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Part BCSIR NET June 2025the-two-cells-proportional-to-theta-pool-into-a-single-exponent

The two cells proportional to theta pool into a single exponent

Consider a discrete random variable X with the probability mass function P(X=0)=θ/3,P(X=1)=1−θ/2,P(X=2)=θ/6P(X = 0) = \theta/3, P(X = 1) = 1 - \theta/2, P(X = 2) = \theta/6, where θ∈(0,1)\theta \in (0, 1) is an unknown parameter. In a random sample of size 90 from this distribution, the observed counts for X = 0, 1 and 2 are 20, 60 and 10, respectively. Then, the maximum likelihood estimate of θ\theta is

  1. A.1/3
  2. B.1/2
  3. C.2/3✓ Correct
  4. D.3/4

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 execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: The MLE is unbiased

More on this topic

The chapter behind this: Maximum likelihood and the method of moments — free to read

From Estimation › MLE and method of moments

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