CDF of $ Y = 2X + 3$

Question

Suppose $X$ is a random variable whose c.d.f $F$ satisfies $F(x) = x^3$ for $0 < x < 1$. Calculate the c.d.f of $Y = 2X + 3$.

Thought process: Consider the function $G(y) = P(Y \le y) = P(2X + 3 \le y) = P(X \le (y-3)/2).$

Then the c.d.f. should be represented by the piecewise function with the following conditions: $0$ if $y \le 3$, $(y-3)^3/8$ if $3 < y < 5$, and 1 if $y \ge 5$.

Is this the correct function modeling the c.d.f of $Y$?

Answer 1

Yes, since you've used the formal CDF definition $$G(y) = Pr(Y \leq y) = \ldots = F_X(\frac{y-3}{2}) = \big( \frac{y-3}{2}\big)^3 \qquad 3 < x< 5$$ with the appropriate interval scaling. Then, your steps are correct.