testing if it's an unbiased estimator of the variance. Once you get your RHS answer, you subtract it by np(1-p) to find the bias.

Spoiler

Reply 10
The answer shoukd be n over n-1 V

Where have i gone wrong

Also arent we looking to find E(v). I thought Var(H)is np(1-p)
Reply 11
Original post by CTArsenal
Okay,

So V = H(1-H/n)
V = H - (H^2)/n
E(V) = E(H) - E(H^2/n)
E(V) = E(H) - 1/n E(H^2)

You know that Var(H) = E(H^2) - [E(H)]^2
Therefore E(H^2) = Var(H) + [E(H)]^2
As a result, E(V) = E(H) - 1/n[Var(H) + (E(H))^2]

From there you should know what to do, as the expectation of V is just np(1-p) because you're testing if it's an unbiased estimator of the variance. Once you get your RHS answer, you subtract it by np(1-p) to find the bias.

Spoiler



I think ive worked out how to do it myself now cheers for your help before though

Here is what i did if youre interested

Quick Reply

Latest