# Finding the density function $f_X$ from the distribution function

Can someone help me find the density function $f_X$ for $X$ and hence find $E(X)$ and $Var(X)$ of the following distribution function $F_X$ given by:

$F_X(x)=\begin{cases} 1-(1+x)e^{-x} & x>0 \\ 0 & otherwise. \end{cases}$

$X$ is a continuous random variable.

From memory, do I have to integrate $1-(1+x)e^{-x}$ or something similar? I can't recall on what to do, I get mixed up with the range in which I must integrate these sort of things (I am not even sure if I must integrate it but I know that when going from the probability density function to the distribution function, I must integrate it).

Differentiate the cumulative distribution function $F_X(x)$ to find $f_X(x)$.
You will get $xe^{-x}$ (for $x\gt 0)$.
For $E(X)$, you need $\int_0^\infty x^2e^{-x}\,dx$. This is usually done using a couple of integrations by parts, but you know an antiderivative of $xe^{-x}$.
For the variance, using $E(X^2)-(E(X))^2$, you will need $\int_0^\infty x^3e^{-x}\,dx$. In principle this requires three integrations by parts, but in reality because of your work on \int_0^\infty x^2e^{-x}\,dx$, you will only need one. The variable$X$follows a Gamma distribution$\Gamma(\alpha,\beta)$with$\alpha=2,\beta=1$, i.e.$X\sim\Gamma(2,1)$. So the mean and the variance are $$\Bbb{E}(X)=\frac{\alpha}{\beta}=2\qquad \operatorname{Var}(X)=\frac{\alpha}{\beta^2}=2.$$ If$X$ha support$[0,\infty)$, X is memoryless with respect to$t$if for any non-negative real numbers$t\ne 0 $and$s$, we have $$\Bbb{P}(X>t+s \mid X>t)=\Bbb{P}(X>s).$$ Call$G(t)=\Bbb{P}(X>s)=1-F(x)$the survival function of$X$(note that$G(t)$is then monotonically decreasing). Then$X$is memoryless if $$G(t+s)=G(t)G(s).$$ In your case$X\$ hasn't this property because $$(1+t+s)\ne (1+t)(1+s).$$