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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

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    **Densities** don't have to be $\le 1$. Cumulative distribution functions do.2012-10-11
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    @AndréNicolas i understand that part. but do you know what does $f(0.9)$ yield?2012-10-11
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    @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$2012-10-11
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    @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 $3(0.9)^2h$.2012-10-11

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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.

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    ok, when $x =0.9$ then $f(0.9)= 3(0.9)^2 which \neq 0$. That's what I'm trying to ask2012-10-11
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    But $\int_.9^.9 f(x) dx = 0$2012-10-11
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    @chris ok, i see. then what do i get when I plug in 0.9 for x?2012-10-11
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    @user133466 the actual value of the density function at a point has no meaning2012-10-11
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    @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 $P(X \in A) \approx f(0.1)\cdot \Delta = 0.03\Delta$.2012-10-11
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    It's a Radon-Nikodym derivative.2012-10-11
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    @DilipSarwate that's certainly fair and a useful approximation for continuous densities. Trouble starts popping up if you use stuff like that with discontinuous densities (for example piecewise constant densities) or if you (like me) tend to work with things that are only defined a.e. Your response is probably better for him though :)2012-10-11