$F_x(x)=\begin {cases} \frac1 2e^X &\text {if $X<0$} \\ X+\frac 1 2 &\text{if $0\leq X<\frac1 2$} \\ 1&\text{if $\frac1 2\leq X$} \end {cases}$
Find $P({X≤-1/16})$, $P({-1/16<X≤1/16})$, $P({X=0})$, and $P({X=1/2})$.
I'm having a bit of trouble understanding how to calculate these probabilities. I found $f_x(x)$.
$f_x(x)=\begin {cases} \frac1 2e^x &\text {if $x<0$} \\ 1 &\text{if $0\leq x<\frac1 2$} \\ 0&\text{if $\frac1 2\leq x$} \end {cases}$
$f_x(x) = \frac d {dx}F_x(x)$. I wasn't sure how the integrating rules worked here. For $P({X≤-1/16})$ would I have to integrate two separate portions?
$(1/2)e^x \text{ for $x<0$}$
then
$1 \text{ for } 0 ≤ x < 1/2$
The probability of $P({X≤-1/16})$ falls under two of the equations in the piecewise functions. I.e.
$(1/2)e^x$ from $0$ to $-\infty$ and $1$ at $0 = P({X≤-1/16})$
Could someone help explain this to me?
For any piecewise function, if you cross from one boundary to another (i.e. any integral in you equation that crosses over zero), you divide the integral into two separate integrals using $\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx$, where $c$ is the boundary point. You then integrate the piecewise function that specifically corresponds to your bounds.