- #1
Zazubird
- 2
- 0
If Y~N(mu,sigma) and y=logX, with X~LN(mu,sigma),
with a*=exp{ybar+1/2*theta*sample variance of y}, where ybar=sample mean of y and a=E[X]=exp{mu+1/2*sigma^2}, theta is constant.
If theta=1, a* is consistent but biased and we can reduce the bias by choosing a different value of theta. Use a large n-approximation in the expression E[a*] to find a value that reduces the bias, as compared to when theta=1.
During my attempt to do this, I ended up with E[a*]=E[geometric mean of x * geometric variance of x]. Knowing that y=logx and thus ybar=log(x1*x2*...*xn)/n, I ended with such an expression. I have a feeling that this is most likely incorrect and am thus completely lost. I was thinking along the lines of possibly using CLT or Weak Law of Large Numbers, with the n-approximation detail in the question, but still don't know where to go from there.
with a*=exp{ybar+1/2*theta*sample variance of y}, where ybar=sample mean of y and a=E[X]=exp{mu+1/2*sigma^2}, theta is constant.
If theta=1, a* is consistent but biased and we can reduce the bias by choosing a different value of theta. Use a large n-approximation in the expression E[a*] to find a value that reduces the bias, as compared to when theta=1.
During my attempt to do this, I ended up with E[a*]=E[geometric mean of x * geometric variance of x]. Knowing that y=logx and thus ybar=log(x1*x2*...*xn)/n, I ended with such an expression. I have a feeling that this is most likely incorrect and am thus completely lost. I was thinking along the lines of possibly using CLT or Weak Law of Large Numbers, with the n-approximation detail in the question, but still don't know where to go from there.