Expectation of Rayleigh Distribution Conditional, or Given, X Less Than Y

Question

Suppose the pdf and cdf (Rayleigh Distribution):

$$ f_X(x) = \frac{2x}{\alpha}e^{-x^2/\alpha} \implies F_X(x) = 1-e^{-x^2/\alpha}$$

I calculate with integration by parts that:

$$E[X|X<y] = \frac{1}{F_X(y)} \int_0^y\frac{2x^2}{\alpha}e^{-x^2/\alpha}dx = \frac{1}{F_X(y)} \left (-ye^{-y^2/\alpha} + \frac{\sqrt{\alpha \pi}}{2}erf(\frac{y}{\alpha}) \right )$$

Next I am looking for:

$$E[X|z<X<y] = \frac{F_X(y)E[X|X<y]-F_X(z)E[X|X<z]}{F_X(y)-F_X(z)}$$

But this does not give sensible values, i.e. I find that $x$ is outside of the range $[z,y]$. Is this a calculation error or theory error?

Answer 1

Found the answer.. $erf(y/\alpha)$ should be $erf(y/\sqrt{\alpha})$ .. painful.