# In Bayesian, what is the effect of prior on the curvature of posterior?

Just wonder about the simplest example, I have a simple Normal prior on my scalar parameter $w$.

$$P(w|D) \sim P(D|w) P(w)$$

$$P(w) = \mathcal{N}(0,\alpha^2)$$

When I increase $\alpha$ from 0 to $\infty$, how would the curvature of $P(w|D)$ changes?

• When you ask about the "curvature", are you interested specifically in the curvature in the strict sense (related to the second derivative of the function), or do you mean something else?
– Ben
Jul 15, 2018 at 11:59
• Yes! The second order derivative, in the 1D case is fine! Jul 15, 2018 at 13:27
• In general wouldn't this depend on $P(D \mid w)$? Jul 15, 2018 at 17:37
• @KevinLi, yes. But I want to know, when P(D|w) is fixed, how does regularization prior affect the curvature. Jul 16, 2018 at 0:10

From your clarification in the comments, it appears that you are interested in the second derivative of the posterior, not the curvature in its strict sense. Obtaining this is just a matter of basic calculus.

To facilitate our analysis, let's use the alternative notation $\pi_D(w) \propto L_D(w) \pi_0 (w)$ so that we are not using generic function notation. From your specification we have $\pi_0 (w) = \text{N}(w|0,\alpha^2)$. With a bit of simple calculus on the normal density, the derivative of the specified prior density is:

\begin{aligned} \frac{d \pi_0}{dw}(w) = \frac{d \text{N}}{dw}(w|0,\alpha^2) &= - \frac{w}{\alpha^2}\cdot \text{N}(w|0,\alpha^2) = - \frac{w}{\alpha^2}\cdot \pi_0(w). \end{aligned}

Then, using repeated application of the product rule you get:

\begin{aligned} \frac{d \pi_D}{dw}(w) &= \frac{d L_D}{dw}(w) \cdot \pi_0(w) + L_D(w) \cdot \frac{d \pi_0}{dw}(w) \\[6pt] &= \Bigg[ \frac{d L_D}{dw}(w) - \frac{w}{\alpha^2} \cdot L_D(w) \Bigg] \pi_0(w), \\[10pt] \frac{d^2 \pi_D}{dw^2}(w) &= \Bigg[ \frac{d^2 L_D}{dw^2}(w) - \frac{1}{\alpha^2} \cdot L_D(w) - \frac{w}{\alpha^2} \cdot \frac{dL_D}{dw}(w) \Bigg] \pi_0(w) + \Bigg[ \frac{d L_D}{dw}(w) - \frac{w}{\alpha^2} \cdot L_D(w) \Bigg] \frac{d\pi_0}{dw}(w) \\[6pt] &= \frac{1}{\alpha^2} \Bigg[ \alpha^2 \cdot \frac{d^2 L_D}{dw^2}(w) + \Big( \frac{w^2}{\alpha^2}-1 \Big) L_D(w) - 2w \cdot \frac{dL_D}{dw}(w) \Bigg] \pi_0(w). \\[6pt] \end{aligned}

Taking the limit as $\alpha \rightarrow \infty$ you get:

\begin{aligned} \lim_{\alpha \rightarrow \infty} \frac{d \pi_D}{dw}(w) &= \frac{d L_D}{dw}(w) \cdot \pi_0(w), \\[8pt] \lim_{\alpha \rightarrow \infty} \frac{d^2 \pi_D}{dw^2}(w) &= \frac{d^2 L_D}{dw^2}(w) \cdot \pi_0(w). \end{aligned}