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Let \(\{X_{n}, n \ge 1\}\) be a sequence of independent identically distributed random variables with mean 0 and finite variance \(\sigma^{2} \gt 0\) :

Let \(S_{n} = \dfrac{X_{1} + X_{2} + \cdots + X_{n^{2}}}{n^{2}}\)

Then, which of the following is false ?

  • A.\(P\left[\lim_{n\to\infty} S_{n}e^{S_{n}} = 0\right] = 1\)
  • B.As \(n \to \infty\), \(S_{n}e^{S_{n}}\) converges to a normal random variable with mean '0' and variance \(\sigma^{2}\)
  • C.As \(n \to \infty\), \(S_{n}\) converges to '0' in probability
  • D.As \(n \to \infty\), \(e^{S_{n}}\) converges to 1 with probability 1

Answer: B

The false statement is \(S_{n}e^{S_{n}}\) converges to a normal random variable with mean 0 and variance \(\sigma^{2}\).

\(S_{n}\) is the average of the first \(n^{2}\) terms of an iid sequence with mean 0, so by the strong law of large numbers \(S_{n} \to 0\) almost surely. By continuity, \(S_{n}e^{S_{n}} \to 0\) and \(e^{S_{n}} \to 1\) almost surely, and a.s. convergence implies convergence in probability. That makes A, C and D true.

A degenerate limit 0 is not normal with variance \(\sigma^{2} \gt 0\). (A normal limit would appear only after rescaling, for \(\sqrt{n^{2}}\,S_{n} = nS_{n}\).)

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