# Multiplication of a random variable with constant

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.

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I see you are a new member to MSE. A Hearty Welcome!. Please make a note to accept (if it is acceptable to you) the answers to your previous questions. This motivates (mostly) MSE to help you more. – dineshdileep Jan 11 '13 at 6:08

For a random variable $X$ with finite first and second moments (i.e. expectation and variance exist) it holds that $\forall c \in \mathbb{R}: E[c \cdot X ] = c \cdot E[X]$ and $\mathrm{Var}[c\cdot X] = c^2 \cdot \mathrm{Var} [ X]$
However the fact that $c\cdot X$ follows the same family of distributions as does $X$ is not trivial and has to be shown seperately. (Not true e.g. for the Beta distribution, which is also in the exponential family). You can see it if you look at the characteristic function of the product $c\cdot X$: $\exp\{i\mu c t - \frac{1}{2} \sigma^2 c^2 t^2\}$ which is the characteristic function of a normal distribution wih $\mu'= \mu\cdot c$ and $\sigma' = \sigma \cdot c$.
The first paragraph is a perfectly correct answer (except for a missing $E$ in $E[c\cdot X]=c\cdot E[X]$) but I am not sure I understand the issue in the second paragraph. If $X$ is a continuous random variable with probability density function $f_X(x)$, then, for $c\neq 0$, so is $c\cdot X$ a continuous random variable with probability density function $$f_{c\cdot X}(a)=\frac{1}{|c|}f_X\left(\frac{a}{c}\right)$$ which, if not belonging to the same nuclear family of distributions, is at least a kissing cousin of the family, and of course, $c\cdot X$ is normal if $X$ is normal. – Dilip Sarwate Jan 11 '13 at 14:38
Thanks for pointing out the missing "$E$". – user1965813 Jan 11 '13 at 15:34
Use the definition of expectation of function of a random variable and variance of function of a random variable. If $g(X)=KX$, what is its mean and variance? This result is very general and is not unique to normal distributions alone.