Math problem: derivative of correlation coefficient

[DoubtingThomas29]

Thanks alot for the help. I really need it.

Anyways, the thing you have to know about the Avg is that it's only constant while calculating the coefficient. The set X is the result of feeding input data through an array of weight variables and an activation function. The input values are a constant set during training, but the weight values are not, so any change there would affect the value of the average of the entire set.

Here is the way I see it, the average of x is x bar, meaning, you have your data, you added them all up and divided by the number that were there. That gives you a fixed number that isn't changing, it suppose to be zero when you take the derivative, there is no way you could make x bar a variable, X is the variable, little x is suppose to be vlaues of the variable. Here I'll do a calculation and put what I got up here, but I don't see why you would be wrong, becaue you just have to follow the rules for taking the derivative.

 

[DoubtingThomas29]

Note to Mods: Sorry if this is in the wrong section. There doesn't seem to be any better place to post in on the forums but I really need help so I'm posting anyways.

I need to find the partial derivative with regard to x for a normal correlation coefficient function. The function is:




C(x,y) =

Sum[ (x - Avg(x)) * (y - Avg(y)) ] ​

Sqr[ Sum[ x - Avg(x) ]^2 ] * Sqr[ Sum[ y - Avg(y) ]^2 ]​

What I have so far is:

Numerator​

let u = Sum[ (x - Avg(x)) * (y - Avg(y)) ]​

partial derivative of numerator: sum rule, constant multiple, chain rule​

du/dx = Sum[ (y - Avg(y)) * (x' - dAvg(x)/dx) ]​

denominator​

let v = Sqr[ Sum[ x - Avg(x) ]^2 ] * Sqr[ Sum[ y - Avg(y) ]^2 ]​

partial derivative of denominator: constant multiple, chain rule, sum rule​

dv/dx = (1/2) * Sqr[ Sum[ y - Avg(y) ]^2 ] * (Sum[ x - Avg(x) ]^2)^(-1/2) * 2* Sum[ x - Avg(x) ] * Sum[ x' - dAvg(x)/dx ]​

I think you are not far off really, I believe the sqr[Sum[Y - Avg(y)] would be in the top not the bottom because sqr(x)*c' = x^(-1/2) *c The c doesn't go to the bottom.

With the sum[x' - davg(x)/dx] I think you don't need the chain rule there.

Sum[x - Avg(x)]^2' = 2Sum[x - Avg(x)]dx

Because the derivative is distrubting over addition, and you are taking the derivative of directly a function of x, the squaring function of x, so you stop right there and add up all the x - avg(x), since they are all being multiplied by the derivate of x, which is one really.

[quote]
final partial derivative: quotient rule




dC/dx =

du/dx*v - u*dv/dx ​

v^2​

[/quote]

Alright you are going to have to ask your professor about this to be sure. But those are the only suggestions I would make. Good luck, and study for actuary exams, pass two or four of them, before you finish your education. Actuarys have a good job.

 

[Sojourner<><]

is this...

2shared.com/file/3231662/140d231c/CXY.html

you c(x,y)?

put 3 w's if front of the link...I cant post a link since i have fewer than 15 posts....ug

The link on that page isn't working.

 

[graph]

what do you get when you compute


d/dx [avg(x)]

this makes no sense to me.


what is your avg(x)?

is it ...writing it in LaTeX form...


\sum_{i=1}^n\frac{x_i}{n}

"the sum of the observations divided by the sample size"

?

Is x some function of another variable? like x=x(t)

is x a function from R to R^n

x:R->R^n

like x(t)={x_1(t),...,x_n(t)}

?

 

[Sojourner<><]

what do you get when you compute


d/dx [avg(x)]

this makes no sense to me.


what is your avg(x)?

is it ...writing it in LaTeX form...


\sum_{i=1}^n\frac{x_i}{n}

"the sum of the observations divided by the sample size"

?

Is x some function of another variable? like x=x(t)

is x a function from R to R^n

x:R->R^n

like x(t)={x_1(t),...,x_n(t)}

?

Yes, X is a function.

OK, here's what I think the issue is...

For the purposes of my question in the OP only, I'm looking for the partial derivatives dC/dx (to keep things simple). In this case, the extent to which I am calculating the derivatives almost looks unneeded. For example, the derivative of n^2 equals 2nn', but in most cases n' = 1 so its usually just 2n.

IRL though what I really need is dC/dw, w being a variable within the function represented by X. So for example again, if n = f(w), the derivative of n^2 equals 2n(df(w)/dw). This is why I need to see the full partial derivative, even the the little bits that are equivalent to 1.

As far as it applies to the Avg of X, as w changes, the set of X changes and therefore the average would also change.

Does that answer your question?

 

[graph]

Yes, X is a function.

OK, here's what I think the issue is...

For the purposes of my question in the OP only, I'm looking for the partial derivatives dC/dx (to keep things simple). In this case, the extent to which I am calculating the derivatives almost looks unneeded. For example, the derivative of n^2 equals 2nn', but in most cases n' = 1 so its usually just 2n.

IRL though what I really need is dC/dw, w being a variable within the function represented by X. So for example again, if n = f(w), the derivative of n^2 equals 2n(df(w)/dw). This is why I need to see the full partial derivative, even the the little bits that are equivalent to 1.

As far as it applies to the Avg of X, as w changes, the set of X changes and therefore the average would also change.

Does that answer your question?

I think we are getting closer.

since you are taking sums and averages I would think that X represents more than just one function. I would think that it represents a sequence of functions..otherwise what is there to sum over or take an average.

So do we have functions X1(w),X2(w),...,Xn(w)?

then avg(X(w))=sum{i=1 to n}Xi(w)/n

and d[avg(X(w))]/dw=sum{i=1 to n}[dXi(w)/dw]/n

 

[JonF]

C(x,y) =

Sum[ (x - Avg(x)) * (y - Avg(y)) ] ​

Sqr[ Sum[ x - Avg(x) ]^2 ] * Sqr[ Sum[ y - Avg(y) ]^2 ]​

What I have so far is:

Numerator​

let u = Sum[ (x - Avg(x)) * (y - Avg(y)) ]​

partial derivative of numerator: sum rule, constant multiple, chain rule​

du/dx = Sum[ (y - Avg(y)) * (x' - dAvg(x)/dx) ]​

denominator​

let v = Sqr[ Sum[ x - Avg(x) ]^2 ] * Sqr[ Sum[ y - Avg(y) ]^2 ]​

partial derivative of denominator: constant multiple, chain rule, sum rule​

dv/dx = (1/2) * Sqr[ Sum[ y - Avg(y) ]^2 ] * (Sum[ x - Avg(x) ]^2)^(-1/2) * 2* Sum[ x - Avg(x) ] * Sum[ x' - dAvg(x)/dx ]​

Yoru way is an ugly way:

do a change of variables and z = x-av(x),

note that z is a funciton of x.

take the derivative (you will end up with a ordinary DE from the chain rule) and back substitute

also do you know in general what d(sum[f(x)])/dx is equal to, for unconditional convergent series (if this is a conditionally convergent series we may have deeper issues than you'd think with this question)

edit:
at one point you have your denominator as: Sqr[ Sum[ x - Avg(x) ]^2 ] and in another post it is Sqr[ Sum[ x - Avg(x)^2 ]]. If it is the second (and it probably is) this is still a simple question, instead of doing a substitution you simply need to remember the avg of a function over [a,b] is 1/(b-a)int{a,b}(f(x)dx) and your FTC