And once again, I'm not going to do a rigorous proof here, this is really just a property of variances, but I'm going to use this to establish what the variance of our random variable A is. So if this squared distance on average is some variance, and this one is completely independent, and its squared distance on average is some distance, then the variance of their sum is actually going to be the sum of their variances. So this is going to be equal to the variance of random variable X plus the variance of random variable Y. So another way of thinking about it is that the variance of Z, which is the same thing as the variance of X plus Y, is equal to the variance of X plus the variance of random variable Y. And hopefully that makes some sense, I'm not proving it too rigorously, and you'll see this in a lot of statistics books. Now, what I want to show you is that the variance of random variable A is actually this exact same thing. And that's the interesting thing, because you might say, hey, why wouldn't it be the difference?