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Mathematics Asked by M1996rg on August 8, 2020

For a special case of the gamma distribution:

$$f(x)=frac{x}{theta^2}e^{-x/ theta}$$

$$E(x) = 2theta\ V(x) = 2theta^2$$

I find MLE of $hat{theta}$ to be $sum_{i=1}^n frac{x_i}{2n}$

To show consistency, I’d like to show that:

  • $lim_{nto infty} E(hat{theta}) = theta$
  • $lim_{nto infty} V(hat{theta}) = 0$

Focusing on $lim_{nto infty} V(hat{theta}) = 0$:

$V(sum_{i=1}^n frac{x_i}{2n}) = frac{1}{4n^2}V(sum_{i=1}^n x_i)$

I’m somewhat stuck here, it seems obvious that this limit approaches, $0$, however, I’m not sure if I need to break the variance further (say $V(x) = E(x^2)-E(x)^2$) to prove this

I’d appreciate any help!

One Answer

I assume that $x_1,dots,x_n$ are independent, resulting in: begin{align*} lim_{n to infty} V(hat{theta}) = lim_{n to infty} V(sum_{i = 1}^n frac{x_i}{2n}) = lim_{n to infty} frac{1}{4n^2}sum_{i = 1}^n V(x_i) = lim_{n to infty} frac{1}{4n^2}sum_{i = 1}^n 2theta^2 = lim_{n to infty} frac{theta^2}{2n} = 0 end{align*}

Correct answer by delivery101 on August 8, 2020

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