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This question is from DeGroot's "Probability and Statistics" :

Unbounded p.d.f.’s. Since a value of a p.d.f.(probability density function) is a probability density, rather than a probability, such a value can be larger than $1$. In fact, the values of the following p.d.f. are unbounded in the neighborhood of $x = 0$:$$f(x) = \begin{cases} \frac{2}{3}x^{-\frac{1}{3}} & \text{for 0<$x$<1,} \\ 0 & \text{otherwise.} \\ \end{cases}$$

Now, I don't know how the p.d.f. can take value larger than $1$.Please let me know the difference between the probability and probability density.

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2 Answers 2

up vote 2 down vote accepted

Simply put:

$\rho(x) \delta x$ is the probability of measuring $X$ in $[x,x+\delta x]$. With

$\rho(x):=$ probability density.

$\delta x:=$ interval lenght.

A probability will be obtained by computing the integral of $ \rho(x) $ over a given interval (i.e. the probability of getting $X\in [a,b] $ is $\int_a^b \rho(x) dx$. While $\rho(x)$ can diverge, the integral itself will not, and this is due to the fact that we ask that $\int_\mathbb{R}\rho(x) dx=1$, which means that the probability of measuring any outcome is 1 (we are sure that we will observe something). If the integral over the whole range gives 1, the integral over a smaller portion will give less than 1, because p.d.f. can't be negative (a negative probability is meaningless).

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The specific values $f(x)$ of the density function $f$ are the probability densities, and they express "relative probabilities", and the main point is that for a (measurable) subset $A$ of possible values (now $A\subseteq\Bbb R$), we have $$\int_Af\ =\ P(X\in A)$$ if the random variable $X$ has distribution described by $f$. In particular, $\int_{\Bbb R}f=1$, though its specific values, as shown by the given unlimited example, can be greater than $1$.

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Will you please explain "relative probability" term? –  Silent Oct 10 '13 at 16:26

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