Hi Guys,

Can anybody tell me why it holds that:

$\displaystyle

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

$

I got it from my teacher, who left for vacation. I'm really confused by the change in indices.

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- Sep 13th 2010, 07:44 PMleoemilIndex change in Double summation
Hi Guys,

Can anybody tell me why it holds that:

$\displaystyle

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

$

I got it from my teacher, who left for vacation. I'm really confused by the change in indices. - Sep 14th 2010, 02:52 AMHallsofIvy
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) $\displaystyle \left(\sum_{y= 1}^1 1\right)x_{i1}= x_{i1}$

t= 2) $\displaystyle \left(\sum_{y= 1}^2 1\right)x_{i2}= (1+ 1)x_{i2}= 2x_{i2}$

t= 3) $\displaystyle \sum_{y= 1}^3 1\right)x_{13}= (1+ 1+ 1)x_{i3}= 3x_{i3}$

(Adding "1" from y= 1 to t just gives "t")

That entire sum is $\displaystyle \sum_{t= 1}^3\left(\sum_{y= 1}^t\right)x_{it}= x_{1i}+ 2x_{i2}+ 3x_{i3}$.

On the right side, again with T= 3, y takes on values of 1, 2, 3 and the inner sum there is

y= 1)$\displaystyle \sum_{t= 1}^3 x_{it}= x_{i1}+ x_{i2}+ x_{i3}$

y= 2)$\displaystyle \sum_{t= 2}^3 x_{it}= x_{i2}+ x_{i3}$

y= 3)$\displaystyle \sum_{t= 3}^3 x_{it}= x_{i3}$

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

For any T, both right and left hand sides give $\displaystyle \sum_{t= 1}^T tx_{it}$. - Sep 14th 2010, 03:07 AMOpalg
$\displaystyle \setlength{\unitlength}{5mm}

\begin{picture}(10,10)

\put(0,1){\vector(1,0){10}}

\put(1,0){\vector(0,1){10}}

\put(7,2){\line(0,1){5}}

\put(5,5){\line(1,0){4}}

\thicklines

\put(2,2){\line(1,0){7}}

\put(2,2){\line(1,1){7}}

\put(9,2){\line(0,1){7}}

\put(1.8,0){$1$}

\put(6.8,0){$t$}

\put(8.8,0){$T$}

\put(0.5,1.8){$1$}

\put(0.4,4.8){$y$}

\put(0.3,8.8){$T$}

\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. - Sep 14th 2010, 08:16 AMleoemil
Brilliant....!