Gaussian

154. Gaussian

𝑓 (𝑥 | 𝜇, 𝜎^{2}) = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} 𝑒^{- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}}

We want $𝑓^{'} (𝑥) = \frac{d}{d 𝑥} 𝑓 (𝑥)$ , treating $𝜇, 𝜎$ as constants.

The constant factor pulls out, leaving the exponential:

𝑓^{'} (𝑥) = \frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \cdot \frac{d}{d 𝑥} exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})

Chain rule. Let $𝑢 (𝑥) = - \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}}$ , so the exponential is $exp (𝑢 (𝑥))$ and

\frac{d}{d 𝑥} exp (𝑢 (𝑥)) = exp (𝑢 (𝑥)) \cdot 𝑢^{'} (𝑥)

Differentiate $𝑢$ . Pull out $- \frac{1}{2 𝜎^{2}}$ , then chain-rule ${(𝑥 - 𝜇)}^{2}$ :

𝑢^{'} (𝑥) = - \frac{1}{2 𝜎^{2}} \cdot 2 (𝑥 - 𝜇) = - \frac{𝑥 - 𝜇}{𝜎^{2}}

Assemble. The last two factors reconstruct $𝑓 (𝑥)$ itself:

𝑓^{'} (𝑥) = \underset{𝑓 (𝑥)}{\underset{⏟}{\frac{1}{\sqrt{2 𝜋 𝜎^{2}}} \cdot exp (- \frac{{(𝑥 - 𝜇)}^{2}}{2 𝜎^{2}})}} \cdot (- \frac{𝑥 - 𝜇}{𝜎^{2}})

𝑓^{'} (𝑥) = - \frac{𝑥 - 𝜇}{𝜎^{2}} \cdot 𝑓 (𝑥)

Setting $𝜇 = 0$ , $𝜎 = 1$ collapses $𝑓$ to the standard normal $𝜙$ :