TITLE: Multiplication of a random variable with constant
QUESTION [47 upvotes]: Suppose $X$ is a random variable which follows standard normal distribution then how is $KX$ ($K$ is constant) defined. Why does it follow a normal distribution with mean $0$ and variance $K^2$. 
Thank You.

REPLY [22 votes]: Another way of characterizing a random variable's distribution is by its distribution function, that is, if two random variables have the same distribution function then they are equal.
In our case, let $X \sim N(\mu,\sigma^2)$ then set $Y = c X$ with $c > 0$ and call $F$ the distribution function of $X$ and $G$ the distribution function of $Y$. Then:
$G(y) = P[Y \le y] = P[cX \le y] = P\Big[X \le \frac  yc\Big] = F\Big(\frac yc\Big)$
Now we differentiate and we get:
$g(y) = f(\frac yc) \frac1c$
where $g$ is the density function for $Y$ and $f$ is the density function for $X$. Then we just try to express this as a normal density:
$g(y) = f(\frac yc ) \frac1c = \frac{1}{\sqrt{2\pi}(c\sigma)} e^{\frac{-(cx-c\mu)^2}{2(c\sigma)^2}} = \frac{1}{\sqrt{2\pi}(c\sigma)} e^{\frac{-(y-c\mu)^2}{2(c\sigma)^2}}$
But this last is the density of a $N(c\mu,(c\sigma)^2)$
This is a calmed formulation of what Dilip Sarwate pointed out in the comments before.
The case c < 0
$G(y) = P[Y \le y] = P[cX \le y] = P\Big[X \ge \frac  yc\Big] = 1 - F\Big(\frac yc\Big)$
differentiating:
$g(y) = -f(\frac yc) \frac1c = f(\frac yc) \frac{1}{|c|} = \frac{1}{\sqrt{2\pi}(|c|\sigma)} e^{\frac{-(y-c\mu)^2}{2(c\sigma)^2}}$
Note that this does not pose difficulties since $\sqrt{(c \sigma)^2} = |c| \sigma$.