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Math Help - Index change in Double summation

  1. #1
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    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.
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  2. #2
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    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}.
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  3. #3
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    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.
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  4. #4
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    Brilliant....!
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