[R] LM with summation function
Peter Ehlers
ehlers at ucalgary.ca
Wed May 23 03:37:15 CEST 2012
Robbie,
Here's what I *think* you are trying to do:
1.
y is a cubic function of x:
y = b1*x + b2*x^2 + b3*x^3
2.
s is the cumsum of y:
s_i = y_1 + ... + y_i
3.
Given a subset of x = 1:n and the corresponding
values of s, estimate the coefficients of the cubic.
If that is the correct understanding, then you should
be able to estimate the coefficients as follows:
a) since s_i = b1 * sum of x_k for k=1, ..., i
+ b2 * sum of (x_k)^2 for k=1, ..., i
+ b3 * sum of (x_k)^3 for k=1, ..., i
we can regress s on the cumsums of x, x^2 and x^3:
using your sample data:
d <- data.frame(x = c(1, 4, 9, 12),
s = c(109, 1200, 5325, 8216))
e <- data.frame(x = 1:12)
e <- merge(e, d, all.x = T)
e <- within(e,
{z3 <- cumsum(x^3)
z2 <- cumsum(x^2)
z1 <- cumsum(x)})
coef(lm(s ~ 0 + z1 + z2 + z3, data = e))
# z1 z2 z3
# 100 10 -1
Peter Ehlers
On 2012-05-22 09:43, Robbie Edwards wrote:
> I don't think I can.
>
> For the sample data
>
> d<- data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
>
> when x = 4, s = 1200. However, that s4 is sum of y1 + y2 + y3 + y4.
> Wouldn't I have to know the y for x = 2 and x = 3 to get the value of y
> for x = 4?
>
> In the previous message, I created two sample data frames. d is what I'm
> trying to use to create df. I only know what's in d, df is just used to
> illustrate what I'm trying to get from d.
>
> robbie
>
>
>
>
>
> On Tue, May 22, 2012 at 12:30 PM, R. Michael Weylandt<
> michael.weylandt at gmail.com> wrote:
>
>> But if I understand your problem correctly, you can get the y values
>> from the s values. I'm relying on your statement that "s is sum of the
>> current y and all previous y (s3 = y1 + y2 + y3)." E.g.,
>>
>> y<- c(1, 4, 6, 9, 3, 7)
>>
>> s1 = 1
>> s2 = 4 + s1 = 5
>> s3 = 6 + s2 = 11
>>
>> more generally
>>
>> s<- cumsum(y)
>>
>> Then if we only see s, we can get back the y vector by doing
>>
>> c(s[1], diff(s))
>>
>> which is identical to y.
>>
>> So for your data, the underlying y must have been c(109, 1091, 4125,
>> 2891) right?
>>
>> Or have I completely misunderstood your problem?
>>
>> Michael
>>
>> On Tue, May 22, 2012 at 12:25 PM, Robbie Edwards
>> <robbie.edwards at gmail.com> wrote:
>>> Actually, I can't. I don't know the y values. Only the s and only for a
>>> subset of the data.
>>>
>>> Like this.
>>>
>>> d<- data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
>>>
>>>
>>>
>>> On Tue, May 22, 2012 at 11:57 AM, R. Michael Weylandt
>>> <michael.weylandt at gmail.com> wrote:
>>>>
>>>> You can reconstruct the y values by taking first-differences of the s
>>>> vector, no? Then it sounds like you're good to go
>>>>
>>>> Best, Michael
>>>>
>>>> On Tue, May 22, 2012 at 11:40 AM, Robbie Edwards
>>>> <robbie.edwards at gmail.com> wrote:
>>>>> Hi all,
>>>>>
>>>>> Thanks for the replies, but I realize I've done a bad job explaining
>> my
>>>>> problem. To help, I've created some sample data to explain the
>> problem.
>>>>>
>>>>> df<- data.frame(x=c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12), y=c(109,
>>>>> 232,
>>>>> 363, 496, 625, 744, 847, 928, 981, 1000, 979, 912), s=c(109, 341, 704,
>>>>> 1200, 1825, 2569, 3416, 4344, 5325, 6325, 7304, 8216))
>>>>>
>>>>> In this data frame, y results from y = x * b1 + x^2 * b2 + x^3 * b3
>> and
>>>>> s
>>>>> is sum of the current y and all previous y (s3 = y1 + y2 + y3).
>>>>>
>>>>> I know I can find b1, b2 and b3 using:
>>>>> lm(y ~ 0 + x + I(x^2) + I(x^3), data=df)
>>>>>
>>>>> yielding...
>>>>> Coefficients:
>>>>> x I(x^2) I(x^3)
>>>>> 100 10 -1
>>>>>
>>>>> However, I need to find b1, b2 and b3 using the s column. The reason
>>>>> being, I don't actually know the values of y in the actual data set.
>>>>> And
>>>>> in the actual data, I only have a few of the values. Imagine this
>> data
>>>>> is
>>>>> being used a reward schedule for like a loyalty points program. y
>>>>> represents the number of points needed for each level while s is the
>>>>> total
>>>>> number of points to reach that level. In the real problem, my data
>>>>> looks
>>>>> more like this:
>>>>>
>>>>> d<- data.frame(x=c(1, 4, 9, 12), s=c(109, 1200, 5325, 8216))
>>>>>
>>>>> Where I need to use a few sample points to help define the parameters
>> of
>>>>> the curve.
>>>>>
>>>>> thanks again and hopefully this makes the problem a bit clearer.
>>>>>
>>>>> robbie
>>>>>
>>>>>
>>>>>
>>>>> On Fri, May 18, 2012 at 7:40 PM, David Winsemius
>>>>> <dwinsemius at comcast.net>wrote:
>>>>>
>>>>>>
>>>>>> On May 18, 2012, at 1:44 PM, Robbie Edwards wrote:
>>>>>>
>>>>>> Hi all,
>>>>>>>
>>>>>>> I'm trying to model some data where the y is defined by
>>>>>>>
>>>>>>> y = summation[1 to 50] B1 * x + B2 * x^2 + B3 * x^3
>>>>>>>
>>>>>>> Hopefully that reads clearly for email.
>>>>>>>
>>>>>>>
>>>>>> cumsum( rowSums( cbind(B1 * x, B2 * x^2, B3 * x^3)))
>>>>>>
>>>>>>
>>>>>>
>>>>>> Anyway, if it wasn't for the summation, I know I would do it like
>> this
>>>>>>>
>>>>>>> lm(y ~ x + x2 + x3)
>>>>>>>
>>>>>>> Where x2 and x3 are x^2 and x^3.
>>>>>>>
>>>>>>> However, since each value of x is related to the previous values of
>> x,
>>>>>>> I
>>>>>>> don't know how to do this. Any help is greatly appreciated.
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>
>>>>>> David Winsemius, MD
>>>>>> West Hartford, CT
>>>>>>
>>>>>>
>>>>>
>>>>> [[alternative HTML version deleted]]
>>>>>
>>>>> ______________________________________________
>>>>> R-help at r-project.org mailing list
>>>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>>>> PLEASE do read the posting guide
>>>>> http://www.R-project.org/posting-guide.html
>>>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>>
>>
>
> [[alternative HTML version deleted]]
>
> ______________________________________________
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