Scootertaj
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1. Essentially what I'm trying to do is find the asymptotic distributions for
a)
Y2
b) 1/Y and
c) eY where
Y = sample mean of a random iid sample of size n.
E(X) = u; V(X) = σ2
a) Y^2=Y*Y which converges in probability to u^2,
V(Y*Y)=\sigma^4 + 2\sigma^2u^2
So, \sqrt{n}*(Y^2 - u^2) converges in probability to N(0,\sigma^4 + 2\sigma^2u^2)
So, Y^2\sim N(u^2,\frac{\sigma^4 + 2\sigma^2u^2}{n})
Is that right?
b) \frac{1}{Y} converges in probability to \frac{1}{E(X)} = \frac{1}{u}
V(\frac{1}{x}) = \frac{1}{σ^2} ??
Thus,
\sqrt{n}(\frac{1}{Y} - \frac{1}{u}) converges in distribution to N(0,V(\frac{1}{x})*\frac{1}{n})
What is V(1/X) ?Am I on the right track?
a)
Y2
b) 1/Y and
c) eY where
Y = sample mean of a random iid sample of size n.
E(X) = u; V(X) = σ2
Homework Equations
a) Y^2=Y*Y which converges in probability to u^2,
V(Y*Y)=\sigma^4 + 2\sigma^2u^2
So, \sqrt{n}*(Y^2 - u^2) converges in probability to N(0,\sigma^4 + 2\sigma^2u^2)
So, Y^2\sim N(u^2,\frac{\sigma^4 + 2\sigma^2u^2}{n})
Is that right?
b) \frac{1}{Y} converges in probability to \frac{1}{E(X)} = \frac{1}{u}
V(\frac{1}{x}) = \frac{1}{σ^2} ??
Thus,
\sqrt{n}(\frac{1}{Y} - \frac{1}{u}) converges in distribution to N(0,V(\frac{1}{x})*\frac{1}{n})
What is V(1/X) ?Am I on the right track?
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