Paper 3, Section II, 27 K27 \mathrm{~K}

Principles of Statistics | Part II, 2011

Random variables X1,X2,…X_{1}, X_{2}, \ldots are independent and identically distributed from the exponential distribution E(θ)\mathcal{E}(\theta), with density function

pX(x∣θ)=θe−θx(x>0),p_{X}(x \mid \theta)=\theta e^{-\theta x} \quad(x>0),

when the parameter Θ\Theta takes value θ>0\theta>0. The following experiment is performed. First X1X_{1} is observed. Thereafter, if X1=x1,…,Xi=xiX_{1}=x_{1}, \ldots, X_{i}=x_{i} have been observed (i⩾1)(i \geqslant 1), a coin having probability α(xi)\alpha\left(x_{i}\right) of landing heads is tossed, where α:R→(0,1)\alpha: \mathbb{R} \rightarrow(0,1) is a known function and the coin toss is independent of the XX 's and previous tosses. If it lands heads, no further observations are made; if tails, Xi+1X_{i+1} is observed.

Let NN be the total number of XX 's observed, and X:=(X1,…,XN)\mathbf{X}:=\left(X_{1}, \ldots, X_{N}\right). Write down the likelihood function for Θ\Theta based on data X=(x1,…,xn)\mathbf{X}=\left(x_{1}, \ldots, x_{n}\right), and identify a minimal sufficient statistic. What does the likelihood principle have to say about inference from this experiment?

Now consider the experiment that only records Y:=XNY:=X_{N}. Show that the density function of YY has the form

pY(y∣θ)=exp⁡{a(y)−k(θ)−θy}p_{Y}(y \mid \theta)=\exp \{a(y)-k(\theta)-\theta y\}

Assuming the function a(⋅)a(\cdot) is twice differentiable and that both pY(y∣θ)p_{Y}(y \mid \theta) and ∂pY(y∣θ)/∂y\partial p_{Y}(y \mid \theta) / \partial y vanish at 0 and ∞\infty, show that a′(Y)a^{\prime}(Y) is an unbiased estimator of Θ\Theta, and find its variance.

Stating clearly any general results you use, deduce that

−k′′(θ)Eθ{a′′(Y)}⩾1.-k^{\prime \prime}(\theta) \mathbb{E}_{\theta}\left\{a^{\prime \prime}(Y)\right\} \geqslant 1 .

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