Let's take a look at a more general case. Continuous version of Bayes theorem is that f of theta given y is equal to f of y given theta times f of theta over f of y. Or we can write that of f of y given theta times f of theta or with the interval of f of y given theta times f of theta d theta. We can see that this is the likelihood. Times the prior, divided by a normalizing constant. Likelihood times the prior and then this is just a constant that will make sure that this integrates to one, because we need this at the end of the day to be a proper probability density function. And it's clear that this will integrate to one because we're dividing by exactly what it is on top. In practice, sometimes this integral can be a pain to compute. And so, we may work with looking at saying this is proportional to The likelihood times the prior. And if we can figure out what this looks like and just put the appropriate normalizing constant on at the end, we don't necessarily have to compute this integral. So for example, suppose we're looking at a coin and it has unknown probability theta of coming up heads. Suppose we express ignorance about the value of theta by assigning it a uniform distribution. So we can write f of theta is just the indicator function that theta is between zero and one. Now we flip the coin and we get one head. We want to say, having observed one flip of the coin, what's our posterior probability distribution for theta? So f of theta, given y, is this expression here. In practice, try to compute this in more general form, we can do it but it will be a little bit more work. Let's go ahead and plug in f of theta given we observed ahead, given y = 1. So this will be theta to the 1, 1 minus theta to the 0, times the indicator function that theta is between 0 and 1, divided by the integral of all of this. D theta. So this simplifies on top to theta times the indicator function. And the bottom, this will be the integral now from 0 to 1 of theta d theta. The integral of theta d theta is going to be one half theta squared. We evaluate that on 0, 1 and we just get one-half. So this whole thing simplifies to 2 theta times the indicator function that theta's between 0 and 1. As a side note, we could do this whole thing in terms of proportionality f(theta given y) is proportional to f(y given theta) times f(theta). And so, we can say that's proportional to theta times the indicator function that theta is between 0 and 1. And then, we just need to normalize this so that it will integrate to one. And as we saw earlier, then taking the integral of this, the correct normalizing constant is that we just stick a two on this. So this is the same type of approach, we get to the same answer. We stick the normalizing constant on at the end, if we can recognize what this is at the end. In some cases later, this will turn out much easier to just use the proportionality approach rather than a full equality approach and trying to work out the integral.