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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. – 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
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.

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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 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
@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

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