For a parametric family $\{p_\theta(x)\}$ we know that $\lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2}D(p_\theta||p_{\theta'}) = \alpha J(\theta)$, where $J(\theta)$ is the Fisher information. Use natural logarithm for KL divergence to compute $\alpha$.
user
Sure, I can provide the general formula for computing the betas in a parametric family, which include the alpha parameter you're asking about.

First, let's define the KL divergence (also known as relative entropy) between two probability distributions p and q:

D(p || q) = ∫  LOG(p / q) dq

In this case, we are taking the logarithm in base e (natural logarithm) for convenience.

Next, using Taylor's expansion, we have:

 LOG(1 + x) = x - x^2/2 + x^3/3 - ... (for |x| < 1)

Substituting this into the formula for alpha, we get:

α = lim_{Δθ → 0} D(p_{\theta} || p_{\theta + Δθ}) / (2Δθ)^2 = lim_{x → 0} LOG(1 + x) / x^2

Now, recall that for small values of x, we can approximate LOG(1 + x) using the first two terms in the Taylor expansion:

LOG(1 + x) ≈ x - x^2/2

Plugging this in, we have:

α ≈ lim_{x → 0} (x - x^2/2) / x^2

Simplifying this, we get:

α ≈ 1/2

In summary, for this specific example, the beta parameter associated with KL divergence (in natural logarithm) is 1/2.

Note that this formula assumes certain conditions, such as the existence of the limit and the smoothness of the logarithm. In practice, a numerical approximation may be used to estimate this value. Additionally, depending on the problem context, you might want to use a different form of the exponential function (e.g., natural, Neperian) for the KL divergence computation.
assistant
For a parametric family $\{p_\theta(x)\}$ we know that $\lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2}D(p_\theta||p_{\theta'}) = \alpha J(\theta)$, where $J(\theta)$ is the Fisher information. Use natural logarithm for KL divergence to compute $\alpha$.
user
The Kullback-Leibler (KL) divergence between two distributions $p(x)$ and $q(x)$ is defined as:

$D(p||q) = \int p(x) \log \frac{p(x)}{q(x)} dx$

Given that $\lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2}D(p_\theta||p_{\theta'}) = \alpha J(\theta)$, we want to find $\alpha$.

We start by expanding the KL divergence for the parametric family $\{p_\theta(x)\}$:

$D(p_\theta||p_{\theta'}) = \int p_\theta(x) \log \frac{p_\theta(x)}{p_{\theta'}(x)} dx$

Now, we take the limit as $\theta' \to \theta$:

$\lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2}D(p_\theta||p_{\theta'}) = \lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2} \int p_\theta(x) \log \frac{p_\theta(x)}{p_{\theta'}(x)} dx$

Next, we perform a second-order Taylor expansion of $\log \frac{p_\theta(x)}{p_{\theta'}(x)}$ around $\theta$:

$\log \frac{p_\theta(x)}{p_{\theta'}(x)} \approx (\theta - \theta') \frac{\partial}{\partial \theta} \log p_\theta(x) + \frac{1}{2} (\theta - \theta')^2 \frac{\partial^2}{\partial \theta^2} \log p_\theta(x)$

Now, substitute the Taylor expansion back into the limit:

$\lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2}D(p_\theta||p_{\theta'}) = \lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2} \int p_\theta(x) \left[(\theta - \theta') \frac{\partial}{\partial \theta} \log p_\theta(x) + \frac{1}{2} (\theta - \theta')^2 \frac{\partial^2}{\partial \theta^2} \log p_\theta(x)\right] dx$

Since we are taking the limit as $\theta' \to \theta$, the first term in the integral vanishes, and we are left with:

$\lim_{\theta'\to\theta}\frac{1}{(\theta-\theta')^2}D(p_\theta||p_{\theta'}) = \lim_{\theta'\to\theta}\frac{1}{2} \int p_\theta(x) \frac{\partial^2}{\partial \theta^2} \log p_\theta(x) dx$

By definition, the Fisher information is given by:

$J(\theta) = -\int p_\theta(x) \frac{\partial^2}{\partial \theta^2} \log p_\theta(x) dx$

Comparing this to our last expression, we see that $\alpha = \frac{1}{2}$ since the negative sign cancels out when taking the second derivative with respect to $\theta$. Therefore,

$\alpha = \frac{1}{2}$
assistant