# Help interpreting this equation

• Apr 16th 2012, 08:55 AM
mazz7755
Help interpreting this equation
Hi,

This is my first post here and I hope that I am in the correct subforum (Tongueout).

So, I have this equation: $r_{Gm}(y)=\left \{ Y_{j}|D_{Gm}(Y_{j})\leq D_{Gm}(y),j=1,...,m \right \}/m$

Also there is a # before the curly bracket, I don't know why the equation writer won't allow me to put it.

Now to the question. Disregard what the variables are, I just want to know how this equation works. How can I interpret it? If I have the variables how do I get rGm?

• Apr 16th 2012, 10:22 AM
MathoMan
Re: Help interpreting this equation
Here's what I see in that.

For a given $y$ $r_{Gm}(y)$ is the cardinality of the set $\left\{ Y_j \vert D_{Gm}(Y_j)\leq D_{Gm}(y), \,\,j=1,\ldots , m \right\}$ divided by $m$.

Since you have m values $\left\{Y_1,Y_2,\ldots , Y_m \right\}$ you count how many of them are such that for a given y $D_{Gm}(Y_i)\leq D_{Gm}(y)$ holds. Obviously that number is somewhere between 0 and m. Take that number and divide it by m and you get the $r_{Gm}(y)$ which then obviously is always a number between 0 and 1.

$r_{Gm}(y)$ can be thought of as a percentage provided you multiply it by 100.
• Apr 16th 2012, 11:49 AM
mazz7755
Re: Help interpreting this equation
Thank you Mathoman.

I understand in principle want you have explained me but with the data I have I'm not getting the right values. (Or maybe I did not understand as well as I think.) Maybe you can use an example?( I'm attaching a table where the column r(X) should be the result of the equation)
Attachment 23617
• Apr 16th 2012, 12:31 PM
MathoMan
Re: Help interpreting this equation
Ok I'll make up an example, trivial one.
Let ${\cal Y}=\left\{-10,1,2,10,30,20\right\}$. Hence $m=6$
Let $D_{Gm}(x)=x^2-2$.
For a given $y$, say $y=1$, $D_{Gm}(y)=D_{Gm}(1)=1^2-2=-1$.

Then $r_{Gm}(1)=\frac{\left\vert \left\{Y_j \vert Y_j^2-2 \leq -1,\,\, j=1,\ldots, 6 \right\} \right\vert}{6}$.

Evaluate $D_{Gm}$ for every element in ${\cal Y}$ - apply $D_{Gm}$ elementwise to ${\cal Y}$ - $D_{Gm}({\cal Y})=\left\{98,-1,2,98,898,398 \right\}$.
Now count how many values are $\leq -1$: 1.
$r_{Gm}(1)=\frac{\left\vert \left\{Y_j \vert Y_j^2-2 \leq -1,\,\, j=1,\ldots, 6 \right\} \right\vert}{6}=\frac{1}{6}$.

If y=11, then $D_{Gm}(y)=D_{Gm}(11)=11^2-2=119$.
In ${ \cal Y}$ there are 4 values for which function $D_{Gm}$ evaluates to a value $\leq 119$ so:
$r_{Gm}=\frac{4}{6}=\frac{2}{3}$.

That's my way of looking at it, hope it helps.
• Apr 16th 2012, 01:58 PM
mazz7755
Re: Help interpreting this equation
Thank you for the example. So I will try to follow your example with mine(table in my previous post):
The values for $D_{Gm}$ are on the second column D(X), and we have a set of 80 values so m=80.
If I calculate the $r_{Gm}$ of the first value (0.0028) according to your previous example I should check which values are equal or less than 0.0028. Checking the table I have (if I counted correctly) 36 values. So doing $r_{GM}=\frac{36}{80}$ we get 0.45 and not the 0.082 which is the given result.

Maybe you can see where I'm making the mistake?
• Apr 16th 2012, 02:33 PM
MathoMan
Re: Help interpreting this equation
My guess is that you are counting the wrong values.

The values of $\left\{\ Y_1, Y_2,\ldots,Y_m \right\}$ are some other values not given in that table. That table contains values X, D(X), r(X). Unless I'm mistaking, the first row with X=1, D(X)=0.0028 and r(X)=0.082 should be interpreted as follows:

For $X=1$, based on the definition of the function $D$ (which is not explicitly given), $D(X)=D(1)=0.0028$.
Then $r(X)=r(1)=\frac{\left\vert \left\{ Y_j \vert D(Y_j)\leq 0.0028, j=1,\ldots,m \right\} \right\vert}{m}=0.082$.

Table doesn't tell you how many Y's there are, nor what are their respective values, nor what are the values of D(Y). You just get the final result.
• Apr 16th 2012, 02:58 PM
mazz7755
Re: Help interpreting this equation
The values on the column X are 80 random values of a distribution(these values are not written on the table, just the number from 1 to 80 in order to represent them). With each of this values and another function I get the values in the column D(X). For this case in the equation we can regard X=Y, so the equation of the beginning can be written as $r_{Gm}(x)=\left \{ X_{j}|D_{Gm}(X_{j})\leq D_{Gm}(x),j=1,...,m \right \}/m$

Quote:

Originally Posted by MathoMan
For $X=1$, based on the definition of the function $D$ (which is not explicitly given), $D(X)=D(1)=0.0028$.
Then $r(X)=r(1)=\frac{\left\vert \left\{ Y_j \vert D(Y_j)\leq 0.0028, j=1,\ldots,m \right\} \right\vert}{m}=0.082$.

I agree with your interpretation, but then which values are this $Y_j \vert D(Y_j)$ ? The random values?
• Apr 16th 2012, 03:46 PM
MathoMan
Re: Help interpreting this equation
Maybe the values are repeated, but only counted once. Really don't know anything else that could help. Maybe someone else has a better idea. Its 2am here, and I'm off to bed. Nighty night! :)
• Apr 16th 2012, 08:19 PM
Bingk
Re: Help interpreting this equation
I'm not sure about this, but I think you have two distinct sets of numbers. One of them is the one represented by your first column. I'm not sure what the other set is, but it's probably the one you're asking about ( the $Y_j \vert D(Y_j)$). Now, from your interpretation, you may have tried using the same set of numbers for both, which is clearly not the case, because at least one of the values on the third column should be 1 (i.e. the $X_i$ that generates the biggest $D(X_i)$ should give $r(X_i)=1$.

Another thing to note is that most likely $m \neq 80$, because for most values on the 3rd column, when multiplied by 80, does not give you an integer (which it should be because that represents the cardinality of a set).

So, here's something to look into, you said that the X's you got are from a distribution. I'm guessing that the distribution has more than 80 values, and that the number of values it has should be your m, and that those values are the one's that represent the $Y_j \vert D(Y_j)$ that you're asking about.

Hope that's it :D