Skip to content
Part CCSIR NET December 2025check-what-each-piece-actually-converges-to-before-trusting-the-ratio-simplifies-to-theta

Check what each piece actually converges to before trusting the ratio simplifies to theta

Let X1,X2,…,Xn(n≥2)X_{1}, X_{2}, \dots, X_{n} (n\ge2) be a random sample from an Exponential distribution with the probability density function f(x∣μ,σ)=(1/σ)exp⁡((μ−x)/σ)\mu,\sigma) = (1/\sigma)\exp((\mu-x)/\sigma) if x>μ,0x>\mu, 0 otherwise, where parameters μ\mu and σ\sigma are unknown and positive. Let Xˉn,Sn2{\bar{X}}_{n}, S_{n}^{2} and X1:nX_{1}:_{n} denote the sample mean, the sample variance and the sample smallest order statistic, respectively, and let θ=σ/μ\theta = \sigma/\mu. Then which of the following statements are true?

  1. A.Sn(X1:n)S_{n}(X_{1}:_{n})⁻1^{1} is a consistent estimator of θ\theta✓
  2. B.(Xˉn−X1:n)(X1:n)({\bar{X}}_{n} - X_{1}:_{n})(X_{1}:_{n})⁻1^{1} is a consistent estimator of θ\theta✓
  3. C.(Xˉn−X1:n)(Xˉn−Sn)({\bar{X}}_{n} - X_{1}:_{n})({\bar{X}}_{n} - S_{n})⁻1^{1} is a consistent estimator of θ\theta✓
  4. D.Sn(Xˉn)S_{n}({\bar{X}}_{n})⁻1^{1} is a consistent estimator of θ\theta

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.

See pricing

50 are analysed free — try those first.

Related counterexample: A sufficient statistic is complete

More on this topic

The chapter behind this: Sufficiency, completeness and the UMVUE machine — free to read

From Estimation › Sufficiency, completeness, UMVUE, Cramér–Rao

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

ShareWhatsAppTelegram