I'm given the following probability density function of $x$:
$$f(x)=\frac{1}{2}e^{x}\mathbf{1}_{\{x<0\}}+\frac{1}{2}e^{-x}\mathbf{1}_{\{x\geq 0\}}$$
How do I find the cumulative distribution of the following function? I'm confused on how to handle the indicator function while integrating.
Really appreciate the help! Thank you!
Here $X$ has the standard Laplace (sometimes called "double exponential") distribution. We compute the distribution function for $t<0$: $$ F_X(t) = \int_{-\infty}^t \frac12 e^x\ \mathsf dx = \frac12 e^t $$ and for $t>0$: $$ F_X(t) = \int_{-\infty}^t f(x)\ \mathsf dx = \int_{-\infty}^0 \frac12 e^x\ \mathsf dx + \int_0^t \frac12 e^{-x}\ \mathsf dx = 1 - \frac12e^{-t}. $$ The density is given by $$ f_X(t) = \frac12 e^{-|t|}. $$
Split the integral into two intervals using this property:
$$\int_{a}^{c} f(x) dx = \int_{a}^{b} f(x) dx + \int_b^c f(x) dx $$
Now you can evaluate the indicator functions on their respective intervals.
