1890 Shares

Statistics: Conditional distributions. Any help?

Topic: Help me with my statistics homework
July 16, 2019 / By Delight
Question: This isn't a homework question; I have a final in my statistics class tomorrow, and I need to understand conditional probability. :( Let X and Y be independent and have the same geometric distribution with success probability p. Find the conditional distribution of X given X + Y = n. The book doesn't seem to give a good explanation of this, so any help would be great! Thanks!

Best Answers: Statistics: Conditional distributions. Any help?

Bryanne | 5 days ago
Hope this isn't too late. I'm going to assume that the geometric distribution you have is for the number of failures before you get 1 success. That is, Pr(X=j) = p(1-p)^j One of the features of the geometric distribution is that it has no memory meaning that the conditional probability does not depend on the number of prior failed trials. Using Bayes Theorem Pr(X=j|X+Y=n) = Pr(X+Y=n|X=j)Pr(X=j)/Pr(X+Y=n) Pr(X=j|X+Y=n) = Pr(Y=n-j)Pr(X=j)/Pr(X+Y=n) Pr(X=j|X+Y=n) = p(1-p)^(n-j) p(1-p)^j / Pr(X+Y=n) Now Z = X+Y is, I believe, a negative binomial random variable with parameters 2 and p. Pr(Z=n;r=2) = Γ(n+2)/[n!Γ(2)] p²(1-p)^n Since Γ(m) = (m-1)! then Γ(n+2)/n! = (n+1)!/n! = n+1 So Pr(Z=n;r=2) reduces to p²(n+1)(1-p)^n Substituting Pr(X=j|X+Y=n) = p(1-p)^(n-j) p(1-p)^j / Pr(X+Y=n) Pr(X=j|X+Y=n) = p(1-p)^(n-j) p(1-p)^j / [p²(n+1)(1-p)^n] Pr(X=j|X+Y=n) = 1/(n+1) I can't guarantee that this is correct, but I believe I was correct when I stated Z=X+Y is negative binomial, and that my use of Bayes theorem was correct. Good luck.
👍 258 | 👎 5
Did you like the answer? Statistics: Conditional distributions. Any help? Share with your friends

We found more questions related to the topic: Help me with my statistics homework