R时间序列内插和特定值的外插

Question

I have daily values for 11 different yield curves, that is time series for 11 yield maturities (1yr, 2yr, 3yr, 4yr, 5yr, 7yr, 10yr, 15yr, 20yr, 25yr, 30yr) in the same period of time. Some of the yields in some days are missing (NAs) and I'd like to extrapolate their values knowing the value of the other yields at the same day. This should be done by a first linear interpolation of the available yields in a given day, and a successive extrapolation of the missing yields in the same day, using the maturity duration (1yr, 2yr, etc) as weight. For example, I have the following data set and I'd like to extrapolate the daily value for 5yr yield based on an interpolation of all available yilds at the same day:

Date      1     2     3     4  5  7     10    15    20    25 30
7/4/2007  9.642 9.278 8.899 NA NA 8.399 8.241 8.183 8.117 NA NA
7/5/2007  9.669 9.302 8.931 NA NA 8.44  8.287 8.231 8.118 NA NA
7/6/2007  9.698 9.331 8.961 NA NA 8.437 8.295 8.243 8.13  NA NA
7/9/2007  9.678 9.306 8.941 NA NA 8.409 8.269 8.214 8.092 NA NA
7/10/2007 9.65  9.283 8.915 NA NA 8.385 8.243 8.185 8.065 NA NA
7/11/2007 9.7   9.342 8.976 NA NA 8.445 8.306 8.249 8.138 NA NA
7/12/2007 9.703 9.348 8.975 NA NA 8.448 8.303 8.245 8.152 NA NA
7/13/2007 9.69  9.334 8.965 NA NA 8.439 8.294 8.24  8.145 NA NA
7/16/2007 9.683 9.325 8.964 NA NA 8.442 8.299 8.244 8.158 NA NA
7/17/2007 9.712 9.359 8.987 NA NA 8.481 8.33  8.277 8.192 NA NA
7/18/2007 9.746 9.394 9.018 NA NA 8.509 8.363 8.311 8.22  NA NA
...

Does anyone have suggestions on how to do it? Thanks.

Answer 1

This is one of the ways to build a linear model for each Date based on the available info you have and use it to predict/estimate the value at year 5.

Run the process step by step to see how it works. Check the estimations to make sure they make sense.

dt = read.table(text=
"Date      1     2     3     4  5  7     10    15    20    25 30
7/4/2007  9.642 9.278 8.899 NA NA 8.399 8.241 8.183 8.117 NA NA
7/5/2007  9.669 9.302 8.931 NA NA 8.44  8.287 8.231 8.118 NA NA
7/6/2007  9.698 9.331 8.961 NA NA 8.437 8.295 8.243 8.13  NA NA
7/9/2007  9.678 9.306 8.941 NA NA 8.409 8.269 8.214 8.092 NA NA
7/10/2007 9.65  9.283 8.915 NA NA 8.385 8.243 8.185 8.065 NA NA
7/11/2007 9.7   9.342 8.976 NA NA 8.445 8.306 8.249 8.138 NA NA
7/12/2007 9.703 9.348 8.975 NA NA 8.448 8.303 8.245 8.152 NA NA
7/13/2007 9.69  9.334 8.965 NA NA 8.439 8.294 8.24  8.145 NA NA
7/16/2007 9.683 9.325 8.964 NA NA 8.442 8.299 8.244 8.158 NA NA
7/17/2007 9.712 9.359 8.987 NA NA 8.481 8.33  8.277 8.192 NA NA
7/18/2007 9.746 9.394 9.018 NA NA 8.509 8.363 8.311 8.22  NA NA", header=T)


library(dplyr)
library(tidyr)


dt %>%
  gather(time, value, -Date) %>%                               # reshape dataset
  filter(!is.na(value)) %>%                                    # ignore NA values
  mutate(time = as.numeric(gsub("X","",time))) %>%             # get rid of the X created by importing data
  group_by(Date) %>%                                           # for each date
  do({model = lm(value~time, data=.)                              # build a linear model
      data.frame(pred = predict(model, data.frame(time=5)))})     # use model to predict at time = 5


# Source: local data frame [11 x 2]
# Groups: Date [11]
# 
#          Date     pred
#        (fctr)    (dbl)
# 1  7/10/2007 8.920932
# 2  7/11/2007 8.979601
# 3  7/12/2007 8.981383
# 4  7/13/2007 8.970571
# 5  7/16/2007 8.968542
# 6  7/17/2007 8.999584
# 7  7/18/2007 9.032026
# 8   7/4/2007 8.917645
# 9   7/5/2007 8.950605
# 10  7/6/2007 8.970669
# 11  7/9/2007 8.946661

I'm not suggesting that the linear model is the best fit, as I didn't spend time on checking that. But, you can use a quadratic model instead of a linear, which might give you a better estimation.

In case you want to check the model output and get info about the models you built and used for each Date you can do this:

library(dplyr)
library(tidyr)
library(broom)


dt %>%
  gather(time, value, -Date) %>%                               # reshape dataset
  filter(!is.na(value)) %>%                                    # ignore NA values
  mutate(time = as.numeric(gsub("X","",time))) %>%             # get rid of the X created by importing data
  group_by(Date) %>%                                           # for each date
  do({model = lm(value~time, data=.)                              # build a linear model
      tidy(model)})                                               # check model output


# Source: local data frame [22 x 6]
# Groups: Date [11]
# 
#         Date        term    estimate  std.error statistic      p.value
#       (fctr)       (chr)       (dbl)      (dbl)     (dbl)        (dbl)
# 1  7/10/2007 (Intercept)  9.29495818 0.19895389 46.719158 8.485928e-08
# 2  7/10/2007        time -0.07480530 0.01875160 -3.989275 1.043399e-02
# 3  7/11/2007 (Intercept)  9.34942937 0.19823019 47.164509 8.093526e-08
# 4  7/11/2007        time -0.07396561 0.01868339 -3.958897 1.075469e-02
# 5  7/12/2007 (Intercept)  9.35001022 0.20037595 46.662337 8.537618e-08
# 6  7/12/2007        time -0.07372537 0.01888563 -3.903781 1.136592e-02
# 7  7/13/2007 (Intercept)  9.33730855 0.19974786 46.745476 8.462114e-08
# 8  7/13/2007        time -0.07334758 0.01882643 -3.895989 1.145551e-02
# 9  7/16/2007 (Intercept)  9.33045446 0.19856561 46.989276 8.245272e-08
# 10 7/16/2007        time -0.07238243 0.01871501 -3.867615 1.178869e-02
# ..       ...         ...         ...        ...       ...          ...