See the question above. I am not quite sure, if my result is correct, because I do not have any solutions. I tried with the following formula:

$$ E[X] = e^{\mu+\frac{1}{2} \sigma^{2}} \cdot \Phi\left(\frac{\ln \left(\frac{1}{\alpha}\right)+\mu+\sigma^{2}}{\sigma}\right)-\alpha \cdot \Phi\left(\frac{\ln \left(\frac{1}{\alpha}\right)+\mu}{\sigma}\right) $$

What is the correct result?

  • $\begingroup$ Forgot to mention, my result is 24.403 $\endgroup$ – gvncore Mar 18 at 18:37
  • $\begingroup$ What's $\alpha$? $\endgroup$ – Glen_b Mar 19 at 6:11
  • $\begingroup$ $\alpha = 100$ (given in title) $\endgroup$ – gvncore Mar 19 at 7:33
  • $\begingroup$ Thanks; there's a 100 in the title but no indication there that it's $\alpha$ $\endgroup$ – Glen_b Mar 19 at 9:11

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