Tweet
Let X and Y be continous random variables with joint density function f(x,y) = 24xy for o<x<1 and 0<y<1-x. Calculate P[Y<X|X = 1/3]. The solution I found in this manual confuses me. It says that the marginal density of X at X = 1/3 is fx(1/3) = integral from 0 to 2/3 of 24(1/3) y dy = 1/3. Now wouldnlt this equal 16/9. I believe this is a misprint since in the following line the conditional denisity of f y|x(y|x = 1/3) = 8y / (16/9). So my main question is can a marginal density at a specific point be greater than 1 as it is in this case? I assumed that the marginal density of fx(1/3) is interpreted as P[X= 1/3] which I know cannot be greater than 1.
-
-
Sure! It is the cumulative distribution function that has to be less or equal to 1. The probability density function can be greater than 1.Originally posted by Hobogov View Post
ctperng -
... but...
Don't forget that density functions have no meaning at points since for each point x the Pr(X=x) is 0.
- junkComment
-
I don't think your assumption is correct! since this is a continuous distribution, we cannot interpret f(x) as a probability.As said in upstairs thread, at this point P(x=1/3) or at particular point is zero.Originally posted by Hobogov View PostComment
-
Interesting...I had the same question.Comment
-
What's your requestion? Can f(x,...) > 1? Yes. Read this for example: http://mathworld.wolfram.com/Probabi...yFunction.html nowhere does it place an upper bound on a pdf.Originally posted by vanckzhu View PostComment
-
Not really a requestion...but
Thanks anyways. I was gonna go look for the conditions a function must satisfy to be a continuous, multivariate PDF, but the link does a pretty good jobOriginally posted by NoMoreExams View Post
Comment
Comment