Question

Suppose X ~ N(μ,σ ^2). Find the distribution of Z = (X - μ)/σ using one dimensional change of variable.

Answer 1

I don't know what a "dimensional change of variable" means, but you can do it this way. Suppose the distribution of X is the function\[F_X(x).\]Lets find the distribution of Z in terms of\[F_X(x).\]
\begin{eqnarray*}F_Z(x) &=& \mathbb{P}(Z \leq x) \\&=& \mathbb{P}\left(\frac{X - \mu}{\sigma} \leq x\right)\\&=& \mathbb{P}(X \leq \sigma x + \mu) \\&=& F_X(\sigma x + \mu).\end{eqnarray*}Then,\[F_Z'(x) = F_X'(\sigma x + \mu) \sigma = f_X(\sigma x + \mu) \sigma = \frac{1}{\sqrt{2\pi} \sigma}e^{-\frac{((\sigma x + \mu) - \mu)^2}{2\sigma^2}} \sigma = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\]which is the distribution of a standard normal, so \[Z \leadsto \mathcal{N}(0, 1).\]

Answer 2

hmmm thats also a nice way to prove!

Answer 3

also for this ques, whats P(|Z| > 1) ?

Answer 4

\begin{eqnarray*}\mathbb{P}(|Z| > 1) &=& \mathbb{P}(Z < -1) + \mathbb{P}(Z > 1)\\&=&\mathbb{P}(Z \leq 1) + (1 - \mathbb{P}(Z \leq 1))\\&=& 1 + \mathbb{P}(Z \leq -1) - \mathbb{P}(Z \leq 1)\\&=& 1 + F_Z(-1) - F_Z(1)\\&=& 1 + \int_{-\infty}^{-1} \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx - \int_{-\infty}^{1} \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx\\&=& 1 - \int_{-1}^1 \frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx \approx 0.317311.\end{eqnarray*}

Answer 5

good!