3
$\begingroup$

Not sure if "linear transformation" is the correct terminology, but...

Let $X$ be a random variable with a normal distribution $f(x)$ with mean $\mu_{X}$ and standard deviation $\sigma_{X}$: $$f(x) = \frac{1}{\sigma_{X}\sqrt{\tau}}\exp{\left[\frac{-1}{2}\left(\frac{x-\mu_{X}}{\sigma_{X}}\right)^2\right]}$$

(Here, $\tau=2\pi$)

Let $Y$ be a random variable defined by the linear transformation $$Y = u(X) = aX+b$$

Let $v(y) = u^{-1}(y) = \frac{y-b}{a}$. Then $v^\prime(y) = \frac{1}{a}$.

Prove: $Y$ is normally distributed, with density function $$g(y)=f\bigl(v(y)\bigr)\,\bigl|v^\prime(y)\bigr|$$ $$= \frac{1}{\sigma_{X}\sqrt{\tau}}\exp{\left[\frac{-1}{2}\left(\frac{\frac{y-b}{a}-\mu_{X}}{\sigma_{X}}\right)^2\right]}\,\left|\frac{1}{a}\right|$$ $$= \frac{1}{\left|a\right|\sigma_{X}\sqrt{\tau}}\exp{\left[\frac{-1}{2}\left(\frac{y-(a\mu_{X}+b)}{a\sigma_{X}}\right)^2\right]}$$ with mean $\mu_{Y} = a\mu_{X}+b$ and standard deviation $\sigma_{Y} = \left|a\right|\sigma_{X}$.

  • 0
    Are you sure about this problem? $$h(x) = \frac{1}{\sigma\sqrt{2\pi}}e^{\frac{-1}{2}\left(\frac{(mx+b)-\mu}{\sigma}\right)^2}$$ is a function $R \to (0,1/\sigma\sqrt{2\pi})$ and so, with $X$ being a normal random variable, $h(X)$ is _definitely_ not a normal random variable.2012-09-21
  • 0
    Did you give your own interpretation of the problem? The normal (unintended pun) question is to say show that $g(X)$ has normal distribution (and find the mean, variance of $g(X)$). The question as posed makes no sense.2012-09-21
  • 0
    @TestSubject528491: I am typo-prone, but I **mean** $g(X)$. The function $g(t)=mt+b$ is indeed a linear function. But $g(X)$ is a normally distributed random variable, mean $m\mu+b$, variance $m^2\sigma^2$. Very standard stuff, that I have taught and used many times.2012-09-21
  • 0
    ::Foot-in-mouth:: comment deleted2012-09-21
  • 0
    There is an elegant proof of this using generalized polynomial chaos that no one seems to appreciate. --wistful sigh--2012-11-04

1 Answers 1