Way to find CDF of a function with pdf similar to exponential distribution

Question

image link to the pdf of the function

$$ f(x;\theta) = \begin{cases} \frac{1}{4} e^{-\frac{1}{4} (x - \theta)} & \theta \lt x\\ 0 & \theta \gt x \end{cases} $$

How would you calculate the CDF of such a function? I am having trouble determining the domain after integrating the function. I know that, if $x$ must be bigger than $\theta$, then the CDF would be $0$ if $x$ smaller than theta. I got the integral to be $1-\exp[-(x-\theta)/4]$. Thank you in advance:)

Answer 1

Your way of phrasing your question makes me wonder if by "integrating the function" you mean just finding its antiderivatives. The cdf is \begin{align} F(x) & = \Pr(X\le x) = \int_{-\infty}^x f(t) \, dt \text{ where $f$ is the density function} \\[10pt] & = \begin{cases} \displaystyle\int_{-\infty}^x 0 \, dx & \text{if } x\le \theta, \\ \\ \displaystyle \left( \int_{-\infty}^\theta + \int_\theta^ x \right) f(t) \,dt = \int_{-\infty}^\theta 0\,dt + \int_\theta^x \frac 1 4 e^{-\frac 1 4 (t-\theta)} \, dx & \text{if }x>\theta, \end{cases} \\[20pt] & = \begin{cases} 0 & \text{if } x\le\theta, \\ \\ \displaystyle \int_\theta^x \frac 1 4 e^{-\frac 1 4 (t-\theta)} \, dt & \text{if } x>\theta. \end{cases} \end{align}

If $x>\theta$ then you have $$ \int_\theta^x \frac 1 4 e^{-\frac 1 4 (t-\theta)} \,dt = \left. -e^{-\frac 1 4 (t-\theta)} \vphantom{\frac{\displaystyle\sum}1}\right|_{t\,:=\,\theta}^{t\,:=\,x} = -e^{-\frac 1 4 (x-\theta)} - (-1) = 1 - e^{-\frac 1 4(x-\theta)}. $$

Answer 2

That is correct; $$F_X(x) = \begin{cases} 1 - e^{-(x-\theta)/4}, & x > \theta \\ 0, & x \le \theta. \end{cases}$$ You answered your own question.