It's very common in probability problems to give a probability density function and in the end you have to equal the integral of the pdf to one, since by definition $\int_{-\infty }^{\infty }f(x)dx =1$ . And therefore you have to break it into two pieces, from ${-\infty }$ to $0$ and from $0$ to ${\infty }$. So, what's the role the area from ${-\infty }$ to $0$ play? Should I ignore it? It always must be computed? Does negative values have probability? Thanks!
1 Answer
It depends on the distribution. Negative values can definitely have a probability associated with them. For example, when drawing from the standard normal distribution (the one with mean $0$ and variance $1$), the probability of getting a negative number is $1/2$.
And that is why the "general" formulas are defined in terms of integration from $-\infty$ to $\infty$, as density functions are always zero outside of their support, anyway. So if you were computing the expectation of the unit uniform, you would integrate:
$$ \int_{-\infty}^\infty x f(x) dx = \int_{0}^1 x f(x) dx\\ = \int_0^1 x dx $$
since $f$ is zero outside of $[0,1]$.