 So let's consider the expected value of a Poisson experiment. If x has a Poisson distribution, the probability that we obtain k successes is given by lambda to the k, e to the negative lambda, divided by k factorial, where lambda is the expected number of occurrences of the event. And that's a huge hint that the expected value of x should be lambda. But let's verify this. Now we can verify this using some calculus. So we'll set up our formula for the computation of the expected value. That's the number of occurrences times the probability that we observed the event occurring that many times. And we'll note that because we have a factorial in the denominator, some of these terms simplify. First of all, the first term, since it's multiplied by zero, vanishes. Now since every term has a factor of e to the negative lambda, let's factor that out. Now there's not a lot we can really do with the first term. So we'll just copy that one down. As for the second term, we note that 2 divided by 2 factorial can simplify 2, and our second term becomes lambda squared. 3 divided by 3 factorial, if we expand the 3 factorial, remove the common factor, and simplify we get 1 divided by 2 factorial, which means the next term simplifies to lambda cubed divided by 2 factorial. 4 divided by 4 factorial will be, and so our next term simplifies as lambda to the fourth divided by 3 factorial, and so on, and in general n divided by n factorial will be 1 divided by n factorial. And so our terms inside the parentheses can simplify, and everything inside the parentheses has at least one factor of lambda, so we can remove that factor as well. And at this point it's rather unavoidable. We do have to introduce a little bit of calculus. Fortunately, it's not that much, and if you haven't been introduced to this yet, it's not all that important for our purposes, but it's a matter of recognizing that the Taylor series for e to the power x looks like this, and so the terms inside the parentheses can be simplified to e to the power of lambda. And we can simplify further to get lambda, our expected value, which is what we expected it should be. How about the variance? We can find the expected value of x squared. Again, our first term has k equals 0, so we can omit it and start our summation at k equal to 1. We can simplify our sum, and k squared lambda to the k e to the minus k divided by k factorial as, and we can remove the constant factor of e to the negative lambda, and so that gives us a sum, and we can re-index our series using i equal to k minus 1, and so our new re-index series will be, and we can split the series, and the first term of the first series vanishes, so we can simplify. Now we'll remove an inconvenient factor of lambda squared from the first series, then re-indexing using k equal i minus 1, and here we see our old familiar series expansion for e to the lambda, and so this first series will become lambda squared e to the lambda. Similarly, we can remove an inconvenient factor of lambda from the second series, and once again we see our familiar e to the lambda, and so our expected value of x squared will be, and consequently the variance of the Poisson distribution will be