ASM 68.13

Chris100 · #1 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 · #2 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 · #3 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.

Thread Tools
Show Printable Version Email this Page
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode