Moving average in R

Question

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?

Answer 1

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).

Answer 2

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])
...