Basic question on proportionality in Bayesian Inference for Normal distribution

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?

• $e^{ax}=(e^x)^a \ne e^xe^a$ – user158565 Dec 2 '18 at 19:04
• Ok but is it not true that $e^{ax} \propto e^x$? As $e^x$ gets larger/ smaller then $e^{ax}$ should also get larger or smaller. @user158565 – samvoit4 Dec 2 '18 at 20:01
• $e^{ax}∝e^x$ means $e^{ax} = ke^x$. So try to find $k$. For $x>0$, $x$ and $x^2$ get larger/ smaller together. But we do not think they are proportional to each other. getting larger/ smaller together and proportional to each other are different stories. – user158565 Dec 2 '18 at 20:09
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{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}
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.