Distribution of Log of Variance

[Yagoda]

Homework Statement

If $Y_1, Y_2, ...$ are iid with cdf $F_Y$ find a large sample approximation for the distribution of $\log(S^2_N)$, where $S^2_N$ is the sample variance.

Homework Equations

The Attempt at a Solution

The law of large numbers states that for large N $S^2_N$ converges in probability to $\sigma^2$. However, because I don't know the distribution of the Y's I don't know what $\sigma^2$ is.

Also $\log S^2_N = \log(\frac{1}{N} \sum_{i=1}^{N} (Y_i - \mu) ^2) = \log(\frac{1}{N}) + \log(\sum_{i=1}^{N} (Y_i - \mu) ^2)$, but I am not sure if this helps me find an approximation for the distribution.

 

[mighty2000]

To approximate the distribution of \log(S^2_N), we can use the central limit theorem. This theorem states that the sample mean of a large sample from any distribution will be approximately normally distributed. Therefore, we can approximate the distribution of \log(S^2_N) by a normal distribution.

Let's denote the sample mean of the Y's as \bar{Y} = \frac{1}{N} \sum_{i=1}^{N} Y_i. Then, we have \log(S^2_N) = \log(\frac{1}{N} \sum_{i=1}^{N} (Y_i - \bar{Y})^2). We can rewrite this as \log(S^2_N) = \log(\frac{1}{N} \sum_{i=1}^{N} (Y_i - \mu + \mu - \bar{Y})^2). By expanding the square and using the fact that \frac{1}{N} \sum_{i=1}^{N} (Y_i - \mu) = \sigma^2, we get \log(S^2_N) = \log(\frac{1}{N} \sum_{i=1}^{N} (Y_i - \mu)^2 + \frac{1}{N} \sum_{i=1}^{N} (\mu - \bar{Y})^2 + \frac{2}{N} \sum_{i=1}^{N} (Y_i - \mu)(\mu - \bar{Y})).

Now, as N \to \infty, the first and second terms on the right-hand side approach \sigma^2 and 0, respectively. Also, the third term approaches 0 since \mu is a constant and \bar{Y} is the sample mean. Therefore, we can approximate \log(S^2_N) by \log(\sigma^2) = \log(\sigma^2) + \log(\frac{1}{N} \sum_{i=1}^{N} (Y_i - \mu)^2) + \log(\frac{2}{N} \sum_{i=1}^{N} (Y_i - \mu)(\mu - \bar{Y})).

Using the properties of logarithms, we can rewrite this as \log(S^2_N) =