Hi, Can someon please explain the following: Bernoulli random variable. multivariate random variables.
Sure. Bernoulli: You are interested in success/failure. Let p= prob. of success. Since the sum of probabilities for a random variable must equal one, we will let the probability of failure be q = 1-p. Next you need to understand random variables. Basically, we let X take on a specific value that depends on the outcome of the expiriment. In the case of Bernoulli, X takes on value of one if success, and value of zero if failure. So, probability X=1 is p, and prob X=0 is q=1-p. Example: Fair Coin flip, let heads be success, and tails be failure. P(X=1)=P(X=0)=0.5. The distribution in this case is discrete, so geometrically speaking, it is a piecewise function. Binomial random variables are simply the sum of multiple independent Bernoullis. Multivariate is a bit harder to explain, and not something you need to know for this test.
Wow… Quite an explanation… Think I get the rough idea which is all I was looking for… thanks
It’s a binary outcome probability. What is the probability that is rains tomorrow… the outcome is binary… it rains or it doesn’t What is the probability that a certain baseball team wins x number of games… for each game, either they will or lose.
thanks, I sort of got that but i couldn’t understand what the difference between bernoullis and binomial was… I get it now that binomial is the sum of the individual bernoullis random variables.
mambovipi Wrote: ------------------------------------------------------- > thanks, I sort of got that but i couldn’t > understand what the difference between bernoullis > and binomial was… I get it now that binomial is > the sum of the individual bernoullis random > variables. I should point out that it is not just the sum. Binomial takes into account all possible combinations. For example, if p=0.6 and q=0.4, and we do 5 trials (n=5), we now define our random variable slightly different. X can now take on 0,1,2,3,4, or 5. Binomial distribution will give probability of each of these by taking into account different possible combinations. So, P(X=3)=(5!/(3!*2!)(.6^3)(.4^2)=.3456