The c.d.f. of a function of a random variable

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if you have a continuous random variable X whose c.d.f. is F(x) for any x, and you have Y = 2X, then we know that the p.d.f. of Y is g(y) = $\frac{1}{2}g(\frac{1}{2}y)$. I would like to know if we can say that the c.d.f. of Y is G(y) = $F(\frac{1}{2}y)$, without any "plus constant C" that follows. Thank you very much!