1)X and Y are independent normal random variables suchthat E(X)=0, E(X^2)=1, E(Y)=1, E(Y^2)=5 (a) What is the distribution of (X-Y) ?(b) Find P{Y > X+1). 2) A basketball team plays a series of games and stops playing when it has won two games. It has 60% chance to win each game. The outcomes of different games are independent. Let X be the total number of games played. (a) What is the chance that the team will play at most 4 games ?(b) What is the name of the distribution of X (binomial, poisson, geometric,......) ?

Looks like, homework, I suggest you look at SUSANCHELSEA's posts, for answers.

QuoteOriginally posted by: MCMario1)X and Y are independent normal random variables suchthat E(X)=0, E(X^2)=1, E(Y)=1, E(Y^2)=5 (a) What is the distribution of (X-Y) ?(b) Find P{Y > X+1). 2) A basketball team plays a series of games and stops playing when it has won two games. It has 60% chance to win each game. The outcomes of different games are independent. Let X be the total number of games played. (a) What is the chance that the team will play at most 4 games ?(b) What is the name of the distribution of X (binomial, poisson, geometric,......) ?MEAN OF X-Y IS e(x)-e(y)=-1var(X-y)=varx +vary=5(x-y+1)/sqrt(5) is std normal with mean 0 and var 1...easy from herelet x be prob of winning 1 game...answer is x^2 +3x^2*(1-x)+6x^2*(1-x)^2 where x is 0.6

i'd recommend getting either a schaum's book or Ross is you're still having problems with these...

thanks for the suggestion phirangm......think I will.DT, thanks for the answer. I standardized X-Y z=-1- (-1) / sqrt(5) = 0 P(z<0)= 0.5 correct ?

Hi,The second question is a bit tricky if I understand it correctly. At first it appears to be a Binomial distribution, but in fact it is not strictly so because it contains the property of the geometric distribution; because for X=4, the last games HAS to have been won.DT, your answer should thus read:x^2 +2x^2*(1-x)+3x^2*(1-x)^2 where x is 0.6You can actually think of the problem as a binomial distribution if you fix the last outcome to be a 'win'... this removes the geometric distribution trait and what is left is just a binomial distribution...P(X=n) = 0.6* P(Y=1).... where Y ~ Bin(n-1, p), p=0.6Regards,Sam

does anyone know a good web-reference for understanding derivation of joint distribution functions. i'm self-studying freund's math stats and hes got one example which confuses more than explain. specifically, i'm a bit confused about the limits of the integrals. for ex: f(x,y)=x+y for 0<x<1, 0<y<1 and 0 elsewhere. he's got four regions he's integrating over, but aside from the first quadrant (as above), i don't understand where/how he gets the limits of the others from ? also i don't quite get the point of integrating over the quadrant x>1, 0<y<1 when x is defined only at 0<x<1 as above.