Basic question on proportionality in Bayesian Inference for Normal distribution - Cross Validated most recent 30 from stats.stackexchange.com 2019-09-18T22:29:19Z https://stats.stackexchange.com/feeds/question/379949 https://creativecommons.org/licenses/by-sa/4.0/rdf https://stats.stackexchange.com/q/379949 1 Basic question on proportionality in Bayesian Inference for Normal distribution samvoit4 https://stats.stackexchange.com/users/198966 2018-12-02T18:49:54Z 2018-12-20T09:52:06Z <p>I have a nagging question regarding the Normal distribution and maintaining proportionality in Bayesian Inference. Say for example that:</p> <p><span class="math-container">$\pi(\theta|Y) \propto L(Y|\theta)\pi(\theta)$</span></p> <p><span class="math-container">$Y | \theta \sim N(\theta, \sigma^2)$</span></p> <p><span class="math-container">$\theta \sim N(\mu, \tau^2)$</span></p> <p>If we want to calculate the posterior distribution for <span class="math-container">$\theta$</span>, we can drop any terms that do not include the parameter of interest (i.e. normalizing constant). This results in the following:</p> <p><span class="math-container">$\pi(\theta|Y) \propto \exp(-\frac{1}{2 \sigma^2}\sum_{i = 1}^{n} (Y_i - \theta)^2) \exp(-\frac{1}{2 \tau^2}(\theta - \mu)^2)$</span></p> <p>My question is why is it that we do not drop the <span class="math-container">$\frac{1}{2}$</span> in the exponential terms? Why would this no longer be proportional to the distribution? Or do we avoid dropping it for another reason?</p> <p>Thank you for your help!</p> https://stats.stackexchange.com/questions/379949/-/379995#379995 3 Answer by Ben for Basic question on proportionality in Bayesian Inference for Normal distribution Ben https://stats.stackexchange.com/users/173082 2018-12-02T23:48:37Z 2018-12-20T09:52:06Z <p>As is pointed out in the comments, exponentials factor via <span class="math-container">$e^{ab} = (e^a)^b$</span>. Taking proportionality with respect to <span class="math-container">$\theta$</span>, your posterior kernel is of the form:</p> <p><span class="math-container">\begin{equation} \begin{aligned} \pi (\theta|y) &amp;\propto \exp (- \tfrac{1}{2} \cdot f(\theta, y)) \cdot \exp (- \tfrac{1}{2} \cdot g(\theta, y)) \\[6pt] &amp;= \exp (- f(\theta, y))^{1/2} \cdot \exp (- g(\theta, y))^{1/2} \\[6pt] &amp;= \sqrt{\exp (- f(\theta, y)) \cdot \exp (- g(\theta, y))} \\[6pt] &amp;{\propto\kern-8pt \diagup} \exp (- f(\theta, y)) \cdot \exp (- g(\theta, y)). \\[6pt] \end{aligned} \end{equation}</span></p> <p>Since proportionality does not hold between a function and its square-root, it is not possible to remove the factors <span class="math-container">$\tfrac{1}{2}$</span> from the posterior in this case.</p>