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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].

Abraham Weishaus
03-14-2009, 09:34 PM
The 100 is a typo.

What's Y? But you'll ask me what's X? By X, I mean an exponential random variable with mean 100.

Chris100
03-14-2009, 10:20 PM
yes i see now - thanks.

since we have changed the limits of integration due to the substitution, that last expression is the second raw moment of an exponential with mean 100.