Chris100
03-14-2009, 08:26 PM
Losses follow an exponential distribution with mean 100. Calculate the semi-deviation of the distribution.
The answer given goes like this:
\sigma^{2}_{SV} = \frac{1}{100}\int_{100}^{\infty}(x-100)^{2}e^{-x/100}dx
To evaluate this, substitute y = x - 100:
\sigma^{2}_{SV} = \frac{1}{100}\int_{0}^{\infty}y^{2}e^{-(y+100)/100}100dy
[I am not sure where that 100 right before dy comes from, but it disappears in the next step]
= \frac{1}{100e}\int_{0}^{\infty}y^{2}e^{-y/100}dy
= \frac{E[X^{2}]}{e}
What I do not understand is why the numerator in this last step is E[X] and not E[Y].
The answer given goes like this:
\sigma^{2}_{SV} = \frac{1}{100}\int_{100}^{\infty}(x-100)^{2}e^{-x/100}dx
To evaluate this, substitute y = x - 100:
\sigma^{2}_{SV} = \frac{1}{100}\int_{0}^{\infty}y^{2}e^{-(y+100)/100}100dy
[I am not sure where that 100 right before dy comes from, but it disappears in the next step]
= \frac{1}{100e}\int_{0}^{\infty}y^{2}e^{-y/100}dy
= \frac{E[X^{2}]}{e}
What I do not understand is why the numerator in this last step is E[X] and not E[Y].