# Math Help - Normal Distribution...

1. ## Normal Distribution...

Given that the random variable, X, follows a normal distribution with mean μ and variance (sigma)^2 , show that the random variable Z = (x - μ) / sigma , that is a linear transformation of X, has a mean of 0 and variance equal to 1..

Thanks for the help!

2. Hello,
Originally Posted by Vedicmaths
Given that the random variable, X, follows a normal distribution with mean μ and variance (sigma)^2 , show that the random variable Z = (x - μ) / sigma , that is a linear transformation of X, has a mean of 0 and variance equal to 1..

Thanks for the help!

If you just have to prove that the mean is 0 and the variance is 1 :
$\mathbb{E}(Z)=\mathbb{E}\left(\frac{X-\mu}{\sigma}\right)$
Then use the fact that $\mathbb{E}(aX+b)=a\mathbb{E}(X)+b$

Use the fact that $\text{Var}(aX+b)=a^2 \text{Var}(X)$

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If you want to prove that Z has a normal distribution with mean 0 and variance 1 :

Have you ever dealt with characteristic functions ?

They characterise a distribution.

$\psi_X(t)=\mathbb{E}(e^{itX})$

And for a normal distribution, $\psi_X(t)=e^{it\mu-\frac{\sigma^2t^2}{2}}$

So
\begin{aligned}
\psi_{\frac{X-\mu}{\sigma}}(t)
&=\mathbb{E}\left(\exp\left(it \cdot \frac{X-\mu}{\sigma}\right)\right) \\
&=\mathbb{E} \left(\exp\left(i \cdot \frac t \sigma \cdot X\right) \exp\left(\frac{-it\mu}{\sigma}\right)\right) \\
&=\exp\left(\frac{-it\mu}{\sigma}\right) \mathbb{E}\left(\exp\left(i \cdot \frac t\sigma \cdot X\right)\right)
\end{aligned}

But $\mathbb{E}\left(\exp\left(i \cdot \frac t\sigma \cdot X\right)\right)=\psi_X \left(\frac t \sigma\right)=\exp\left(\frac{it\mu}{\sigma}-\frac{\sigma^2t^2}{2 \sigma^2}\right)=\exp\left(\frac{it\mu}{\sigma}-\frac{t^2}{2}\right)$

Thus $\psi_{\frac{X-\mu}{\sigma}}(t)=\exp\left(\frac{-it\mu}{\sigma}\right) \exp\left(\frac{it\mu}{\sigma}-\frac{t^2}{2}\right)=\exp\left(\frac{-t^2}{2}\right)$

and this is the characteristic function for a normal distribution with mu=0 and sigma=1.

That was a little extra