Distribution of sum of i.i.d random variables

Matonya

Matonya

Open question

2022-08-18

Assume that we have a sequence of n realizations x 1 , x 2 , . . . , x n of an i.i.d random variable X with cdf F X ( x ) and pdf f X ( x ) . Now define Y as the sum of k consecutive realizations of X
Y i = j = i k x j
which generates the sequence Y 1 , . . . , Y n k + 1 where Y i contains k 1 elements of Y i 1 (and Y i + 1 ).
What is the disribution F Y ( x ) of Y?

Answer & Explanation

Brogan Navarro

Brogan Navarro

Beginner2022-08-19Added 24 answers

Step 1
I will denote Y k = j = i k X j , as the dependence on k is what matters.
As you said, the pdf f Y k ( x ) is given by the k-fold convolution of f Y k ( x ) . You can prove this using the characteristic function of Y k , ϕ Y k ( t ) = e i t Y k Y k = e i t j = i k X j X i , , X k + i
You don't need to distinguish these two cases, as the pdf of Y k depends only on k (not on n). In fact, you can prove that the distribution of Y k does not depend on i. For large k, you have a central limit theorem for Y k .
That is true also for finite n

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