how to find the cdf of X in terms of Z when $X=2Z+1$

Question

Consider Z a Normal (Gaussian) random variable with mean 0 and variance 1.
It has density $$f_Z(z)=\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}} \text{for all x real numbers}$$ We consider $X=2Z+1$. Write the CDF of X in terms of the one of Z and take the derivative to get that the density of X is $$f_X(x)=\frac{1}{2\sqrt{2\pi}}e^{\frac{-(x-1)^2}{8}} \text{for all x real numbers}$$
I know that I have to take the integral of $f_Z(z)$ in terms of X to get the cdf, I just do not know how to get it in terms of X. The bounds of the integral should be $-\infty$ to $\infty$ right?

Answer 1

To get the pdf of $X$, let us first get the cdf of $X$:

$$F_X(x) = P(X \le x) = P(2Z + 1 \le x) = P(Z \le \frac{x-1}{2})$$

Now $P(Z \le \frac{x-1}{2}) = F_Z(\frac{x-1}{2})$ so:

$$F_X(x) = F_Z(\frac{x-1}{2})$$

$$\to f_X(x) = f_Z(\frac{x-1}{2}) (\frac{x-1}{2})'$$

$$\to f_X(x) = f_Z(\frac{x-1}{2}) (\frac{1}{2})$$

Answer 2

The easier way to proceed is to note that $2Z+1 \leq x$ if and only if $Z \leq \frac{x-1}{2}$. This gives you the CDF, which you can differentiate as necessary.