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Let $Y_1, \ldots , Y_n$ be a random sample from the uniform distribution on the interval $(0, \theta)$ with an unknown $\theta > 1$.

Suppose we only observe for $i = 1, \ldots , n$

$$X_i= \begin{cases} Y_i & \text{if } Y_i \leq 1, \\ 1 & \text{if } Y_i > 1. \end{cases}$$

Find the Method of Moments estimator of $\theta$ and derive its asymptotic distribution

So I found the expected value like usual when finding an MME, however I was unable to solve for $\theta.$ This is what I got:

$$ \operatorname{E}(X_i) = \frac{\theta-1} 2 +\frac 1 \theta$$

is my $\operatorname{E}(X_i)$ wrong? Because I can't seem to solve for $\theta$ when equating it to the sample mean

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You have $$ X_i= \begin{cases} Y_i & \text{if } Y_i \leq 1, \\ 1 & \text{if } Y_i > 1. \end{cases} $$ Let $$ I_i= \begin{cases} 0 & \text{if } Y_i \leq 1, \\ 1 & \text{if } Y_i > 1. \end{cases} $$ Then $$ \operatorname{E}(X_i \mid I_i) = \left. \begin{cases} \dfrac 1 2 & \text{if } I_i=0 \\[6pt] 1 & \text{if } I_i=1 \end{cases} \right\} = \left. \begin{cases} \dfrac 1 2 & \text{with probability } \dfrac 1 \theta \\[6pt] 1 & \text{with probability } 1 - \dfrac 1 \theta \end{cases} \right\}. $$ Therefore $$ \operatorname{E}(X_i) = \operatorname{E}(\operatorname{E}(X_i\mid I_i)) = \frac 1 2 \cdot \frac 1 \theta + 1 \cdot \left( 1 - \frac 1 \theta \right) = \frac {1 + (2\theta - 2)} {2\theta} = \frac{2\theta -1}{2\theta} = 1 - \frac 1 {2\theta}. $$ Setting the sample mean equal to that, we have $$ \bar X = 1 - \frac 1 {2\theta} $$ $$ \frac 1 {2\theta} = 1-\bar X $$ and $$ \theta = \frac 1 {2(1-\bar X)}. $$

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    $\begingroup$ Sorry I must have typed it out wrong, its supposed to read: $$ X_i= \begin{cases} Y_i & \text{if } Y_i \geq 1, \\ 1 & \text{if } Y_i < 1. \end{cases} $$ however the only thing I really need help with is: $E(X_i | Y_i \geq 1)$ would it be $(1+\theta)/2$ $\endgroup$ – Alex Chavez Nov 24 '16 at 7:18
  • $\begingroup$ @AlexChavez : That is correct. Except that I would write $\operatorname{E}(X_i\mid Y_i<1)$ rather than $\operatorname{E}(X_i|Y_i<1)$, with \mid rather than a simple solidus. You can see that those look different. $\qquad$ $\endgroup$ – Michael Hardy Nov 24 '16 at 15:37
  • $\begingroup$ Would you know anything about the asymptotic distribution ? $\endgroup$ – Alex Chavez Nov 24 '16 at 16:39
  • $\begingroup$ @AlexChavez : The central limit theorem should give you the asymptotic distribution of $\overline X,$ and then the delta method should find the asymptotic distribution of $1/(2(1-\overline X). \qquad$ $\endgroup$ – Michael Hardy Nov 25 '16 at 15:36
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Your $E[X_i]$ is indeed wrong

  • $Y_i \gt 1$ with probability $\frac{\theta-1}{\theta}$ and $E[X_i \mid Y_i \gt 1]=E[1 \mid Y_i \gt 1]=1$
  • $Y_i \le 1$ with probability $\frac{1}{\theta}$ and $E[X_i \mid Y_i \le 1]=E[Y_i \mid Y_i \le 1]=\frac12$
  • So $E[X_i]= \frac{\theta-1}{\theta}+\frac{1}{2\theta}=1-\frac{1}{2\theta}$

This will lead to an estimator for $\theta$ of $\dfrac{1}{2(1-\overline{X})}$

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  • $\begingroup$ Sorry I must have typed it out wrong, its supposed to read: $$ X_i= \begin{cases} Y_i & \text{if } Y_i \geq 1, \\ 1 & \text{if } Y_i < 1. \end{cases} $$ however the only thing I really need help with is: $E(X_i | Y_i \geq 1)$ would it be $(1+\theta)/2$ $\endgroup$ – Alex Chavez Nov 24 '16 at 7:18
  • $\begingroup$ Yes it would. You would then have $E[X_i]=\dfrac{\theta^2+1}{2\theta}$ leading to an estimator for $\theta$ of $\overline{X}+\sqrt{\overline{X}^2-1}$ $\endgroup$ – Henry Nov 24 '16 at 8:50
  • $\begingroup$ would you know how to find the asymptotic distribution of that ? $\endgroup$ – Alex Chavez Nov 24 '16 at 8:57

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