Results 1 to 4 of 4

Thread: Index change in Double summation

  1. #1
    Banned
    Joined
    Sep 2010
    Posts
    15

    Index change in Double summation

    Hi Guys,

    Can anybody tell me why it holds that:

    <br />
\sum_{t=1}^T \left( \sum_{y=1}^t 1 \right) x_{it} = \sum_{y=1}^T 1 \sum_{t=y}^T x_{it}<br />

    I got it from my teacher, who left for vacation. I'm really confused by the change in indices.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,188
    Thanks
    2820
    Look at some simple examples, with, say, T= 3. The outer sum then goes t= 1, t= 2, t= 3: for each such t, the term being added is
    t= 1) \left(\sum_{y= 1}^1 1\right)x_{i1}= x_{i1}
    t= 2) \left(\sum_{y= 1}^2 1\right)x_{i2}= (1+ 1)x_{i2}= 2x_{i2}
    t= 3) \sum_{y= 1}^3 1\right)x_{13}= (1+ 1+ 1)x_{i3}= 3x_{i3}
    (Adding "1" from y= 1 to t just gives "t")
    That entire sum is \sum_{t= 1}^3\left(\sum_{y= 1}^t\right)x_{it}= x_{1i}+ 2x_{i2}+ 3x_{i3}.

    On the right side, again with T= 3, y takes on values of 1, 2, 3 and the inner sum there is
    y= 1) \sum_{t= 1}^3 x_{it}= x_{i1}+ x_{i2}+ x_{i3}
    y= 2) \sum_{t= 2}^3 x_{it}= x_{i2}+ x_{i3}
    y= 3) \sum_{t= 3}^3 x_{it}= x_{i3}

    Multiplying each of those by "1" and adding gives \sum_{y= 1}^T 1 \sum_{t= y}^T x_{it}= (x_{i1}+ x_{i2}+ x_{i3})+ (x_{i2}+ x_{i3})+ (x_{i3})= x_{i1}+ 2x{i2}+ 3x_{i3}, exactly the same.

    For any T, both right and left hand sides give \sum_{t= 1}^T tx_{it}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    10
    Quote Originally Posted by leoemil View Post
    Hi Guys,

    Can anybody tell me why it holds that:

    <br />
\sum_{t=1}^T \left( \sum_{y=1}^t 1 \right) x_{it} = \sum_{y=1}^T 1 \sum_{t=y}^T x_{it}<br />

    I got it from my teacher, who left for vacation. I'm really confused by the change in indices.
    \setlength{\unitlength}{5mm}<br />
\begin{picture}(10,10)<br />
\put(0,1){\vector(1,0){10}}<br />
\put(1,0){\vector(0,1){10}}<br />
\put(7,2){\line(0,1){5}}<br />
\put(5,5){\line(1,0){4}}<br />
\thicklines<br />
\put(2,2){\line(1,0){7}}<br />
\put(2,2){\line(1,1){7}}<br />
\put(9,2){\line(0,1){7}}<br />
\put(1.8,0){$1$}<br />
\put(6.8,0){$t$}<br />
\put(8.8,0){$T$}<br />
\put(0.5,1.8){$1$}<br />
\put(0.4,4.8){$y$}<br />
\put(0.3,8.8){$T$}<br />
\end{picture}

    The summation is over the integer points in the triangular region in the picture. If you start by fixing y, then you sum over t (along a horizontal line) with t going from y to T (and then you take the results and sum them over y going from 1 to T). But if you start by fixing t, then you sum over y (along a vertical line) with y going from 1 to t (and then you take the results and sum them over t going from 1 to T).

    Edit. This is just another way of saying what HallsofIvy has explained, but using a graphical picture rather than an algebraic illustration.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Banned
    Joined
    Sep 2010
    Posts
    15
    Brilliant....!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Change/simplify the index of a radical
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 20th 2011, 03:09 PM
  2. Double summation
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: Aug 23rd 2009, 06:09 PM
  3. index of summation
    Posted in the LaTeX Help Forum
    Replies: 1
    Last Post: May 23rd 2009, 07:18 PM
  4. double summation
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Apr 28th 2009, 09:15 PM
  5. double summation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: Mar 30th 2009, 12:11 PM

Search Tags


/mathhelpforum @mathhelpforum