Standard equation for covariance is given as
Cov(X,Y) = E(XY)- E(X)E(Y)
A study card has the following:
We perform n indep. trials & p(sub i) is the prob. of outcome i & p(sub j) is the prob. of outcome j. What is the cov. of the number of times outcome i occurs and the number of times outcome j occurs?
Solution: Cov[X(sub i),X(sub j)] =-np(sub i)p(sub j), i not= j
What is this problem addressing and how does it relate to standard cov? Is it saying, "Never mind the expected values, if you know the probability of the events and the number of trials you can find the cov. this way"?
Cov(X,Y) = -n(p of x)(p of y)?
Is this only because they are independent?
Cov(X,Y) = E(XY)- E(X)E(Y)
A study card has the following:
We perform n indep. trials & p(sub i) is the prob. of outcome i & p(sub j) is the prob. of outcome j. What is the cov. of the number of times outcome i occurs and the number of times outcome j occurs?
Solution: Cov[X(sub i),X(sub j)] =-np(sub i)p(sub j), i not= j
What is this problem addressing and how does it relate to standard cov? Is it saying, "Never mind the expected values, if you know the probability of the events and the number of trials you can find the cov. this way"?
Cov(X,Y) = -n(p of x)(p of y)?
Is this only because they are independent?
Covariance clarificiation