Symbolic manipulation with unevaluated sums

Question

If I have two expressions with sums in them, like this:

$$\begin{align*} b&=\frac{\sum_{i} (x_i - \bar{x})(y_i -\bar{y})}{\sum_{i}(x_i -\bar{x})^2}\\ r&=\frac{\sum_{i} (x_i - \bar{x})(y_i -\bar{y})}{\sqrt{\sum_{i}(x_i -\bar{x})^2\sum_{i}(y_i -\bar{y})^2}} \end{align*}$$

and I wanted to produce a simplified expression of $\frac{b}{r}$, how would I do it? The problem is the sums. It seems Mathematica doesn't like the unspecified, unevaluated sums.

Edit: I was expecting to end up with something like this:

$$\sqrt{\frac{\sum_{i}(y_i -\bar{y})^2}{\sum_{i}(x_i -\bar{x})^2}}$$

I've been playing around with Expand, Simplify, FullSimplify. There may just be a way to apply it that I'm missing.

(Sum[(Subscript[x, i] - OverBar[x]) (Subscript[y, i] - OverBar[y]), i]/Sum[(Subscript[x, i] - OverBar[x])^2, i])/(Sum[(Subscript[x, i] - OverBar[x]) (Subscript[y, i] - OverBar[y]), i]/ Sqrt[Sum[(Subscript[x, i] - OverBar[x])^2, i] Sum[(Subscript[y, i] - OverBar[y])^2, i]]) works on my end... — J. M.'s missing motivation, Aug 09 '12 at 12:37
@Verde, Yes I'm well aware of how linear regressions are derived, but since Mean[] doesn't know how to handle something that isn't explicitly a list... — J. M.'s missing motivation, Aug 09 '12 at 12:41
@J.M. Really? Another clue I can add to my little black book... — Sjoerd C. de Vries, Aug 09 '12 at 13:17
@SjoerdC.deVries Do you take notes of all embarrassing confessions? — Dr. belisarius, Aug 09 '12 at 18:56
@SjoerdC.deVries Thanks god! I am getting so perfect these days that I am afraid of forgetting my life as a mere mortal being. Your notes will be useful for my autobiography. — Dr. belisarius, Aug 09 '12 at 19:10

score 8 · Accepted Answer · answered Aug 09 '12 at 15:34

8

The absence of a terminal $n$ in the sums suggests you are looking for a combination of symbolic reduction and typography. Let's separate the two, then, by using symbols for the sums, performing the reduction, and then replacing the symbols by whatever we like:

MMA screen shot

answered Aug 09 '12 at 15:34

whuber

20,544
2
59
111

1

Or, one can use PowerExpand[]... – J. M.'s missing motivation Aug 09 '12 at 15:47
@J.M. Interestingly (but understandably), using PowerExpand without explicit assumptions yields a different answer: the square roots are no longer distributed over the quotient. – whuber Aug 09 '12 at 15:56

score 5 · Answer 2 · answered Aug 10 '12 at 00:13

Here is a completely symbolic manipulation:

ClearAll[x, y, xMean, yMean, n];

varianceX = Sum[(x[i] - xMean)^2, {i, n}]

$\sum _i^n (x(i)-\text{xMean})^2$

varianceY = Sum[(y[i] - yMean)^2, {i, n}]

$\sum _i^n (y(i)-\text{yMean})^2$

coVariance = Sum[(x[i] - xMean) (y[i] - yMean), {i, n}]

$\sum _i^n (x(i)-\text{xMean}) (y(i)-\text{yMean})$

b = coVariance/varianceX;

r = coVariance/Sqrt[varianceX varianceY];

b/r /. {Sqrt[a_ b_] :> Sqrt[a] Sqrt[b]}

$\frac{\sqrt{\sum _i^n (y(i)-\text{yMean})^2}}{\sqrt{\sum _i^n(x(i)-\text{xMean})^2}}$

All I had to do in order to help Mathematica get to the simplified result is to state the rule Sqrt[a_ b_] :> Sqrt[a] Sqrt[b].

Symbolic manipulation with unevaluated sums

2 Answers2