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    Post 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!

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    Hello,
    Quote Originally Posted by Vedicmaths View Post
    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)


    -------------------------------------------------------------
    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}<br />
\psi_{\frac{X-\mu}{\sigma}}(t)<br />
&=\mathbb{E}\left(\exp\left(it \cdot \frac{X-\mu}{\sigma}\right)\right) \\<br />
&=\mathbb{E} \left(\exp\left(i \cdot \frac t \sigma \cdot X\right) \exp\left(\frac{-it\mu}{\sigma}\right)\right) \\<br />
&=\exp\left(\frac{-it\mu}{\sigma}\right) \mathbb{E}\left(\exp\left(i \cdot \frac t\sigma \cdot X\right)\right)<br />
\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
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