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.
Best Answer
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 length.
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).