Originally Posted by

**tompa909** Ok, I am allowed to standardize P(Y>0) and get the probability using Z-tables.

With the integral I have,

$\displaystyle \int_{0}^{+\infty} y \, \frac{1}{4\,\sqrt{2\pi}} \, e^{-\frac{1}{2}(\frac{y+3}{4})^2} \, dy$

Using substitution z = (y+3)/4, we get y = 4z-3 dy = 4dz

$\displaystyle \int_{3/4}^{+\infty} \, \frac{4z-3}{\sqrt{2\pi}} \, e^{-\frac{1}{2}z^2} \, dz$

$\displaystyle \int_{3/4}^{+\infty} \, \frac{4z}{\sqrt{2\pi}} \, e^{-\frac{1}{2}z^2} \, dz \, - \int_{3/4}^{+\infty} \, \frac{3}{\sqrt{2\pi}} \, e^{-\frac{1}{2}z^2} \, dz$

$\displaystyle \frac{4}{\sqrt{2\pi}}\int_{3/4}^{+\infty} \, z \, e^{-\frac{1}{2}z^2} \, dz \, - 12\int_{3/4}^{+\infty} \, \frac{1}{4\sqrt{2\pi}} \, e^{-\frac{1}{2}z^2} \, dz$ *****

$\displaystyle \frac{4}{\sqrt{2\pi}} \, (-e^{-\frac{1}{2}(+\infty)^2}--e^{-\frac{1}{2}(\frac{3}{4})^2}) - 12 * P(Z > 3/4)$

$\displaystyle \frac{4}{\sqrt{2\pi}} \, (0.75484) - 12 * 0.2266$

Thus $\displaystyle \int_{0}^{+\infty} y \, \frac{1}{4\,\sqrt{2\pi}} \, e^{-\frac{1}{2}(\frac{y+3}{4})^2} \, dy = -1.51465$

Going back to original question

$\displaystyle E(Y|Y>0) = $$\displaystyle \frac{\int_{0}^{+\infty} y \, f(y) \, dy}{\Pr(Y > 0)}= \frac{-1.51465}{P(Y>0)} = \frac{-1.51465}{P(Z>3/4)} =\frac{-1.51465}{0.2266} = -6.68425$

Thanks for your help