Assume that the random variable $X$ has the distribution $f(x; \theta) = e^{−(x − \theta)} , x \ge\theta$, while $f(x; \theta) = 0$ for $x < \theta$; $\theta$ denotes a constant parameter.
Let $Y = 2X + 3$. Determine the probability density of the random variable Y. Rain out the probability that $Y> \theta + 1$.
What I can't figure out is how to solve this, when I know the answer to the last part $Y> \theta + 1$ should be 1.
for the first part, I got that the PDF of $Y$ was $\frac{1}{2}e^{-(y-3)/2 + \theta}$ but cannot get the answer for the second part to add up to $1$.