Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

By convention$ P[X=x] = 0$ for all x. How would you explain probability density function $f(x) = 3x^2$ (where x is between 0 and 1), probability is 0 otherwise. Then when x =0.9. f(x) > 1 which does not equal to 0

share|improve this question
    
Densities don't have to be $\le 1$. Cumulative distribution functions do. –  André Nicolas Oct 11 '12 at 1:17
    
@AndréNicolas i understand that part. but do you know what does $f(0.9)$ yield? –  user133466 Oct 11 '12 at 1:29
    
@user133466 I think that Dilip Sarwate's comment is the best response you're going to get to that. If your density function is continuous (as it is here), then for very small intervals $I$ with $x_0 \in I$ and so that the length of $I$ is $\Delta$ you have that $P(X \in I) \simeq f(x_0) \cdot \Delta$ –  Chris Janjigian Oct 11 '12 at 1:32
    
@user133466: Chris has given the same answer I would have given. If $h$ is small, then the probability that $0.9\le x\le 9+h$ is about $3(0.9)^2h$. –  André Nicolas Oct 11 '12 at 1:36
add comment

1 Answer

up vote 2 down vote accepted

When you have a probability density $f_X$ describing the distribution of a random variable $X$, you have $$P(X=x) = 0$$ for any $x$. There are not point-masses in such a distribution. In fact, any denumerable subset of the line has probability zero.

Such a function renders this service. $$P(A) = \int_{A} f_X(x)\, dx $$ for a decent(measurable) set of real numbers.

share|improve this answer
    
ok, when $x =0.9$ then $f(0.9)= 3(0.9)^2 which \neq 0$. That's what I'm trying to ask –  user133466 Oct 11 '12 at 0:17
    
But $\int_.9^.9 f(x) dx = 0$ –  Chris Janjigian Oct 11 '12 at 0:29
    
@chris ok, i see. then what do i get when I plug in 0.9 for x? –  user133466 Oct 11 '12 at 0:30
    
@user133466 the actual value of the density function at a point has no meaning –  Chris Janjigian Oct 11 '12 at 0:32
2  
@Chris I would not go so far as to say the actual value of the density function has no meaning. The probability that $X$ lies inside a small interval $A$ of length $\Delta$ depends on the location of the interval. If $A$ is centered at $x = 0.9$, $P(X \in A) \approx f(0.9)\cdot\Delta = 2.43\Delta$ while if $A$ is centered at $0.1$, then $P(X \in A) \approx f(0.1)\cdot \Delta = 0.03\Delta$. –  Dilip Sarwate Oct 11 '12 at 1:10
show 2 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.