Hello together, got a question that is a bit about the exam level but maybe somebody can answer it… We have got a huge production of tires, of which we take a sample of 4 tires. Each tire has a probabilty of 50 % to be defect and of 50 % to be ok. X is the random variable that reproduces the number of defect tires in the sample Considering this big sample(n=4) we look at the first two of its’ tires (n=2) Y is the random variable that shows the number of defect tires in the small sample. What is the correlation coefficient between X and Y? Have fun, Your cfaisok
O sorry, my english is so bad. I ment “above the exam level” not “about the exam level” cfaisok Wrote: ------------------------------------------------------- > Hello together, > > got a question that is a bit about the exam level > but maybe somebody can answer it… > > We have got a huge production of tires, of which > we take a sample of 4 tires. > Each tire has a probabilty of 50 % to be defect > and of 50 % to be ok. > > X is the random variable that reproduces the > number of defect tires in the sample > > Considering this big sample(n=4) we look at the > first two of its’ tires (n=2) > Y is the random variable that shows the number of > defect tires in the small sample. > What is the correlation coefficient between X and > Y? > > Have fun, > > Your cfaisok
Do you need this question answered or are you just posting a question? It’s pretty straightforward question - the only thing even mildly tough is E(XY) but you can just do this with sum(E(X|y) P(Y=y))
Joey, can you provide with the solution. Thanks.
0.6
It’s a trivial problem. let’s denote z1, z2, z3, z4 independent and have binomial distribution = 1 (defective) with prob .5, and 0 (not defective) with prob. .5. X = z1+z2+z3+z4, Y = z1+z2. A couple of obvious things: E(z) = E(z^2) = .5, sigma(z) = sqrt(E(z^2)-[E(z)]^2) = sqrt(.5-.25) = .5 Let’s denote T = z3+z4. Obviously, T and Y are independent and have the same distribution. Obviously, E(T) = E(Y) = 2*E(z) = 1. sigma(T) = sigma(Y) = sqrt(2)*sigma(z) = sqrt(2)/2, sigma(X) = 2*sigma(z) = 1. Then cov(X,Y) = cov(Y+T, Y) = E[(Y+T)*Y]-E(Y+T)*E(Y)=E(Y^2)+E(YT)-2*E(Y)*E(Y)= //Y and T independent -> E(YT) = E(Y)*E(T) = 1, E(Y^2) = sigma(Y)^2 + E(Y)^2= 1/2+1=3/2 // = 3/2 -1 = 1/2 cor(X,Y) = cov(X,Y)/[sigma(X), sigma(Y)]=1/2/(1*sqrt(2)/2)=sqrt(2)/2. geometric solution. Y and T have the same length and perpendicular (because they are independent), then the agnle between Y+T and T is pi/4. then correlation is cos(pi/4) = sqrt(2)/2. I should’ve just provided the geometric solution but I had a few min to spare.
Maratikus solution is right, well done… Here as gadget the probability function of this “trivial problem” Y/X 0 1 2 3 4 0 0,0625 0,125 0,0625 0 0 1 0 0,125 0,25 0,125 0 2 0 0 0,0625 0,125 0,0625 I must say I am positively surprised by the quant power in this chat room. @maratikus: What’s your academic/job background? You have some quant background (as me) that’s for sure. I can see that how you did it… Yours, cfaisok
cfaisok Wrote: ------------------------------------------------------- > I must say I am positively surprised by the quant > power in this chat room. > There are freaking math professors in the room…
cfaisok Wrote: > I must say I am positively surprised by the quant > power in this chat room. I am positively surprised with the talent on AF. Everyone here is very capable and self-motivated. I truly enjoy talking to people on AF. > @maratikus: What’s your academic/job background? > You have some quant background (as me) that’s for > sure. I can see that how you did it… I got masters financial math (controlled stochastic processes). Then I did some theoretical work after I was done with school and now work as a quant. Welcome to our online community!