First, let Z~f(z) and we can define $z_\alpha$ s.t. $\alpha$=P($Z>z_{\alpha}$). P($Z>z_{\alpha}$)is the integral from $z_{alpha}$ to infinity of $f(z) \ dz$. Show that if X is a random variable with PDF (1/$\sigma$)f((x-$\mu$/$\sigma$) and $x_{\alpha}=\sigma z_{\alpha}+\mu$, then $P(X>x_{\alpha })=\alpha$
So far, what I've done is substitute the relationship of P($X>x_{\alpha}$) and $x_\alpha$ in place of $z_{\alpha}$. I ended up with something like the limit of x approaching infinity of F(x-$\mu$/$\sigma$)-F($\sigma z_{\alpha}$. I'm not so sure what to do from here. Can I get some help?
We know that, $$\int_{z_{\alpha}}f(z)dz=\alpha $$ since this follows from the definition of $P(Z>z_{\alpha})=\alpha$. We want to show somehow by transforming $Z$ to $X$ that this implies that $P(X>x_{\alpha})=\alpha$. If were a bit clever we see that if we let $Z=\frac{X-\mu}{\sigma}$ we can use transformation of variables. If you remember how to transform pdf's from a linear combination of one variable to another we in general that, given transformation of a variable $y=g(x)$, with inverse $g^{-1}(y)=x$ we get,
$$f(x)=\int_{g^{-1}(a)}^{g^{-1}(b)}f(g(x))\frac{dx}{dy}dy$$ so we get in our case that
$$\int_{z_{\alpha}\sigma + mu}f((x-\mu)/\sigma)(1/\sigma)dx=\alpha $$ (notice that $dz=(1/\sigma)dx$). Now the expression inside the integral is the pdf for X. So the integral says the same thing as $P(X>z_{\alpha}\sigma + mu)=\alpha$, and since $x_\alpha=z_{\alpha}\sigma + mu$ we are done!