Moving average in R-CodePudding

So I learn to calculate moving averages in R with this snippet of code

# just to create a distribution
x <- x <- rnorm(n=100, mean = 0, sd = 10)

mn <- function(n) rep (1/n,n)

filter(x, mn(5))

When a plot only the result of mn(5), i see that a get 1/5 repeated 5 times. Why using filter (x, mn(5)) calculates the average of the five values? Where are the part that the mean is calculated?

CodePudding user response：

1) The mean is the average of the values so assuming x has 5 elements we can write the second line and that is the same as the third line and the fourth line so using coefficients of (1/5, 1/5, 1/5, 1/5, 1/5) in the sum is equivalent to taking the mean.

mean(x) 
= (x[1]   x[2]   x[3]   x[4]   x[5])/5
= x[1]/5   x[2]/5   x[3]/5   x[4]/5   x[5]/5
= sum(x * c(1/5, 1/5, 1/5, 1/5, 1/5))

2) Another way to understand this is to note that mean is linear. That is if x and y are two vectors of the same length then mean(x y) = mean(x) mean(y) and if a is any scalar then mean(a * x) = a * mean(x). Now it is known that any linear function that returns a scalar is representable as the inner product of some vector times the input. That is there is a vector v such that

mean(x)
sum(v * x)

are equal for all x. Now since it is true for all x it must be true for x <- c(1, 0, 0, 0, 0) so these are equal

mean(c(1, 0, 0, 0, 0)
v[1] * x[1]

but the second line equals v[1] since x[1] is 1 and the mean of c(1, 0, 0, 0, 0) in the first line equals 1/5 and similarly for

mean(c(0, 1, 0, 0, 0))
v[2] * x[2]

etc. so v must equal c(1/5, 1/5, 1/5, 1/5, 1/5).

CodePudding user response：

Where are the part that the mean is calculated?

See ?filter for argument sides. The default value sides = 2 means "center". You probably want sides = 1 to use past values only.

y <- filter(x, c(1/5, 1/5, 1/5, 1/5, 1/5), sides = 1)

is doing

y[1:4] = NA
y[5] = (x[1]   x[2]   x[3]   x[4]   x[5]) / 5 = mean(x[1:5])
y[6] = (x[2]   x[3]   x[4]   x[5]   x[6]) / 5 = mean(x[2:6])
...