Calculating Variances of Functions of Sample Mean

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SUMMARY

This discussion focuses on calculating the asymptotic distributions for functions of the sample mean, specifically for Y², 1/Y, and e^Y, where Y represents the sample mean of a random independent and identically distributed (iid) sample of size n. The variance calculations utilize the properties of convergence in probability, leading to the conclusions that Y² converges to a normal distribution with mean u² and variance (σ⁴ + 2σ²u²)/n. Additionally, the discussion confirms that 1/Y converges to 1/u, with its variance derived using the delta method, resulting in convergence to a normal distribution as well.

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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

Homework Equations


a) [tex]Y^2=Y*Y[/tex] which converges in probability to [tex]u^2[/tex],

[tex]V(Y*Y)=\sigma^4 + 2\sigma^2u^2[/tex]

So, [tex]\sqrt{n}*(Y^2 - u^2)[/tex] converges in probability to [tex]N(0,\sigma^4 + 2\sigma^2u^2)[/tex]

So, [tex]Y^2\sim N(u^2,\frac{\sigma^4 + 2\sigma^2u^2}{n})[/tex]

Is that right?
b) [tex]\frac{1}{Y}[/tex] converges in probability to [tex]\frac{1}{E(X)} = \frac{1}{u}[/tex]
[tex]V(\frac{1}{x}) = \frac{1}{σ^2}[/tex] ??

Thus,
[tex]\sqrt{n}(\frac{1}{Y} - \frac{1}{u})[/tex] converges in distribution to [tex]N(0,V(\frac{1}{x})*\frac{1}{n})[/tex]
What is V(1/X) ?Am I on the right track?
 
Last edited:
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Solved. Just had to remember the delta method.
 

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