# How can a probability density be greater than one and integrate to one

Wikipedia says:

The probability density function is nonnegative everywhere, and its integral over the entire space is equal to one.

and it also says.

Unlike a probability, a probability density function can take on values greater than one; for example, the uniform distribution on the interval $[0, \frac{1}{2}]$ has probability density $f(x) = 2$ for $0 ≤ x ≤ \frac{1}{2}$ and $f(x) = 0$ elsewhere.

How are these two things compatible?

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It can take a value greater than $1$, but only over a region with measure less than $1$. – Qiaochu Yuan Feb 3 '12 at 22:22
Compute the integral in your example. Is it 1? – GEdgar Feb 3 '12 at 22:33
Think of it from a physics perspective: the probability density function is like the density of some compressible fluid, whereas it's integral is like the mass. If you take a fixed mass of liquid and compress certain parts of it, you can make the local density as high as you like without changing the total mass. – Nick Alger Feb 4 '12 at 0:07
Yeah, I feel a bit silly now. – zenna Feb 4 '12 at 1:05

Consider the uniform distribution on the interval from $0$ to $1/2$. The value of the density is $2$ on that interval, and $0$ elsewhere. The area under the graph is the area of a rectangle. The length of the base is $1/2$, and the height is $2$ $$\int\text{density} = \text{area of rectangle} = \text{base} \cdot\text{height} = \frac 12\cdot 2 = 1.$$

More generally, if the density has a large value over a small region, then the probability is comparable to the value times the size of the region. (I say "comparable to" rather than "equal to" because the value my not be the same at all points in the region.) The probability within the region must not exceed $1$. A large number---much larger than $1$---multiplied by a small number (the size of the region) can be less than $1$ if the latter number is small enough.

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Remember that the 'PD' in PDF stands for "probability density", not probability. Density means probability per unit value of the random variable. That can easily exceed $1$. What has to be true is that the integral of this density function taken with respect to this value must be exactly $1$.

If we know a PDF function (e.g. normal distribution), and want to know the "probability" of a given value, say $x=1$, what will people usually do? To find the probability that the output of a random event is within some range, you integrate the PDF over this range.

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Note that for a PDF, the probability that $x=1$ (or any other value) is $0$ because any point on the $x$-axis has $0$ length, so $\text{base} × \text{height} = 0$. For continuous probabilities it only makes sense to talk about probabilities near a point. For example, the probability that $x ∈ [0.9, 1.1]$ (i.e. $x=1±.1$) would be: $$∫_{0.9}^{1.1} f_X(x)\ dx$$ – Zaz Apr 4 at 23:31

I would like to point out an extreme example. Consider a probability density $$f(x) = \begin{cases} \frac{1}{2d},& -d \leq x\leq +d \\ 0, & \text{elsewhere} \end{cases}$$ Now, $\forall ~ d\neq0$
$$\int_{-\infty}^{+\infty} f(x)dx = 1$$ Start with a smaller $d$ and find the integral again. It will be the same. It gets really interesting when we take the limit when $d \rightarrow0$ . The integral is again 1, but how does $f(x)$ looks like? $$\lim_{d \to 0} f(x) = \begin{cases} \infty, & x=0 \\ 0, & \text{elsewhere} \end{cases}$$ This limit is termed as the Dirac delta function $\delta(x)$, which is not really a function but the limit of a function. So the bottom line here

Not only the probability density can go greater than $1$, it can assume even bigger values (the biggest one is noted here) as long as the area under it is $1$.

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