<s>[INST] Computation of maximum likelihood estimates Question
Solve for the maximum likelihood estimates of the distribution with parameters $\alpha$ and $\beta$ and the following pdf, where $x > 0$: $$f(x; \alpha, \beta) = \frac {\alpha} {\sqrt {2\pi \beta}} x^{-\frac 3 2} \exp\left[-\frac {(\alpha - \beta x)^2} {2\beta x}\right].$$
Hint: You should be able to write the parameter estimates in terms of $\overline{x}$ and $\overline{1/x}$, where $\overline{1/x} = \frac 1 n \sum^n_{i = 1} \frac 1 {x_i}$.
My working
$$\begin{aligned}
l(\alpha, \beta) & = n\ln\frac {\alpha} {\sqrt{2\pi \beta}} - \frac 3 2 \sum^n_{i = 1} \ln x_i - \sum^n_{i = 1} \frac {(\alpha - \beta x_i)^2} {2\beta x_i}\\
& = n\ln \alpha - \frac 1 2 n \ln(2\pi) - \frac 1 2 n \ln \beta - \frac 3 2 \sum^n_{i = 1} \ln x_i - \sum^n_{i = 1} \frac {(\alpha - \beta x_i)^2} {2\beta x_i}\\
\implies \frac {\mathrm{d}l} {\mathrm{d}\alpha} & = \frac n {\alpha} - \sum^n_{i = 1} \frac {2(\alpha - \beta x_i)} {2\beta x_i}\\
& = n \left(\frac 1 {\alpha} + 1\right) - \sum^n_{i = 1} \frac {\alpha} {\beta x_i}\\
& = n \left(\frac 1 {\alpha} + 1\right) - \frac {n\alpha} {\beta} \overline{1/x}\\
& = n\left(\frac 1 {\alpha} + 1 - \frac {\alpha} {\beta} \overline{1/x}\right)\\
\frac {\mathrm{d}l} {\mathrm{d}\beta} & = -\frac n {2\beta} - \sum^n_{i = 1} \frac {2(\alpha - \beta x_i)(-x_i)(2\beta x_i) - (\alpha - \beta x_i)^2(2x_i)} {4\beta^2 x^2_i}\\
& = -\frac n {2\beta} + \sum^n_{i = 1} \frac {(\alpha - \beta x_i)(2\beta x_i) + (\alpha - \beta x_i)^2} {2\beta^2 x_i}\\
& = -\frac n {2\beta} + \sum^n_{i = 1} \frac {(\alpha - \beta x_i)(\alpha + \beta x_i)} {2\beta^2 x_i}\\
& = -\frac n {2\beta} + \sum^n_{i = 1} \frac {(\alpha^2 - \beta^2 x^2_i)} {2\beta^2 x_i}\\
& = -\frac n {2\beta} + \frac {n\alpha^2} {2\beta^2} \overline{1/x} - \frac n 2 \overline{x}\\
& = \frac n 2 \left(\frac {\alpha^2} {\beta^2} \overline{1/x} - \frac 1 {\beta} - \overline{x}\right)
\end{aligned}$$
When
$$\begin{aligned}
\frac {\mathrm{d}l} {\mathrm{d}\alpha} & = 0,\\
\frac {\beta(\alpha + 1)} {\alpha^2} & = \overline{1/x}.
\end{aligned}$$
When
$$\begin{aligned}
\frac {\mathrm{d}l} {\mathrm{d}\beta} & = 0,\\
\alpha^2 \overline{1/x} - \beta & = \beta^2 \overline{x}.
\end{aligned}$$

Assuming my score equations are correct and I have not gone wrong anywhere, I am now stuck here. In particular, I am unable to see how I can further manipulate the two equations above to return just $\alpha$ and $\beta$.
Any intuitive explanations will be greatly appreciated :)
 [/INST] Your calculations look right to me.
You can solve the first equation for $\beta$:
$$
\beta=\frac{\alpha^2}{\alpha+1}\overline{1/x}\;.
$$
Substituting that into the second equation yields
$$
\alpha^2\overline{1/x}-\frac{\alpha^2}{\alpha+1}\overline{1/x}=\left(\frac{\alpha^2}{\alpha+1}\overline{1/x}\right)^2\overline x\;.
$$
Now multiply by $\frac{\alpha+1}{\alpha^2\overline{1/x}}$ (i.e. divide by $\beta$) to obtain
$$
\alpha+1-1=\frac{\alpha^2}{\alpha+1}\overline{1/x}\cdot\overline x\;.
$$
Multiply by $\frac{\alpha+1}\alpha$:
$$
\alpha+1=\alpha\overline{1/x}\cdot\overline x\;.
$$
Now solving for $\alpha$ yields
$$
\alpha=\frac1{\overline{1/x}\cdot\overline x-1}\;.
$$
Finally, substituting into the equation for $\beta$ above yields
\begin{eqnarray}
\beta&=&\left(\overline{1/x}\cdot\overline x-1\right)^{-2}\frac{\overline{1/x}\cdot\overline x-1}{\overline{1/x}\cdot\overline x}\overline{1/x}
\\
&=&
\frac1{\left(\overline{1/x}\cdot\overline x-1\right)\overline x}\;.
\end{eqnarray}
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