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December 4th, 2012, 07:00 PM  #1 
Newbie Joined: Dec 2012 Posts: 4 Thanks: 0  probability at a point for a continuous random variable
Hello all, I am new to this forum, and also to the field of random variables. I have a doubt regarding the existence of the probability that a continuous random variable x to take a particular value X. I read that, this probability is equal to 0, which was proved by considering the probability, P(Xe<x<=X) = F(X)  F(Xe), where F(x) is cumulative distribution function, and e is a small positive number. Since F(x) is continuous, when the limits are taken, to tend e>0, the LHS becomes P(x=X) , and RHS, goes to zero, as the left limit Lt(e>0) F(Xe) = F(X). But i cant interpret this result. Wont the probability density function p(x) at the point x=X, give the probability that x takes the value X? Why would this be equal to 0?. Plz help me out... 
December 5th, 2012, 12:24 PM  #2 
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689  Re: probability at a point for a continuous random variable
Probability density function is the derivative of the probability distribution function. Unless the distribution has a jump at a particular point, the probability of that point = 0.

January 7th, 2013, 02:27 AM  #3 
Newbie Joined: Dec 2012 Posts: 4 Thanks: 0  Re: probability at a point for a continuous random variable
Hello mathman, thanks for the reply... Is the definition of density function is : the probability that the random variable takes the value? In that case, i dont understand why the probability a random variable takes a particular value is zero? Or is my definition of density function wrong ? Kindly help me out... 
January 7th, 2013, 03:25 PM  #4  
Global Moderator Joined: May 2007 Posts: 6,730 Thanks: 689  Re: probability at a point for a continuous random variable Quote:
The only case where probability is nonzero at a point is where F(x) has a jump at that point and the probability is the value of the jump.  

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