Basic question on proportionality in Bayesian Inference for Normal distribution

Question

I have a nagging question regarding the Normal distribution and maintaining proportionality in Bayesian Inference. Say for example that:

$\pi(\theta|Y) \propto L(Y|\theta)\pi(\theta)$

$Y | \theta \sim N(\theta, \sigma^2)$

$\theta \sim N(\mu, \tau^2)$

If we want to calculate the posterior distribution for $\theta$, we can drop any terms that do not include the parameter of interest (i.e. normalizing constant). This results in the following:

$\pi(\theta|Y) \propto \exp(-\frac{1}{2 \sigma^2}\sum_{i = 1}^{n} (Y_i - \theta)^2) \exp(-\frac{1}{2 \tau^2}(\theta - \mu)^2)$

My question is why is it that we do not drop the $\frac{1}{2}$ in the exponential terms? Why would this no longer be proportional to the distribution? Or do we avoid dropping it for another reason?

Thank you for your help!

Answer 1

As is pointed out in the comments, exponentials factor via $e^{ab} = (e^a)^b$. Taking proportionality with respect to $\theta$, your posterior kernel is of the form:

$$\begin{equation} \begin{aligned} \pi (\theta|y) &\propto \exp (- \tfrac{1}{2} \cdot f(\theta, y)) \cdot \exp (- \tfrac{1}{2} \cdot g(\theta, y)) \\[6pt] &= \exp (- f(\theta, y))^{1/2} \cdot \exp (- g(\theta, y))^{1/2} \\[6pt] &= \sqrt{\exp (- f(\theta, y)) \cdot \exp (- g(\theta, y))} \\[6pt] &{\propto\kern-8pt \diagup} \exp (- f(\theta, y)) \cdot \exp (- g(\theta, y)). \\[6pt] \end{aligned} \end{equation}$$

Since proportionality does not hold between a function and its square-root, it is not possible to remove the factors $\tfrac{1}{2}$ from the posterior in this case.