Converting Probability Density Functions to Distribution Functions

Question

So I'm a bit confused as to how this works.

Say we have a probability density function $f(x) = 3e^{-3x}$ and we wish to convert this into a probability distribution function. It would result in $1-e^{-3x}$, so basically one minus the integral of the probability density function. Yet, if we take the probability density function $\frac{43x^5}{7} - \frac{x^6}{6}$ we actually get the integral of it, instead of one minus the integral, which in this case is $\frac{43x^6}{42} - \frac{x^7}{42}$.

EDIT: Forgot to mention, for the first example ($f(x) = 3e^{-3x}$), the bounds are $x \ge 0$ and for the second ($f(x) = \frac{43x^5}{7} - \frac{x^6}{6}$) its $0 < x < 1$

Can someone explain to me why this is? Thanks in advance!

Answer 1

What you are converting to is usually called "cumulative distribution function" or cdf $F(x)$, which is $F(x)=\Pr(X<x)$, the probability that the random variable takes on a value less than the given value.

To compute the cdf for these examples given, calculate $F(x)=\begin{cases}0&x<a\\\int_a^x f(t)dt&a\le x<b\\ 1&x\ge b\end{cases}$.

So for the first one, we have $\int_0^x 3e^{-3t}dt=-e^{-3t}\big|_{t=0}^x=-e^{-3x}-(-1)=1-e^{-3x}$. This is in fact the cdf of an Exponential(3) distribution, and $x$ can range from 0 to $\infty$. Below 0, the cdf is 0, and the cdf approahces 1 as $x\rightarrow\infty$. Formally, we would write $F(x)=\begin{cases}0&x\le0\\1-e^{-3x}&x>0\end{cases}$, but the below 0 part is often left out.

For the second one, we similarly integrate $\int_0^x\left(\frac{43t^5}{7}-\frac{t^6}{6}\right)dt=\frac{43x^6}{42}-\frac{x^7}{42}$ and x can range from 0 to 1. Formally, we have $F(x)=\begin{cases}0&x\le0\\\frac{43x^6}{42}-\frac{x^7}{42}&0<x<1\\ 1&x\ge1\end{cases}$, but we often omit the part that is 0 or 1 if it is obvious.

Answer 2

By definition, the probability distribution function is $F(y) = P(X \leq y) = \int_{-\infty}^y f(x)\,\mathrm dx$, with $f$ the probability density function. For your function you obtain for $y \geq 0$:

$$\begin{align}F(y) &= \int_{-\infty}^y f(x)\,\mathrm dx \\[1ex]&= \int_{-\infty}^0 0\,\mathrm dx + \int_0^y \dfrac{43}{7}x^5 - \dfrac{1}{6}x^6\,\mathrm dx \\[1ex]&= \left[ \dfrac{43}{42}x^6 - \dfrac{1}{42}x^7 \right]_{x=0}^{~~~~~y}\\[1ex]&=\dfrac{43}{42}y^6 - \dfrac{1}{42}y^7\end{align}$$