Can a normal density be proportional to another normal density?

Question

Suppose I have $X \sim \mathcal{N}(\mu, \sigma^2) = f_1$. Basically, in plain English, I have a density call it $f_1$ which is normal with mean $\mu$ and variance $\sigma^2$.

$$\Pr(\mu-\sigma < X < \mu + \sigma) = \int_{\mu-\sigma}^{\mu+\sigma}f_1(x) dx \tag{1}\label{1}.$$

Is it possible to have another normal density call it $f_2$ which is proportional to $f_1$ such that:

$$f_2(x) = \frac{f_1(x)}{k}, \quad \forall x:\mu-\sigma < x < \mu + \sigma,$$

where $k \neq 1$ is a constant.

Or is it possible for all Natural Exponential family members?

Answer 1

Another way to see this is impossible is that if $f_2$ and $f_1$ are both normal and are proportional over $[\mu-\sigma,\mu+\sigma]$, then they both have maximum at the same point $x=\mu$, hence both have the same mean. Now we have to solve $$\frac{1}{k\sigma_1}e^{-\frac{(x-\mu)^2}{2\sigma_1^2}}\equiv\frac{1}{\sigma_2}e^{-\frac{(x-\mu)^2}{2\sigma_2^2}}\text{.}$$

At $x=\mu$ this gives $k=\frac{\sigma_2}{\sigma_1}$, but then everywhere on the interval we have

$$-\frac{(x-\mu)^2}{2\sigma_1^2}\equiv-\frac{(x-\mu)^2}{2\sigma_2^2}\text{,}$$ hence $\sigma_1=\sigma_2$

Answer 2

This is not possible. The question boils down to:

Do there exist positive constants $\sigma_1,\sigma_2,k$ and constants $\mu_1,\mu_2$ satisfying: $$\frac{k}{\sigma_1}\exp\left(-\frac{1}{2\sigma_1^2}(x-\mu_1)^2\right)\equiv\frac{1}{\sigma_2}\exp\left(-\frac{1}{2\sigma_2^2}(x-\mu_2)^2\right)$$Over $x\in\Bbb R$?

Suppose this were the case. Take derivatives to obtain the auxiliary equation: $$\frac{k}{\sigma_1^3}(x-\mu_1)\exp\left(-\frac{1}{2\sigma_1^2}(x-\mu_1)^2\right)\equiv\frac{1}{\sigma_2^3}(x-\mu_2)\exp\left(-\frac{1}{2\sigma_2^2}(x-\mu_2)^2\right)$$Which implies: $$\sigma_2^2(x-\mu_1)\equiv\sigma_1^2(x-\mu_2)$$By comparing coefficents it must be true that $\sigma_2=\sigma_1$ and $\mu_1=\mu_2$. Using this, we easily find $k=1$ either by evaluating the original equation at $x=0$ or by calculating means/variances.