How to prove something at Normal distribution

Question

$X\sim N(\mu,\sigma^2)$.
$A,B$ are constants and $A\ne0$.

How to prove that $AX+B\sim N(A\mu+B,A^2\sigma^2)$ ?

Thank you!

Answer 1

The standard method:

Let $Y \sim \mathcal{N}(A\mu + B, A^2\sigma^2)$. Write the CDF of $Y$:

$$F_Y(y) = \frac{1}{\sqrt{2\pi A^2\sigma^2}}\int_{-\infty}^\infty \exp \left( -\frac{(x-(A\mu+B))^2}{2A^2\sigma^2}\right)\ dx.$$

Compute the expectation of $AX+B$ using the standard methodology.

$$E[AX+B] = \frac{1}{\sqrt{2\pi \sigma^2}}\int_{-\infty}^\infty (Ax+B)\exp \left( -\frac{(x-(\mu))^2}{2\sigma^2}\right)\ dx.$$

Show that the result leads to the mean of $Y$. Do the same for the variance.

Clever way:

Use the fact that $\operatorname{erf}$ and the CDF of a Gaussian are related. Let the CDFs of $AX+B$ and $Y$ be equal to the same uniform random variable (inverse transform method). Use $\operatorname{erf}^{-1}$ to write $AX+B$ in terms of $Y$. Show that the result leads directly to $X$.

Answer 2

You can manipulate $P(aX+b \leq c)$ into $P(X \leq \frac{c-b}{a})$ for $a>0$. Then, you know the CDF of $X$, so plug this into the CDF of $X$. You also know the CDF of $N(a \mu + b , a^2 \sigma^2)$, so compare them to see that they're equal, thus the distributions are the same.

Alternatively you can use characteristic functions/moment generating functions by noting $E[e^{t (aX+b)}] = e^{bt} E[e^{(at)X}] = e^{bt} M_X(at)$, and comparing that to the moment generating function of a $N(a \mu + b , a^2 \sigma^2)$ r.v.