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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!

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    $\begingroup$ $e^{ax}=(e^x)^a \ne e^xe^a$ $\endgroup$ – user158565 Dec 2 '18 at 19:04
  • $\begingroup$ 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 $\endgroup$ – samvoit4 Dec 2 '18 at 20:01
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    $\begingroup$ $e^{ax}∝e^x$ 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. $\endgroup$ – user158565 Dec 2 '18 at 20:09
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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.

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