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Why does $p(x) =1$ for a distribution $U(0,1)$? Shouldn't the probability be $0$ for any one point? I don't understand the intuition behind this.

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    $\begingroup$ $p$ is the density, not the probability. And yes, the probability of any specific one point occurring is zero. $\endgroup$ – copper.hat Mar 6 '13 at 3:24
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For a continuous distribution, the area under the $pdf$ should be $1$ (Probaility axiom). Probaility for any particular is zero. $$\int_0^1 f(x) dx =1$$ and $$\int_a^a f(x) =0$$ where $a\in(0,1)$.

In continuous distribution, probability is measured for an intervals and not for points.

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  • $\begingroup$ I see. I figured it out $\endgroup$ – lord12 Mar 6 '13 at 3:26
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For a continuous real random variable $X$, you can think of the definition of $p_X(x)$ for real $x$ as: $$p_X(x) = \frac{d}{dx} P(X\leq x)$$ from which it follows that for any two real numbers $a,b$ we have $$ P(a\leq X\leq b) = \int_a^b p_X(x)dx$$. For $X = U(0,1)$ we have $P(X\leq x) = x$ for $x\in [0,1]$ so $p_X(x) = 1$, but $$ P(a\leq X\leq a) = \int_a^a p_X(x)dx = 0$$

Hope this helps.

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