So we have a random continuously variable $R$ with density function $f_R(x) = 1/2e^{-|x|}$. First I need to sketch the distribution function of $R$. So
$$F_R(x) = \int_{-\infty}^x f_R(x) dx$$, but what do I do with this, since as far as I know it isn't integrable with elementary means?
Also, suppose I want to find probability of $\{|R| \leq 2\} = \{-2 \leq R \leq 2\}$. So we need
$$\int_{-2}^2 f_R(x) dx = 2\int_0^2 f_R(x) dx = 2(1 – e^{-2}),$$
because $f_R(x)$ is symmetrical along $x=0,$ but this obviously can't be right because it is greater than zero.
I'm guess what is really confusing me is the $|x|$ in my density function and how to deal with that little bugger. Any help appreciated!
Thanks
Best Answer
Just break it into positive and negative parts. To get an expression for $F_R(x)$, if $x<0$, then $$F_R(x) = \int_{-\infty}^x \frac{e^t}{2}\,dt = \left. \frac{e^t}{2}\right|_{-\infty}^x = \frac{e^x}{2}.$$ If $x \geq 0$, then $$F_R(x) = \int_{-\infty}^x \frac{e^{-\lvert t \rvert}}{2}\, dt = \int_{-\infty}^0 \frac{e^t}{2}\, dt + \int_0^x \frac{e^{-t}}{2}\, dt = \frac{1}{2} - \left. \frac{e^{-t}}{2} \right|_0^x = 1 - \frac{e^{-x}}{2}.$$ Also, $$P\left( \lvert R \rvert \leq 2 \right) = F_R(2) - F_R(-2) = 1 - e^{-2}$$ Hopefully this answers your question.