Can a probability density function be derived from any function?

Question

I would like to confirm it is posible that a probability density function can be created ranging from $a$ to $b$.

A PDF's area must be one within it's range so I just divide the function by it's area over that range:

$$\mathbb{h}(x)=\frac{f(x)}{\int\limits_{a}^{b}f(x)dx}$$

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My question is simple, is $\mathbb{h}(x)$ a probability density function ranging from $a$ to $b$? If yes, how can I get a random variate?

Answer 1

There are several issues:

1. The integral must exist and be nonzero to make sense of your expression.
2. In order for the distribution you write to "range from $a, b$" , $f$ must be zero "almost everywhere" outside of $[a,b]$. For continuous $f$, this is the same as asking that $f$ is zero outside of $[a,b]$. For more general$f$, this is the same as asking that $\int_I f dx = 0$ for any interval $I$ disjoint from $[a,b]$.
3. Since probabilities are non-negative, you should also require that $f \geq 0$. (Again this is in the "almost everywhere" sense, which you can interpret as meaning that all of the integrals are non-negative if you don't know measure theory...)

(Maybe I am missing another issue in haste.)

If these are satisfied, then you do get a distribution on $[a,b].$ You will use: $\int c f(x) dx = c \int f(x)dx$, which is true when $c$ is a constant.

I don't understand what you mean by "a random value of given probabilities."