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Warren Schmidt
11-03-2006, 12:38 AM
There was one question where it asked for the risk load where I believe the equation is r = (yz/(1+y)) * (change in standard deviation). The problem gave the value for y and z, and gave the standard deviation before the risk was added and after the risk was added. Was this problem as easy as simply plugging the numbers into the equation, or did I miss something?

KidCA
11-03-2006, 01:41 AM
I hope it was that straight-forward. :) I got the risk load for x, then for x+y and found the difference, but I think it could be solved in one step, I just get hesitant. I wrote down the general formulas too for V = zS - r, just in case I missed something.

11-03-2006, 07:52 AM
There was one question where it asked for the risk load where I believe the equation is r = (yz/(1+y)) * (change in standard deviation). The problem gave the value for y and z, and gave the standard deviation before the risk was added and after the risk was added. Was this problem as easy as simply plugging the numbers into the equation, or did I miss something?

they gave the standard deviation???

it's laughable.....

oh well.

Avi
11-03-2006, 08:37 AM
There was one question where it asked for the risk load where I believe the equation is r = (yz/(1+y)) * (change in standard deviation). The problem gave the value for y and z, and gave the standard deviation before the risk was added and after the risk was added. Was this problem as easy as simply plugging the numbers into the equation, or did I miss something?

they gave the standard deviation???

it laughable.....

oh well.

Yep, it seemed a gimme. You had to remember to compare Risk load (X) and Risk Load (X+Y) and take the difference as opposed straight calculating Risk Load (Y). At least, I think so. :oops:

11-03-2006, 09:23 AM
Wasn't it a buildup question? In that case, wouldn't you take Standard Devitation of (X + Y) and subtract standard deviation of (X) (since X was there already and they were adding y), and then multiply that by the risk load?

NeedaBreak
11-03-2006, 09:58 AM
Wasn't it a buildup question? In that case, wouldn't you take Standard Devitation of (X + Y) and subtract standard deviation of (X) (since X was there already and they were adding y), and then multiply that by the risk load?

Yes. And it was that simple. Loved it!

Avi
11-03-2006, 01:25 PM
Wasn't it a buildup question? In that case, wouldn't you take Standard Devitation of (X + Y) and subtract standard deviation of (X) (since X was there already and they were adding y), and then multiply that by the risk load?

No difference, because the risk load is a linear function of the SD, so it matters not if you subtract risk dollars or SD dollars and then multiply by risk load. (distribution of multiplication over addition ;) ).

tommie frazier
11-03-2006, 02:26 PM
anyone have a numerical answer here?

Avi
11-03-2006, 03:22 PM
anyone have a numerical answer here?\$280.42