Let $Y$ denote a Gaussian random variable characterized by a mean $\mu$ and a variance $\sigma^2$. Consider $N$ independent and identically distributed (i.i.d.) copies of $Y$, denoted as $Y_1, Y_2, \ldots, Y_N$. Now, let's examine the number of $N$ for which the probability satisfies the inequality:

\begin{align} \mathbb{P}\left[\left|-\frac{1}{N}\sum_{i=1}^{N}\log f(Y_i)-h(Y)\right|\geq\delta\right]\leq\epsilon, \end{align}

Here, $f(y)$ represents the probability density function (pdf) of the Gaussian distribution, and $h(Y)$ represents the differential entropy of $Y$. It is worth noting that the expected value of the expression $-\frac{1}{N}\sum_{i=1}^{N}\log f(Y_i)$ is equal to $h(Y)$:

\begin{align} \mathbb{E}\left[-\frac{1}{N}\sum_{i=1}^{N}\log f(Y_i)\right] = h(Y). \end{align}

To analyze the probability inequality, we can employ a suitable concentration inequality. One approach is to consider the subgaussianity of the term $\log f(Y_i) = \frac12\log(2\pi\sigma^2) - \frac{(Y_i-\mu)^2}{2\sigma^2}$. Is considering the subgaussianity of $\log f(Y_i)$ a valid approach? If so, could you provide guidance on how to compute the precise value of the subgaussianity parameter for $-\frac{1}{N}\sum_{i=1}^{N}\log f(Y_i)$?