# why isn't the the marginal distribution needed when using a conjugate prior?

What is a good explanation as to why you wouldn't have to integrate to find the posterior when you use a conjugate prior. Most examples (like for instance: http://www.youtube.com/watch?v=0XD6C_MQXXE) multiply the prior times the likelihood and end with 'look how nice the (unscaled) posterior is of the same distribution as the prior and we didn't need to integrate'. But they never calculate the marginal distribution and divide by it. That kind of makes the conjugate look like a diversion from the fact that the integration problem still isn't solved at all..so why isn't the marginal distribution needed?

You can do it more explicitly if the "by inspection" trick bothers you. Suppose that $X\mid\theta\sim\mathrm{Bin}(n,\theta)$ with prior $\mathrm{Beta}(a,b)$. Then, the posterior density is proportional to $$\theta^{x+a-1} (1-\theta)^{n-x+b-1} \, .$$ The full posterior density is $$\pi(\theta\mid x) = A\;\theta^{x+a-1} (1-\theta)^{n-x+b-1} I_{[0,1]}(\theta) \, ,$$ in which the normalization constant $A$ must be determined by the condition that the posterior density must integrate to one. Therefore, we have $$1 = \int_{-\infty}^\infty \pi(\theta\mid x) \, d\theta = A \int_0^1 \theta^{x+a-1} (1-\theta)^{n-x+b-1} \, d\theta$$ $$= A \frac{\Gamma(x+a)\Gamma(n-x+b)}{\Gamma(n+a+b)} \int_0^1 \frac{\Gamma(n+a+b)}{\Gamma(x+a)\Gamma(n-x+b)} \theta^{x+a-1} (1-\theta)^{n-x+b-1} \, d\theta \, .$$ The integral above is equal to one, because you are integrating the density of a $\mathrm{Beta}(x+a,n-x+b)$ random variable. Hence, $$A = \frac{\Gamma(n+a+b)}{\Gamma(x+a)\Gamma(n-x+b)} \, ,$$ and the full posterior is $$\pi(\theta\mid x) = \frac{\Gamma(n+a+b)}{\Gamma(x+a)\Gamma(n-x+b)}\theta^{x+a-1} (1-\theta)^{n-x+b-1} I_{[0,1]}(\theta) \, ,$$ meaning that $\theta\mid X=x\sim\mathrm{Beta}(x+a,n-x+b)$.