S3 textbook exercise 3b

A baised six-sided die has probability p of landing on a six.Everyday ,for a period of 25 days ,the die is rolled 10 times and the number of sixes X is recorded giving rise to a sample,X1,X2,...........,X25
(a) Write down E(X) in terms of p
(b)show that the sample mean is a baised estimator of p and find the bias.
(c)suggest a suitable unbiased estimator of p

Reply 1

Nirgilis

Moved to Maths help :h:

. They may be more able to help you :yep:

Reply 2

mathsover50

Original post by KRYSTAL JIA

A baised six-sided die has probability p of landing on a six.Everyday ,for a period of 25 days, the die is rolled 10 times and the number of sixes X is recorded giving rise to a sample,X1,X2,...........,X25
(a) Write down E(X) in terms of p
(b)show that the sample mean is a baised estimator of p and find the bias.
(c)suggest a suitable unbiased estimator of p

I struggled with this too until realising that its asking about the Binomial distribution. If the dice was fair, p(six) is 1/6 and X~B(10, 1/6). So E(X) over 10 trials is 10*1/6 or 5/3. The dice is biased but that doesn't matter: X~B(10,p) and E(X) is 10p which answers part a).

Part b) just asks you to remember that the sample mean is an unbiased estimator of the distribution, or E(X). Since we showed in a) that E(X) is np, it is 10p and so the bias is a factor of 10. Part c) just divide the sample mean by 10 to get an unbiased estimator of p.

HOWEVER, I didn't log on looking for help with this question! I'm hoping for help solving another question a few pages on. No responses for three years isn't encouraging - but maybe there are more Maths people out there on TSR these days ...!?

THE QUESTION is Q6 from Exercise 3C.
2 random samples X1, X2, .., Xn and Y1, Y2, ..., Ym are taken from a population with mean mu and variance sigma^2. X-bar and Y-bar are unbiased esimators of mu. A new unbiased estimator T of mu is sought such that T = rX-bar + sY-bar.

a) Show that since T is unbiased r + 2 = 1. This is easy: substitute mu = T. So rmu + smu = mu = mu(r+s). r+s = 1.

b) By writing T = rX-bar + (1-r)Y-bar show that

Var(T) = sigma^2 [r^2/n + (1-r)^2/m]

Other than the thought that r^2 might come from Var(tT) = r^2Var(T) and likewise (1-r)^2 I have no idea about this!

(edited 6 years ago)

Reply 3

mathsover50

Original post by mathsover50

Hmm, either S3 is one of the least popular A-levels going, or I'm posting in the wrong place, or my question is too dum for TSR.

Well, I managed to figure things out.

2 random samples X1, X2, .., Xn and Y1, Y2, ..., Ym are taken from a population with mean mu and variance sigma^2. X-bar and Y-bar are unbiased esimators of mu. A new unbiased estimator T of mu is sought such that T = rX-bar + sY-bar.

a) Show that since T is unbiased r + 2 = 1. This is easy: substitute mu = T. So rmu + smu = mu = mu(r+s). r+s = 1.

b) By writing T = rX-bar + (1-r)Y-bar show that

Var(T) = sigma^2 [r^2/n + (1-r)^2/m]

All that's needed here is the knowledge that E(X-bar) is mu and that Var(X-bar) is a biased estimator of sigma^2 since it is sigma^2/n. So V(rX-bar) is r^2*(sigma^2/n) and V((1-r)Y-bar) is (1-r)^2*(r^2/m). To get Var(Xbar + Ybar) add the variances.

Take out the common factor sigma^2 and the result is
sigma^2 [r^2/n + (1-r)^2/m], which we are defining as Var(T).