Bizarre behaviour of a probability density function

Question

It is said in my lecture notes that if $f$ is a probability density function, then it must fulfill the condition:

$$\int_{-\infty}^{\infty} f(x)dx = 1$$

But then consider this function $$f(x) = \frac{12.5}{(10+x)^2}$$ when $x \in [0,40]$, and it is evaluated at 0 otherwise.

Then it can be observed when integrating over $[0,40]$, the definite integral is 1. But consider also:

$$12.5\int_{-\infty}^{\infty} f(x)dx = 0$$

Does this disqualify $f$ from being a probability density function? What am I misunderstanding here?

Answer 1

This is a case where good notation might be really helpful. You write

But then consider this function $$f(x) = \frac{12.5}{(10+x)^2}$$ when $x \in [0,40]$, and it is evaluated at 0 otherwise.

It might be easier to understand what this is saying if you precisely write it down in notation as a piecewise defined function, to wit $$ f(x) = \begin{cases} \frac{12.5}{(10+x)^2} & \text{if $x\in[0,40]$, and} \\ 0 & \text{if $x \in (-\infty,0)\cup(40,\infty)$}. \end{cases} $$ This makes it clear that if we want to do anything with this function, we have to consider what it does on two different sets: the interval $[0,40]$ on which it does something interesting, and the union of intervals $(-\infty,0)\cup(40,\infty)$ on which it is zero.

Integrating this over $\mathbb{R}$, we obtain \begin{align} \int_{-\infty}^{\infty} f(x) \,\mathrm{d}x &= \int_{-\infty}^{0} f(x) \,\mathrm{d}x + \int_{0}^{40} f(x) \,\mathrm{d}x + \int_{40}^{\infty} f(x) \,\mathrm{d}x \tag{1}\\ &= \int_{-\infty}^{0} 0 \,\mathrm{d}x + \int_{0}^{40} \frac{12.5}{(10+x)^2} \,\mathrm{d}x + \int_{40}^{\infty} 0 \,\mathrm{d}x \tag{2}\\ &= 0 + 1 + 0 \tag{3} \\ &= 1. \end{align} In step (1), we are using the additivity of the integral to break it up into an integral over several intervals–the actual function we are integrating is irrelevant. In step (2), we are applying the definition of $f$–it is defined by the rational expression on $[0,40]$, and is zero otherwise. Finally, we evaluate the integrals–here, I am assuming that your work was correct, and recalling that $\int_a^b 0 \,\mathrm{d}x= 0$ for any $a \le b$, where either could be infinite.

Answer 2

The requirement is that the integral over the sample space be $1$. If the only data values that matter are in the range $[0,40]$ then only that integral must be $1$. (I take for granted that you calculated it correctly.)

There is no need or reason to look at the integral over the whole line. But if you do, and the function is $0$ outside that interval, then the integral over the whole line is $1$, not $0$.