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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).

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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.

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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 $$ (1+t+s)\ne (1+t)(1+s). $$

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