计算R增长中的信用冲动

Part Finance Part R question

部分财务部分R问题

I have been trying to replicate the below formula in R using the Quantmod package and xts also using the diff function. The code gives me a plot for credit impulse but it doesn't seem to be replicating what I am trying to get at. See link

我一直在尝试使用Quantmod包复制下面的公式，xts也使用diff函数。代码给了我一个信贷冲动的情节，但它似乎并没有复制我想要的东西。见链接

https://www.gam.com/media/1434580/biggs.pdf -

page 2 give the formula for Credit Impulse - where C is the stock of credit at time t

第2页给出了信用冲动的公式 - 其中C是时刻t的信用存量

Credit impulset = (Ct – Ct-1)/GDPt – (Ct-1-Ct-2)/GDPt-1

信用冲击=（Ct-Ct-1）/ GDPt - （Ct-1-Ct-2）/ GDPt-1

page 3 take a look at the graph( This is the graph I am trying to replicate for Credit Impulse

第3页看一下图表（这是我试图为Credit Impulse复制的图表

Am I using the diff function the right way also could I be doing this more efficiently in R?

我是否以正确的方式使用diff功能也可以在R中更有效地使用它？

below is my code

下面是我的代码

#US DEBT [BN][USD][Q]
usd_debt <- getSymbols("CRDQUSAPABIS", src = "FRED", auto.assign=FALSE)

##US GDP [BN][USD][Q]
usd_gdp <- getSymbols("GDP", src = "FRED", auto.assign=FALSE)

#USD Credit Impulse
usd_debt <- usd_debt["2000/2016"] 
usd_gdp <- usd_gdp["2000/2016"]
usd_ratio <- usd_debt/usd_gdp
usd_ci <- diff(usd_ratio)
plot(usd_ci)

1 个解决方案

#1

Looks like you may actually want to use:

看起来你可能真的想要使用：

z <- diff(diff(usd_debt) / coredata(usd_gdp))
plot(z)

assuming Ct can be modeling using your usd_debt series?

假设Ct可以使用你的usd_debt系列进行建模？

Yes you're using diff the right way. diff will call diff.xts when you apply it to an xts object, and in your example usd_ratio is indeed an xts object, so it will be fast (efficient).

是的，你正在以正确的方式使用差异。当你将它应用到xts对象时，diff会调用diff.xts，在你的例子中，usd_ratio确实是一个xts对象，所以它会很快（高效）。

Here, coredata is optional but good practice when dividing by xts objects, as it returns the underlying matrix instead. Division for xts objects can be problematic.

在这里，coredata是可选的，但在划分xts对象时是很好的做法，因为它返回底层矩阵。 xts对象的划分可能有问题。

#1