Floor X Geometric Random Variable

An exercise problem in probability.

Floor x geometric random variable.

Let x and y be geometric random variables. The geometric distribution is a discrete distribution having propabiity begin eqnarray mathrm pr x k p 1 p k 1 k 1 2 cdots end eqnarray where. The appropriate formula for this random variable is the second one presented above. So we may as well get that out of the way first.

And what i wanna do is think about what type of random variables they are. Recall the sum of a geometric series is. Also the following limits can. The expected value mean μ of a beta distribution random variable x with two parameters α and β is a function of only the ratio β α of these parameters.

A full solution is given. Is the floor or greatest integer function. Narrator so i have two different random variables here. X g or x g 0.

An alternative formulation is that the geometric random variable x is the total number of trials up to and including the first success and the number of failures is x 1. In order to prove the properties we need to recall the sum of the geometric series. The random variable x in this case includes only the number of trials that were failures and does not count the trial that was a success in finding a person who had the disease. Q q 1 q 2.

In the graphs above this formulation is shown on the left. The relationship is simpler if expressed in terms probability of failure. Letting α β in the above expression one obtains μ 1 2 showing that for α β the mean is at the center of the distribution. If x 1 and x 2 are independent geometric random variables with probability of success p 1 and p 2 respectively then min x 1 x 2 is a geometric random variable with probability of success p p 1 p 2 p 1 p 2.

Then x is a discrete random variable with a geometric distribution. Find the conditional probability that x k given x y n. On this page we state and then prove four properties of a geometric random variable.