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$(7.3, 10.8, 6.5, 10.3, 15.1)$ is an observed sample of size $n = 5$ from a population with the probability density function $$f_{\theta}(x) = \frac{x^3}{6 \theta^4} e^{\frac{-x}{\theta}}, x > 0, \theta > 0$$

MLE is $\hat\theta = \frac{\bar{x}}{4}$

Calculate the standard error of the estimate.

solution:

$X$ ~ $Gamma(a = 4, \lambda = \frac{1}{\theta})$

$E(X) = \frac{a}{\lambda} = 4 \theta$. $V(X) = \frac{a}{\lambda^2} = 4 \theta^2$

$V(\hat\theta) = V(\frac{\bar{X}}{4}) = \frac{V(X)}{16n}$ (I don't get how they got to this. I understand the constant part since $V(aX) = a^2V(X)$. But I don't understand how they got rid of $\bar{X}$ and got the n in the dnominator).

$\frac{V(X)}{16n} = \frac{4 \theta^2}{16n} = \frac{\theta^2}{4n}$.

So SE(X) = $\sqrt{\frac{2.494^2}{20}}$. (I don't get how they got $\theta = 2.494$. I know $n=5$ because five samples though)

Could someone answer my confusions thank you

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I guess it is assumed the observations to be iid:

\begin{align} V \left ( \frac{\overline{X}}{4} \right ) &= \frac{1}{16}V(\overline{X}) = \frac{1}{16}V \left ( \frac{1}{n}\sum_{i = 1}^{n}X_{i} \right ) = \frac{1}{16n^{2}}V\left ( \sum_{i = 1}^{n}X_{i} \right ) = \frac{1}{16n^{2}}\left ( \sum_{i = 1}^{n}V(X_{i}) \right ) \\ &= \frac{1}{16n^{2}} \left ( nV(X_{i}) \right ) = \frac{1}{16n}V(X_{i}) \end{align}

Additionally, I would estimate $\theta$ remembering that the MLE estimator for the expected value is the mean, and that it has the invariance property:

$$\overline{X} \ MLE \ for \ E(X) = 4\theta \Rightarrow \frac{\overline{X}}{4} \ MLE \ for \ \frac{1}{4}E(X) = \theta$$

and, in my computations, $\frac{\overline{X}}{4}$= $2.500$. Maybe, when doing the computation, the values in the sample are actually not approximated up to the first digit.

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