Homework problem - Ways to test if a density function is cumulative density function I have a problem that states:
Let $F : \mathbb R \to R$ be defined by
$$F(x) =\begin{cases}e^{\frac{-1}{x}} &\text{if } x > 0\\
         0&\text{if } x \leq 0\end{cases}$$
Is $F$ a cumulative distribution function? If yes, what is the associated probability density function?
Obviously just by how the function is denoted with $F(x)$ my answer would be yes, and then I would just take the derivative of the function to get the pdf.  I know for the cumulative $P(X \leq x)$.  Is there a way to show this function is cdf with the information given mathematically?  I'm thinking that this problem can't be that easy just by looking at the size of the letter $f$.
 A: The derivative of $F$ for $x>0\;$ is $$f(x)=F'(x)=\frac{e^{-\frac{1}{x}}}{x^2}$$
It is positive for $x>0,\;$ and you have $\lim_{x\rightarrow0+}f(x)=0,\quad \lim_{x\rightarrow\infty}F(x)=1$. Therefore 
$F$ and $f$ are the CDF and PDF of a continous distribution.
A: Definition: A function $F:\mathbb{R}\rightarrow[0,1]$ is a cdf if it satisfies 3 properties:
1. $F$ is right continuous and has a limit on the left
2. F is non-decreasing
3. $F(-\infty)=0,F(\infty)=1$
Let's check these properties.


*

*$F$ is clearly continuous. Why? Its exponential function on $(0,\infty)$, identically zero on $(-\infty,0]$  and $ \lim_{x \to {0+}}F(x)=0.$  

*$x>y \Rightarrow -x<-y \Rightarrow -1/x > -1/y \Rightarrow 0< e^{-1/x}>e^{-1/y}$. So $F$ is non-decreasing.  

*$F(-\infty)=F(0)=0$. $\lim_{x \to \infty}F(x)=e^{-1/\infty}=e^0=1$. 


All three properties are satisfied. Hence $F$ is a cdf!
As for the pdf, you are correct. Just take the derivative. The form given in gammatester's answer is correct:  
$$f(x)=\frac{e^{-1/x}}{x^2} \qquad x>0 \\ \qquad \qquad 
=0 \qquad \qquad otherwise 
$$
