Help on "Reverse order of Summation"

Hi everyone...

Can any one tell me how the limits changed by reversing the order of summation? I didn't get any information about it on internet....
Please help on this.... Its in chapter two of course...

Thanks,
Rana Hassan.

 

Hi, i assume u are refering to the derrivation of tempory annuity? I was confused by that also, the best way to do it is to write out every term in the summation and add columns rather than rows, the trick is in determining the limits. If you can post an example i may be able to help futher but i don't have CT5 notes with me at present.

 

I know exactly what ranahassen means, I went over this a few days ago.

Check out this link, it's based on the same principle.

the tricky bit is that the limits of 1 sum depends on the other, so you have to be careful when reversing.

try out devon matthews trick, that's basically how i did it too.

 

Thank you for your help...

MissAussie: what link you are talking about...

DevonMatthews: I will try by the method you've told and let you know if i can solve it...

Thanks again.

 

Here is the confusion:

 

I don't think there are any hard and fast rules to learn off.
You need to try to picture what it is that is being summed - writing it out and summing columns instead of rows is one way to see it and be able to swap the sigmas around.

As you noted j<=k-1. This is the key.
Both sum all values of v^j+1 k|qx over all values of j,k where 0<=j<=k-1<infinity.
For k in outer sum, its over all k from 0 to infinity. j in inner loop needs to be restricted so j<=k-1 so we sum over all values j such that 0<=j<=k-1;

For j in outer sum, its over all j from 0 to infinity (restrictions need to be in inner loops to be sensible). Again we want j<=k-1 so for each j (outer loop) we sum over all k such that j<=k-1<infinity. In other words j+1<=k< infinity.

For the avoidance of doubt it makes no difference to put the k|qx term in former or v^j+1 term in latter inside of the brackets of the inner sum. This is because these terms are independent of the parameter (j,k) of the inner sum and are effectively constants which can be factored in or out of the brackets.

 

oops, sorry for not including the link, it was this one.
http://mathworld.wolfram.com/DoubleSeries.html

but the guys above have explained it really well

 

hello,
didster, MissAussie, DevonMatthews - Please can anyone preview me how to do this... i still cant figure it out Can anyone upload a solution like i do (in the form of picture) that will be of great help... this thing is giving me very hard time... please

 

Ok, in general if we have F(j,k) summed over all 0<=j<k. (Here the F(j,k) is the discount factor multiplied by the probability of payment.)

Code:

+F(0,1)
+F(0,2)+F(1,2)
+F(0,3)+F(1,3)+F(2,3)
+F(0,4)+F(1,4)+F(2,4)+F(3,4)
+.....
.
.

Relating this to the summations in your picture,
the former relates to summing the terms row by row (ie adding all the k=1, then all the k=2, etc); and
the latter relates to summing the terms column by column (ie adding all the j=0 first, then all the j=1, etc)