I have the following probability distribution function given by: \begin{equation} \label{eq:function} f(x) = \frac{4a}{x^5} \exp \left[ {- \frac{a}{x^4}} \right] \quad \quad 0 \leq x \leq \infty \quad \text{and} \quad a>0 \end{equation}

for which I have derived the following Quartiles, Moments and Estimates for the parameter $a$.

Quartiles: Let $F_X(z) = \exp \left[ {-\frac{a}{z^4}} \right]$.

1st Quartile:

Solve for $F_X(n_0) = 0.25$

\begin{equation} \implies n_0 = \left[ - \frac{a}{\log(0.25)}\right] ^\frac{1}{4} \approx (1.38629a)^\frac{1}{4} \end{equation}

2nd Quartile:

Solve for $F_X(n_0) = 0.5$

\begin{equation} \implies n_0 = \left[ - \frac{a}{\log(0.5)}\right] ^\frac{1}{4} \approx (0.693147a)^\frac{1}{4} \end{equation}

3rd Quartile:

Solve for $F_X(n_0) = 0.75$

\begin{equation} \implies n_0 = \left[ - \frac{a}{\log(0.75)}\right] ^\frac{1}{4} \approx (0.287682a)^\frac{1}{4} \end{equation}


\begin{equation} \mathbb{E} \left[X \right] = a^\frac{1}{4} \cdot \Gamma \left( \frac{3}{4} \right) \end{equation}

\begin{equation} \mathbb{E} \left[X^2 \right] = \left( a \pi \right) ^\frac{1}{2} \end{equation}

\begin{equation} \mathbb{E} \left[X^3 \right] = 4a^\frac{3}{4} \cdot \Gamma \left( \frac{5}{4} \right) \end{equation}

\begin{equation} \mathbb{E} \left[X^4 \right] = \int_{0}^{\infty} \frac{4a}{x} \exp \left[ {- \frac{a}{x^4}} \right] \end{equation}

where the 4th Moment does not converge on $[0,\infty]$ hence it does not exist.

Estimators Derived:

Estimator 1: Using 1st Moment

\begin{equation} \implies \hat{a}_{MM1} = \left[ \frac{\widehat{\mathbb{E} \left[X \right]}}{\Gamma \left( \frac{3}{4} \right)}\right]^4 = \left[ \frac{\bar{X}}{\Gamma \left( \frac{3}{4} \right)} \right]^4 = \left[ \frac{\frac{1}{n} \sum_{i=1}^\infty X_i}{\Gamma \left( \frac{3}{4} \right)} \right]^4 \end{equation}

Estimator 2: Using 2nd Moment

\begin{equation} \label{eq:Estimator_a_MoM} \implies \hat{a}_{MM2} = \frac{\left[ \widehat{\mathbb{E} \left[X^2 \right]} \right]^2}{\pi} = \frac{\left[ \frac{1}{n} \sum_{i=1}^\infty X_i^2 \right]^2}{\pi} \end{equation}

Estimator 3: Using the Median (2nd Quartile) \begin{equation} \label{eq:Estimator_a_Med} \implies \hat{a}_{MED} = - \left( n_0 \right)^4 \cdot \log(0.5) \end{equation}

where $n_0$ is the sample median.

My issue arises in finding the Bias and Variance of each of the above estimators as I am having trouble deriving these. In taking the Expected Value of each estimator to derive its Bias, I am getting confused because of the large powers, and the same applies for the variance. Any help in obtaining these is greatly appreciated.



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