Is the limit $x \to - \infty$ of the cumulative distribution function always equal to $0$?
I've the probability density function:
$$ f_X(x) = \begin{cases} \frac{1}{2} x^2 \, e^{-x} & \text{if } x \geq 0 \\ 0 & \text{if } x < 0 \end{cases} $$
The CDF is:
$$ CDF_X(x) = 1 - \left( \frac{x^2}{2} + x + 1 \right) \, e^{-x} $$
Now if I want to check the goodness of the CDF I should check the limits for $x \to \pm \infty$; the limit for $x \to + \infty$ is 1 and that's ok. But the limit for $x \to - \infty$ is $- \infty$.
Maybe in this case I should consider the fact that $f_X(x)$ is non zero for $x \geq 0$ so I should consider the limit for $x \to 0$ (in this case the result is $0$ which is correct)?
Solution 1:
Yes, distribution functions must have these properties:
- $\lim_{t\to \infty }F(t)=1$
- $\lim_{t\to-\infty}F(t)=0$
- $F$ must be right continuous and increasing
In your case, you have that
$$ \begin{align*} F_X(c)&=\int_{(-\infty ,c]}f_X(t)\,d t\\ &=\int_{(-\infty ,c]}\mathbf{1}_{[0,\infty )}(t)\frac1{2}t^2 e^{-t}\,d t\\ &=\int_{[0,c]}\frac1{2}t^2 e^{-t}\,d t\\ &=\left(1-e^{-c}\left(\frac{c^2}{2}+c+1\right)\right)\mathbf{1}_{[0,\infty )}(c) \end{align*} $$