I'm new to Mathematica and would like to show that for $x_1, ..., x_k$ independent, standard normal random variables the variance of the sum of their squares, i.e., $var(\sum_{i=1}^kx_k^2)$, is equal to $2k$.

Therefore, I implemented the following:

x= NormalDistribution[mx,stdx];

Variance[TransformedDistribution[Sum[a*a,{i,i_max}], {Distributed[a, x]}]];

The result is not 2i_max. However, the output is:

2(2mx^2stdx^2i_max^2 + stdx^4i_max^2)

What am I doing wrong?

  • $\begingroup$ i_max is not a valid variable name. Use imax, instead. $\endgroup$
    – bbgodfrey
    Nov 20 '21 at 22:36
  • $\begingroup$ The result is the same only with imax replaced by i_max. $\endgroup$
    – user82859
    Nov 21 '21 at 1:40
  • $\begingroup$ Your code only has a single random variable ($a$) so the result is correct for that. You are essentially getting the variance of $imax * a$. What you probably intend (because you state the answer is $2k$) is that you have $k$ independent and identically distributed random variables. Therefore you should use a[i] instead of a. $\endgroup$
    – JimB
    Nov 21 '21 at 1:47
  • $\begingroup$ Then, the result is 0. Or do I also need to change something inside “Distributed“ ? $\endgroup$
    – user82859
    Nov 21 '21 at 1:58
  • 1
    $\begingroup$ var[k_] = FindSequenceFunction[seq, k] where seq is the sequence provided by @JimB Alternatively, you can do a proof by induction. $\endgroup$
    – Bob Hanlon
    Nov 21 '21 at 3:00

Besides the ways given in the comments a more direct symbolic approach is to use moment generating functions (or characteristic functions). One obtains the mgf for the square of a unit normal and the mgf of the sum of n independent and identically distributed random variables is the individual mgf raised to the n-th power. Then find the variance from that resulting mgf.

(* Determine moment generating function for a single random variable *)
mgf = MomentGeneratingFunction[TransformedDistribution[x^2, x \[Distributed] NormalDistribution[0, 1]], t];

(* First raw moment of the sum of n iid random variables *)
m1 = (D[mgf^n, t]) /. t -> 0;

(* Second raw moment *)
m2 = (D[mgf^n, {t, 2}]) /. t -> 0;

(* Find variance *)
variance = m2 - m1^2 // Expand
(* 2 n *)
  • $\begingroup$ Thanks a lot! That works. I'd like to calculate now the variance of $\sum_{i=1}^k x_i(x_i+y_i)$ with $k$ independent and identically distributed random variables $x_i, y_i$. Therefore, I changed your first line to: mgf = MomentGeneratingFunction[TransformedDistribution[x*(x+y), { x \[Distributed] NormalDistribution[0, s1],y \[Distributed] NormalDistribution[0, s2]}], t];. Is that correct? The solution is, unfortunately, not such a short expression and contains different terms with "MomentGeneratingFunction". $\endgroup$
    – user82859
    Nov 21 '21 at 8:11
  • $\begingroup$ @user82859 - your comment is not a clarification of your original question but rather a whole new question. This should be posted as a new and separate question. You should link back here for context. You should also upvote this answer and, since it apparently answers your original question, accept the answer. $\endgroup$
    – Bob Hanlon
    Nov 21 '21 at 15:04
  • $\begingroup$ Thanks, Bob, you’re right! I upvoted your answer. Thanks again, this really helped. I try to adjust to the stackexchange standards as fast as possible. :-) $\endgroup$
    – user82859
    Nov 21 '21 at 15:11
  • $\begingroup$ This is @JimB answer and you have not upvoted it yet since the count only shows 1 which is mine. $\endgroup$
    – Bob Hanlon
    Nov 21 '21 at 16:16
  • $\begingroup$ It says “ Thanks for the feedback! You need at least 15 reputation to cast a vote, but your feedback has been recorded.” $\endgroup$
    – user82859
    Nov 21 '21 at 21:01

Using proof by induction



(* "12.3.1 for Mac OS X x86 (64-bit) (June 19, 2021)" *)

Each individual term of the sum of squares of i.i.d. standard normal variates (i.e., NormalDistribution[0, 1]) are distributed ChiSquareDistribution[1]

 x \[Distributed] NormalDistribution[]]

(* ChiSquareDistribution[1] *)

The problem is then just the sum of k i.i.d. variates each with distribution ChiSquareDistribution[1]

The distribution of the sum of two of these variates is

TransformedDistribution[z1 + z2,
  {z1 \[Distributed] ChiSquareDistribution[1],
   z2 \[Distributed] ChiSquareDistribution[1]}]

(* ChiSquareDistribution[2] *)

Consequently, assuming that the sum of k - 1 of these variates is distributed ChiSquareDistribution[k - 1] then the sum of k variates would be

TransformedDistribution[z1 + z2,
 {z1 \[Distributed] ChiSquareDistribution[k - 1],
  z2 \[Distributed] ChiSquareDistribution[1]}]

(* ChiSquareDistribution[k] *)

This is consistent with the assumption; so if it is true for k - 1 it is also true for k. Since the assumption is already shown to be true for k = 2 then by induction it is true for all k >= 2

The Variance is then


(* 2 k *)
  • $\begingroup$ +1 This answer is more general (a very good thing) and doesn't require "special knowledge" about moment generating functions. $\endgroup$
    – JimB
    Nov 21 '21 at 17:42

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