00:01
We're told that in a week, the number of random variable x of claims coming into an insurance office has a poison distribution with mean 100.
00:13
We're also told that the probability that any particular claim relates to automobile insurance is 0 .6, independent of any other claim.
00:22
We're given a random variable y representing the number of automobile claims, and we're told that y is then binomial with x trials, each with success probability 0 .6.
00:35
In part a, we're asked to determine the expected value of y given x equals x, and the variance of y, given x equals x.
00:52
So, again, we have that x is binomially, or is a poisson random variable, the parameter of 100, and we have that y, given x equals x, is binomially distributed parameters x, and 0 .6.
01:26
In particular, this is going to be small x here.
01:35
So the expected value of y given x equals x, because this is a binomial variable, this is simply going to be a number of trials n times probability p, which is 0 .6 times x.
02:04
Likewise, we have that the variance for a binomial distribution, is n times p times 1 minus p which in this case is going to be x times 0 .6 times 0 .4 so we get 0 .24 x in part b we're asked to use part a to find the expected value of y we have the expected value of y given x from part a is going to be 0 .6x that's a distribution and from the log of total expectation, it follows that the expected value of y is the same as the expected value of the expected value of y given x...