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Part CCSIR NET June 2025the-variance-of-x-squared-is-the-fourth-moment-minus-1-not-the-fourth-moment

The variance of x squared is the fourth moment minus 1 not the fourth moment

Let X1,X2,…X_{1}, X_{2}, \dots be a sequence of independent and identically distributed random variables with E(X1)=0,E(X12)=1,E(X13)=0,E(X14)=3E(X_{1}) = 0, E(X_{1}^{2}) = 1, E(X_{1}^{3}) = 0, E(X_{1}^{4}) = 3. Let Sn=∑iS_{n} = \sum_{i}₌1nXi,Tn=∑i_{1}^{n} X_{i}, T_{n} = \sum_{i}₌1nXi2,Un=∑i_{1}^{n} X_{i}^{2}, U_{n} = \sum_{i}₌1nXi3_{1}^{n} X_{i}^{3} and Vn=∑iV_{n} = \sum_{i}₌1nXi4_{1}^{n} X_{i}^{4}. Then, which of the following statements are true?

  1. A.Sn/nS_{n}/\sqrt{n} converges in distribution to a random variable Z, where Z ~ N(0, 1)✓ Correct
  2. B.(Tn−n)/3n(T_{n} - n)/\sqrt{3n} converges in distribution to a random variable Z, where Z ~ N(0, 1)
  3. C.nSn/Tn\sqrt{n} S_{n}/T_{n} converges in distribution to a random variable Z, where Z ~ N(0, 1)✓ Correct
  4. D.(Tn−n)/Vn(T_{n} - n)/\sqrt{V}_{n} converges in distribution to a random variable Z, where Z ~ N(0, 1)

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests moments and tails.

See pricing

50 are analysed free — try those first.

The trap it tests

Moments and tails

A moment assumed to exist, or tail behaviour assumed to be tame.

Drill statements like this

Related counterexample: Convergence in probability implies almost sure convergence

More on this topic

The chapter behind this: Modes of convergence and the limit theorems — free to read

From Limit Theorems and Markov Chains › Modes of convergence, WLLN, SLLN, CLT

Last revised . Found a mistake? Tell us — corrections are the fastest thing we act on.

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